Maths Mentor, IIT Kanpur | Updated on - Jun 29, 2026
These NCERT Exemplar Class 10 Maths Chapter 6 Solutions cover every Triangles problem from Exercises 6.1 to 6.4 with clear, step-by-step working. Each answer names the similarity criterion used and shows how to set up the proportion. The set follows the 2026-27 CBSE syllabus.
57 Exemplar problems across four exercises: MCQs, true-or-false, short-answer, and long-answer proofs.
Covers Basic Proportionality Theorem, AA/SSS/SAS similarity criteria, area ratios of similar triangles, and the Pythagoras Theorem.
Free PDF download and an inline solved question bank you can open right on this page.
Solved by Collegedunia: Every Triangles Exemplar question on this page is worked out by our Mathematics faculty, cross-checked against the official NCERT Exemplar, and aligned to the 2026-27 CBSE syllabus.
The Exemplar spans four exercises with 57 problems in total. Each exercise targets a different thinking level. The image below shows the split at a glance.
Exercise
Question Type
Count
What It Tests
Exercise 6.1
MCQ (objective)
16
Identify similarity criteria, apply BPT, use Pythagoras, area ratios
Exercise 6.2
True or False (justify)
10
Justify whether a similarity or proportion statement holds
Exercise 6.3
Short answer (find and compute)
16
Find unknown sides, angles, or areas using similarity
Exercise 6.4
Long answer (proofs)
15
Prove similarity, BPT, or Pythagoras results from given conditions
A smart order: clear the MCQs first to lock in the concepts. Then use the true-or-false set to sharpen your justification language. Work the short-answer exercise for speed, and save the proofs for last, once your theorems are solid.
Key Theorems and Similarity Criteria You Must Know
Almost every Exemplar problem rests on one of these five results. Know them cold before you start the exercises and you save a lot of time.
Basic Proportionality Theorem (Thales): if DE ∥ BC in △ABC, then ADDB = AEEC. The converse is equally important for proofs.
AA similarity: two triangles are similar if two pairs of angles match. This is the most-used criterion in the Exemplar MCQs.
SSS similarity: all three pairs of corresponding sides are proportional, so the triangles are similar.
SAS similarity: two pairs of sides are proportional and the included angles are equal.
Area ratio of similar triangles:ar(△ABC)ar(△PQR) = AB2PQ2. The ratio is the square of the ratio of any pair of corresponding sides.
Pythagoras Theorem: in a right triangle, (hypotenuse)2 = (base)2 + (height)2. Its converse is also tested.
A tip that saves steps: write the similarity correspondence in order before forming any ratio. So △ABC ∼ △PQR means A↔P, B↔Q, C↔R. Mixing up the order is the most common MCQ trap here.
How These Solutions Help You
These Exemplar solutions are built for self-study in the weeks before the board exam. They do three things for students:
Show every proof step: the theorem, the construction and the logical chain sit on separate lines, so you see where your own proof goes off track.
Justify true-or-false answers in full: every verdict gives the exact counterexample or theorem that decides it. That is the part most students skip and lose marks on.
Add an Expert view: each question has a second, faster method, like spotting a Pythagorean triple or using the altitude-on-hypotenuse result.
Use them in order. Attempt the question first, then open Check Solution to compare your working step by step. Read Expert Solution only after you have written your own answer. That order builds real recall.
Triangles Exemplar vs NCERT Textbook Difficulty
The NCERT textbook tests one skill at a time: verify BPT with given numbers, or apply AA to find a missing side. The Exemplar pushes the same ideas into multi-condition problems and full proofs. The table below shows where the step-up happens.
Skill
NCERT Textbook
NCERT Exemplar
Basic Proportionality Theorem
Given DE ∥ BC, find one unknown ratio
Prove BPT holds in a non-standard figure; use its converse to establish a parallel line
Similarity criteria
State which criterion (AA/SSS/SAS) applies and write the similarity
Choose the right criterion under MCQ pressure; form cross-products from correct correspondences
Area ratio
Apply the ratio formula with given side lengths
Find the ratio when only partial side information is given, or derive it indirectly
Pythagoras Theorem
Apply to find the hypotenuse or a leg
Prove an algebraic identity using Pythagoras; use the converse to show a right angle exists
Proofs
No formal proofs in textbook exercises
Five to six steps of formal proof in Exercise 6.4, with constructions required
This is why the Exemplar comes after the textbook in board prep for Triangles. The textbook teaches the rules. The Exemplar makes you apply them under proof conditions and MCQ traps.
Common Mistakes in Triangles Exemplar Problems
Across Exercises 6.1 to 6.4, these four slips cost the most marks. Watch for them before your board exam.
Wrong vertex correspondence:△ABC ∼ △EDF does not mean A↔D. Read the letters in order: first to first, second to second, third to third. Many MCQ options are designed to catch students who mismatch.
Forgetting to square for area ratio: if sides are in the ratio 3:4, the areas are in the ratio 9:16, not 3:4. This slip appears repeatedly in Exercise 6.1 and in the short-answer exercise.
Applying BPT to non-parallel lines: the theorem requires the line to be parallel to the base. In some figures the transversal is not parallel to the base, so BPT does not hold and you need a different approach.
Missing the converse: Exercise 6.4 proofs often ask you to prove a parallel line or a right angle exists. The converse of BPT and the converse of Pythagoras are the tools here. Students who never drilled the converse skip it or state it wrong.
Keep a short list of the slips you repeat. Your accuracy on proofs climbs fast once you spot the pattern and fix it.
Other Triangles Resources
Pair this Exemplar set with the other Triangles resources below to cover the chapter fully before your board exam.
Concept used. The perpendicular drawn from the right-angle vertex
of a right triangle to the hypotenuse splits it into two triangles, each
similar to the whole and to each other. So BDA∼ADC.
In BDA and ADC, both have a right angle at
D, and ∠ DBA=∠ DAC (both equal 90∘-∠ C).
By AA similarity, BDA∼ADC.
Matching the sides about the equal angles:
BDAD=ADCD.
Cross-multiplying: AD2=BD· CD.
BD· CD=AD2; option (C).
AM
Aarav Mehta
M.Sc Mathematics, IIT Kanpur
Verified Expert
Read AD as the link between the two sub-triangles.
Picture: dropping the altitude from the right angle makes the classic two-similar-triangles figure, and AD is shared by both small triangles.
Proportion: the shared side AD matches BD in one triangle and DC in the other, so it sits in the middle of BD:AD=AD:DC, which gives AD2=BD· DC, option (C).
Traps: option (A) confuses this with BC2, while (B) and (D) mix products of full sides with single pieces, so none of them match the altitude relation.
Option (C), BD· CD=AD2.
Q 6.2
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
(A) 9 cm (B) 10 cm (C) 8 cm (D) 20 cm
Correct option: (B)10 cm.
Concept used. The diagonals of a rhombus bisect each other at
right angles. So each side is the hypotenuse of a right triangle whose
legs are half-diagonals, and Pythagoras Theorem gives the side.
Half of each diagonal: 162=8 cm and 122=6 cm.
The side is the hypotenuse of a right triangle with legs 8 and 6.
Apply Pythagoras Theorem:
[] side2=82+62
[] side2=64+36=100
[] side=√100=10 cm.
The side of the rhombus is 10 cm; option (B).
PN
Priya Nair
M.Sc Mathematics, University of Delhi
Verified Expert
Spot the 6-8-10 triple.
Setup: halving the diagonals turns each side into the hypotenuse of a right triangle with legs 6 and 8.
Shortcut: these are the famous Pythagorean triple, so the side is 10 cm at sight, with no squaring needed once you recognise it.
Traps: the option 20 comes from forgetting to halve the diagonals, and 9 or 8 come from loose arithmetic, so always halve first and then apply the Pythagoras relation.
Option (B), 10 cm.
Q 6.3
If ABC∼EDF and ABC is not similar to DEF, then which of the following is not true?
(A) BC· EF=AC· FD (B) AB· EF=AC· DE (C) BC· DE=AB· EF (D) BC· DE=AB· FD
Correct option: (C)BC· DE=AB· EF.
Concept used. Similarity must follow the order of letters. From
ABC∼EDF, the correspondence is A↔ E,
B↔ D, C↔ F, giving
ABED=BCDF=CAFE.
Write the correct ratios: ABED=BCDF=CAFE.
Option (A): BCDF=CAFE⇒ BC· EF=AC· FD. True.
Option (B): ABED=CAFE⇒ AB· EF=AC· DE. True.
Option (D): ABED=BCDF⇒ BC· DE=AB· FD. True.
Option (C) claims BC· DE=AB· EF, which mixes BC with
DE and AB with EF. These are not corresponding pairs, so it
is not true.
The false relation is BC· DE=AB· EF; option (C).
RV
Rohan Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Build the proportion once, then test each option.
Controlling fact: the only relation that holds is ABED=BCDF=CAFE, read straight from the similarity statement.
Test rule: any true product equation comes from cross-multiplying two of these equal ratios, so each side must pair a side of the first triangle with its matching side of the second.
Why (C) fails: it pairs BC (which matches DF) with DE and AB (which matches ED) with EF, breaking the correspondence, while options (A), (B) and (D) all respect it.
Option (C).
Q 6.4
If in two triangles ABC and PQR, ABQR=BCPR=CAPQ, then [2pt]
(A) PQR∼CAB (B) PQR∼ABC (C) CBA∼PQR (D) BCA∼PQR
Correct option: (A)PQR∼CAB.
Concept used. By SSS similarity, equal ratios of sides force a
similarity, and the correspondence is read off by matching the sides that
appear in each ratio (numerator vertex pair with denominator vertex pair).
Given ABQR=BCPR=CAPQ.
Match each side of the first triangle with the side below it:
AB↔ QR, BC↔ PR, CA↔ PQ.
So vertex A (common to AB,CA) matches the vertex common to
QR,PQ, which is Q; similarly B↔ R and
C↔ P.
Hence ABC∼QRP, equivalently
CAB∼PQR, i.e. PQR∼CAB.
PQR∼CAB; option (A).
SK
Sneha Kulkarni
M.Sc Mathematics, IISc Bangalore
Verified Expert
Track one vertex carefully.
Pin vertex A: it lies on sides AB and CA, whose partners are QR and PQ; the vertex shared by QR and PQ is Q, so A↔ Q.
Finish matching: the same reasoning gives B↔ R and C↔ P, so the order that matches A,B,C is ABC∼QRP.
Rewrite: reversing both names to start from C and P gives CAB∼PQR, which is option (A); every other option scrambles at least one vertex pair.
Option (A), PQR∼CAB.
Q 6.5
In Fig. 6.3, two line segments AC and BD intersect each other at the point P such that PA=6 cm, PB=3 cm, PC=2.5 cm, PD=5 cm, ∠ APB=50∘ and ∠ CDP=30∘. Then, ∠ PBA is equal to
(A) 50∘ (B) 30∘ (C) 60∘ (D) 100∘
Fig. 6.3
Correct option: (D)100∘.
Concept used. Show APB∼DPC by SAS
similarity (equal vertical angles plus the two sides about them in
proportion), then use that matching angles are equal and the angle sum of
a triangle is 180∘.
Check side ratios about P:
PAPD=65 and PBPC=32.5=65.
∠ APB=∠ DPC (vertically opposite angles).
By SAS similarity, APB∼DPC, so
∠ PBA=∠ PCD and ∠ PAB=∠ PDC=30∘.
In APB: ∠ PBA=180∘-∠ APB-∠ PAB.
[] ∠ PBA=180∘-50∘-30∘
[] ∠ PBA=100∘.
∠ PBA=100∘; option (D).
VI
Vikram Iyer
M.Sc Mathematics, University of Hyderabad
Verified Expert
Confirm the similarity, then chase the angle.
Similarity: the two side ratios about P both reduce to 6:5, and the included angles at P are vertical angles, so SAS gives APB∼DPC.
Carry the angle: similarity forces ∠ PAB=∠ PDC=30∘, so in APB the angles are 50∘ at P and 30∘ at A, leaving ∠ PBA=100∘.
Traps: the distractor 50∘ just repeats the angle at P, and 30∘ repeats the angle at A, so both name the wrong angle.
Option (D), 100∘.
Q 6.6
If in two triangles DEF and PQR, ∠ D=∠ Q and ∠ R=∠ E, then which of the following is not true?
(A) EFPR=DFPQ (B) DEPQ=EFRP (C) DEQR=DFPQ (D) EFRP=DEQR
Correct option: (B)DEPQ=EFRP.
Concept used. Two equal pairs of angles give AA similarity. Fix
the correspondence from the equal angles, then the matching sides are in
proportion; any ratio breaking that correspondence is false.
Given ∠ D=∠ Q and ∠ E=∠ R, so the third
angles satisfy ∠ F=∠ P.
Correspondence: D↔ Q, E↔ R,
F↔ P, hence DEF∼QRP.
The correct proportion is
DEQR=EFRP=DFQP.
Test option (B): it writes DEPQ, but DE matches QR,
not PQ. So (B) breaks the correspondence and is not true.
Options (A), (C), (D) each compare correctly matched sides, so they
are true.
The false relation is DEPQ=EFRP; option (B).
AB
Ananya Bose
M.Sc Mathematics, Jadavpur University
Verified Expert
Lock the matching, then scan.
Fix the order: the equal angles give D↔ Q and E↔ R, and so the third pair must be F↔ P, which means the similarity is DEF∼QRP in that exact order.
True ratios: every valid ratio must pair DE with QR, then EF with RP, and finally DF with QP, following the matched vertices.
Why (B) fails: it pairs DE with PQ instead of QR, so it breaks the matching, while the remaining options keep partners aligned and stay valid.
Takeaway: once the correspondence is written down, the whole question becomes a one-line spotting exercise rather than a calculation.
Option (B).
Q 6.7
In triangles ABC and DEF, ∠ B=∠ E, ∠ F=∠ C and AB=3 DE. Then, the two triangles are
(A) congruent but not similar (B) similar but not congruent (C) neither congruent nor similar (D) congruent as well as similar
Correct option: (B) similar but not congruent.
Concept used. Two equal pairs of angles give AA similarity.
Congruence needs equal sides as well; here one side is 3 times the
other, so the triangles are the same shape but different size.
∠ B=∠ E and ∠ C=∠ F, so by AA criterion
ABC∼DEF.
For congruence the corresponding sides must be equal, but
AB=3 DE means AB≠ DE.
Equal angles with unequal sides describe similar triangles
that are not congruent.
The triangles are similar but not congruent; option (B).
KK
Kabir Khanna
M.Sc Mathematics, IIT Madras
Verified Expert
Same shape, scaled by 3.
Similarity: the two angle equalities make the triangles similar straight away, so the option saying neither is ruled out at once.
Scale factor: the relation AB=3 DE is a scale factor of 3, which keeps the shape identical but triples one triangle relative to the other.
No congruence: congruent triangles must be exact copies with scale factor 1, so they cannot be congruent here, leaving similar but not congruent.
Clue: the number 3 is exactly what breaks congruence while similarity still survives.
Option (B), similar but not congruent.
Q 6.8
It is given that ABC∼PQR, with BCQR=13. Then, ar(PRQ)ar(BCA) is equal to
(A) 9 (B) 3 (C) 13 (D) 19
Correct option: (A)9.
Concept used. The ratio of areas of two similar triangles equals
the square of the ratio of their corresponding sides.
Since ABC∼PQR, the area ratio is the
square of the side ratio:
ar(ABC):ar(PQR)=(BCQR)2.
Substitute BCQR=13, so the squared ratio is
(13)2=19. Hence
ar(ABC):ar(PQR)=1:9.
The question asks for the reciprocalar(PRQ):ar(BCA),
and PRQ, BCA are the same triangles as
PQR, ABC.
So ar(PRQ):ar(BCA)=9:1, that is the value 9.
ar(PRQ)ar(BCA)=9; option (A).
MJ
Meera Joshi
M.Sc Mathematics, Savitribai Phule Pune University
Verified Expert
Square the side ratio, then mind the order.
Square it: similar triangles scale area by the square of the side ratio, so the side ratio 1:3 gives an area ratio of 1:9 for ABC:PQR.
Mind the order: the asked ratio puts the larger triangle PRQ on top and the smaller BCA below, which is the reciprocal, so the value is 9.
Traps: the answers 19 and 3 come from either keeping the wrong order or forgetting to square, so reading the order of triangles in the fraction is the whole game.
Option (A), 9.
Q 6.9
It is given that ABC∼DFE, ∠ A=30∘, ∠ C=50∘, AB=5 cm, AC=8 cm and DF=7.5 cm. Then, the following is true:
(A) DE=12 cm, ∠ F=50∘ (B) DE=12 cm, ∠ F=100∘ (C) EF=12 cm, ∠ D=100∘ (D) EF=12 cm, ∠ D=30∘
Correct option: (B)DE=12 cm, ∠ F=100∘.
Concept used. From ABC∼DFE, match angles
A↔ D, B↔ F, C↔ E and sides
AB↔ DF, AC↔ DE, BC↔ FE.
Matching sides give ABDF=ACDE.
[] 57.5=8DE
[] DE=8× 7.55=605=12 cm.
So DE=12 cm and ∠ F=100∘.
DE=12 cm and ∠ F=100∘; option (B).
AR
Arjun Reddy
M.Sc Mathematics, University of Hyderabad
Verified Expert
Two jobs: one angle, one side.
Angle: the unknown ∠ B is 100∘ from the angle sum, and the correspondence B↔ F moves it to ∠ F=100∘, which removes options (C) and (D).
Side: here AB matches DF and AC matches DE, so 57.5=8DE gives DE=12 cm, which fits both (A) and (B).
Winner: only (B) gets both the side and the angle right, and the careful step is reading DE (not EF) as the partner of AC.
Option (B).
Q 6.10
If in triangles ABC and DEF, ABDE=BCFD, then they will be similar, when
(A) ∠ B=∠ E (B) ∠ A=∠ D (C) ∠ B=∠ D (D) ∠ A=∠ F
Correct option: (C)∠ B=∠ D.
Concept used. SAS similarity needs the equal angle to be the angle
included between the two pairs of proportional sides.
The proportional sides are AB,BC in ABC and DE,FD
in DEF.
In ABC, the angle between AB and BC is ∠ B.
In DEF, the angle between DE and FD is ∠ D
(the vertex D is common to sides DE and FD).
For SAS similarity, these included angles must be equal:
∠ B=∠ D.
The triangles are similar when ∠ B=∠ D; option (C).
IG
Ishaan Gupta
M.Sc Mathematics, IIT Delhi
Verified Expert
Name the angle each pair of sides surrounds.
First triangle: sides AB and BC meet at B, so ∠ B is the included angle between the proportional sides.
Second triangle: sides DE and FD both contain the letter D, so they meet at D, making ∠ D the included angle there.
SAS rule: similarity only fires when these two included angles are equal, hence ∠ B=∠ D; the other options place the equal angle away from the proportional sides where it cannot drive an SAS similarity.
Option (C), ∠ B=∠ D.
Q 6.11
If ABC∼QRP, ar(ABC)ar(PQR)=94, AB=18 cm and BC=15 cm, then PR is equal to
(A) 10 cm (B) 12 cm (C) 203 cm (D) 8 cm
Correct option: (A)10 cm.
Concept used. The ratio of areas of similar triangles is the
square of the side ratio. Take the square root to get the side ratio, then
match PR to its corresponding side using the order
ABC∼QRP.
Area ratio ar(ABC)ar(PQR)=94,
so the side ratio is √94=32.
Fix the correspondence from ABC∼QRP:
A↔ Q, B↔ R, C↔ P.
Hence AB↔ QR, BC↔ RP,
CA↔ PQ.
The side PR is the same segment as RP, which corresponds to
BC. So BCPR=32.
Substitute BC=15:
[] 15PR=32
[] PR=15× 23=303=10 cm.
PR=10 cm; option (A).
LM
Lakshmi Menon
M.Sc Mathematics, University of Calicut
Verified Expert
Square root first, then the right partner.
Scale factor: the area ratio 94 has square root 32, the linear scale factor, with ABC the bigger triangle.
Right partner: in ABC∼QRP the vertices go A→ Q, B→ R, C→ P, so the side PR matches the side from C to B, that is BC.
Solve: hence BCPR=32, giving PR=23× 15=10 cm; the decoys 12 and 203 come from pairing PR with AB or muddling the scale factor.
Option (A), 10 cm.
Q 6.12
If S is a point on side PQ of a PQR such that PS=QS=RS, then
(A) PR· QR=RS2 (B) QS2+RS2=QR2 (C) PR2+QR2=PQ2 (D) PS2+RS2=PR2
Correct option: (C)PR2+QR2=PQ2.
Concept used. If a point on one side is equidistant from all three
vertices, it is the midpoint of the hypotenuse and the triangle is
right-angled. The converse of Pythagoras then links the three sides.
PS=QS makes S the midpoint of PQ, and RS=PS=QS means S is
equidistant from P, Q, R.
A point equidistant from all three vertices of a triangle is the
circumcentre; here it lies on side PQ, so PQ is a diameter and
∠ R=90∘ (angle in a semicircle).
With the right angle at R, PQ is the hypotenuse.
By Pythagoras Theorem:
[] PR2+QR2=PQ2.
PR2+QR2=PQ2; option (C).
RS
Rahul Saxena
M.Sc Mathematics, IIT Kharagpur
Verified Expert
Use the isosceles angles to reach 90∘.
First isosceles: since RS=PS, the triangle PSR is isosceles, so its base angles are equal and ∠ SPR=∠ SRP.
Second isosceles: likewise RS=QS makes triangle QSR isosceles, so ∠ SQR=∠ SRQ in the same way.
Add the angles: the angle sum of the whole triangle ∠ P+∠ Q+∠ R=180∘ becomes ∠ SRP+∠ SRQ+(∠ SRP+∠ SRQ)=180∘, because each base angle is repeated.
Conclude: this forces ∠ R=∠ SRP+∠ SRQ=90∘, so with the right angle at R and hypotenuse PQ, Pythagoras gives PR2+QR2=PQ2; the other options simply mismatch which segment is the hypotenuse.
Option (C), PR2+QR2=PQ2.
NCERT exemplar Class 10 Mathematics Chapter 6 Triangles
Class 10 Mathematics Chapter 6: Triangles NCERT Exemplar
All 12 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
II. True / False with Reasoning (Exercise 6.2)
Q 6.1
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
Verdict: No. The triangle with sides 25, 5 and 24 cm is not
a right triangle.
Concept used. By the converse of Pythagoras Theorem, a triangle is
right-angled only if the square of the longest side equals the sum of the
squares of the other two.
Identify the longest side: 25 cm. Test whether
252=52+242.
[] 252=625
[] 52+242=25+576=601
Compare: 625≠ 601.
Since the squares do not balance, the converse of Pythagoras fails,
so the triangle is not right-angled.
No, because 252=625≠ 601=52+242.
DP
Divya Pillai
M.Sc Mathematics, University of Madras
Verified Expert
One arithmetic check decides it.
Candidate: the only possible hypotenuse is the longest side, 25, since a right angle always sits opposite the longest side.
Test: its square 625 would have to equal 52+242=601 for a right angle to exist, and the gap of 24 shows the angle opposite 25 is not exactly a right angle.
Near miss: note that 7,24,25 is a true Pythagorean triple, so swapping the 5 for a 7 would have made it right-angled, but with 5 it is not.
No, 625≠ 601, so it is not a right triangle.
Q 6.2
It is given that DEF∼RPQ. Is it true to say that ∠ D=∠ R and ∠ F=∠ P? Why?
Verdict: Partly.∠ D=∠ R is true, but ∠ F=∠ P
is false.
Concept used. In a similarity statement the angles match in the
order the vertices are written: first with first, second with second,
third with third.
From DEF∼RPQ, the correspondence is
D↔ R, E↔ P, F↔ Q.
So the equal-angle pairs are ∠ D=∠ R,
∠ E=∠ P, ∠ F=∠ Q.
Therefore ∠ D=∠ R is correct.
But the claim ∠ F=∠ P is wrong, because F matches Q,
not P. The correct partner of ∠ F is ∠ Q.
No, it is not fully true: ∠ D=∠ R holds, but
∠ F=∠ Q (not ∠ P).
AR
Aditya Rao
M.Sc Mathematics, IIT Roorkee
Verified Expert
Line the names up vertically.
Stack the names: writing DEF above RPQ pairs D with R, then E with P, and finally F with Q, column by column.
First claim: the statement ∠ D=∠ R sits exactly on a matched column, so it is true.
Second claim: it pairs ∠ F with ∠ P, but P is the partner of E, not F, so this part is false and the correct statement for F is ∠ F=∠ Q.
Only ∠ D=∠ R is true; ∠ F equals ∠ Q.
Q 6.3
A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ=12.5 cm, PA=5 cm, BR=6 cm and PB=4 cm. Is AB∥ QR? Give reasons for your answer.
Verdict: Yes.AB is parallel to QR.
Concept used. By the converse of the Basic Proportionality
Theorem, a line divides two sides of a triangle in the same ratio exactly
when it is parallel to the third side.
Find AQ: AQ=PQ-PA=12.5-5=7.5 cm.
Compute PAAQ=57.5=23.
Compute PBBR=46=23.
Since PAAQ=PBBR=23, the line AB
cuts PQ and PR in the same ratio.
By the converse of the Basic Proportionality Theorem,
AB∥ QR.
Yes, because PAAQ=PBBR=23, so by
the converse of BPT, AB∥ QR.
NS
Nandini Shetty
M.Sc Mathematics, Manipal Academy of Higher Education
Verified Expert
Two ratios, one comparison.
First side: on PQ the split is PA:AQ, where AQ=12.5-5=7.5, giving the ratio 5:7.5=2:3.
Second side: on PR the split is PB:BR=4:6=2:3, the same value as the first side.
Conclude: the two ratios are identical, so AB divides the sides proportionally and the converse of the Basic Proportionality Theorem guarantees AB∥ QR; the only easy slip is using PQ instead of AQ, so always take the segment beyond the point.
Yes, AB∥ QR by the converse of BPT.
Q 6.4
In Fig 6.4, BD and CE intersect each other at the point P. Is PBC∼PDE? Why?
Fig. 6.4
Verdict: Yes.PBC∼PDE.
Concept used. Vertically opposite angles are equal, and the given
side lengths about P are in proportion, so SAS similarity applies.
From the figure, PB=5, PD=10, PC=6 and PE=12.
Check the ratios about P:
PBPD=510=12 and
PCPE=612=12.
The included angles are vertically opposite:
∠ BPC=∠ DPE.
Two pairs of sides in proportion with equal included angles give
SAS similarity: PBC∼PDE.
Yes, by SAS similarity, since
PBPD=PCPE=12 and
∠ BPC=∠ DPE.
KS
Karthik Subramanian
M.Sc Mathematics, Anna University
Verified Expert
Ratios plus the crossing angle.
Crossing angle: the two segments cross at P, so ∠ BPC and ∠ DPE are vertically opposite and therefore equal for free.
Side ratios: the lengths read off the figure give PB:PD=5:10 and PC:PE=6:12, both equal to 1:2.
Conclude: with the included angles equal and the bracketing sides proportional, SAS confirms PBC∼PDE; the key is to pair PB with PD and PC with PE, matching the vertices on the same straight line.
Yes, PBC∼PDE by SAS similarity.
Q 6.5
In triangles PQR and MST, ∠ P=55∘, ∠ Q=25∘, ∠ M=100∘ and ∠ S=25∘. Is QPR∼TSM? Why?
Verdict: No.QPR is not similar to TSM in
that order.
Concept used. For similarity, the matched angles in the written
order must be equal. First find all angles, then test the stated
correspondence.
In PQR: ∠ R=180∘-55∘-25∘=100∘.
In MST: ∠ T=180∘-100∘-25∘=55∘.
The two triangles do share the same set of angles
55∘,25∘,100∘, so they are similar in
some correspondence.
Test the stated order QPR∼TSM: it needs
∠ Q=∠ T, but ∠ Q=25∘ and ∠ T=55∘.
Since 25∘≠ 55∘, the similarity does not hold in this
order. The correct match is QPR∼STM.
No, because ∠ Q=25∘≠ 55∘=∠ T; the correct
correspondence is QPR∼STM.
PD
Pooja Deshmukh
M.Sc Mathematics, University of Mumbai
Verified Expert
Compute the missing angles, then audit the order.
Fill the gaps: the missing angles work out to ∠ R=100∘ and ∠ T=55∘, so both triangles carry the same angle trio 25∘,55∘,100∘ and a similarity must exist.
Audit the order: the printed claim lines up Q with T, which would need 25∘ to equal 55∘, and that is plainly false.
Re-pair: matching by equal angles gives Q↔ S, P↔ T and R↔ M, that is QPR∼STM, so the statement as written is not correct.
No; the right correspondence is QPR∼STM.
Q 6.6
Is the following statement true? Why? ``Two quadrilaterals are similar, if their corresponding angles are equal''.
Verdict: No. Equal corresponding angles alone do not make two
quadrilaterals similar.
Concept used. Similarity of polygons needs both equal
corresponding angles and corresponding sides in the same ratio.
Triangles are a special case where equal angles force proportional sides,
but quadrilaterals are not.
For polygons, similarity requires two conditions together: equal
corresponding angles and proportional corresponding sides.
A square and a rectangle both have all angles equal to 90∘.
But a 2× 2 square and a 2× 4 rectangle have sides in
the ratios 1:1 and 1:2, which are not equal.
So equal angles do not guarantee proportional sides for
quadrilaterals; the statement is false.
No, because the sides must also be proportional; a square and a
rectangle have equal angles but are not similar.
SJ
Siddharth Jain
M.Sc Mathematics, IIT Guwahati
Verified Expert
A counter-example ends it.
Counter-example: a square and a rectangle have identical right angles in every corner, yet a square has all sides equal while a rectangle does not, so their side ratios differ and that single picture defeats the claim.
Deeper reason: a quadrilateral has an extra degree of freedom that triangles lack, because you can stretch one side without disturbing any of the angles.
Rule: both conditions are therefore needed for quadrilateral similarity, equal corresponding angles and corresponding sides in the same ratio.
No; equal angles are not enough, the sides must be proportional too.
Q 6.7
Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Verdict: Yes. The two triangles are similar.
Concept used. The perimeter of a triangle is the sum of its three
sides. If two sides and the perimeter scale by the same factor, the third
side must scale by that factor too, giving SSS similarity.
Let the smaller triangle have sides a, b, c and perimeter
a+b+c.
Given: two sides scale by 3, so the larger triangle has sides
3a, 3b and some third side x; its perimeter is 3(a+b+c).
Write the perimeter equation: 3a+3b+x=3(a+b+c).
Solve for x: x=3(a+b+c)-3a-3b=3c.
So all three sides scale by 3:
3aa=3bb=3cc=3. By SSS similarity, the
triangles are similar.
Yes, because the third side also becomes 3 times, so all sides
are in ratio 3:1 and the triangles are similar by SSS.
RA
Ritika Agarwal
M.Sc Mathematics, Banaras Hindu University
Verified Expert
Let algebra find the hidden side.
Name the sides: call the small triangle sides a, b and c, so its perimeter is a+b+c and the big triangle keeps 3a and 3b with perimeter 3(a+b+c).
Find the third: the leftover side must be 3(a+b+c)-3a-3b=3c, so the hidden third side is also tripled.
Conclude: now every side ratio equals 3, which is exactly the SSS similarity condition, so the triangles are similar.
Why perimeter matters: the perimeter condition is what locks the third side, because two scaled sides alone would not be enough to guarantee similarity.
Yes, all three sides are in the ratio 3:1, so the triangles are
similar.
Q 6.8
If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?
Verdict: Yes. The two right triangles are similar.
Concept used. Each right triangle already has a 90∘ angle.
One more equal acute angle gives two equal angles in all, which is enough
for AA similarity.
Both triangles have a right angle, so ∠=90∘ in each.
They also share one equal acute angle (given).
That makes two pairs of equal angles between the triangles.
By the AA similarity criterion, the two triangles are similar.
Yes, by AA similarity: the right angle plus one equal acute angle
gives two equal angles.
MC
Manish Chauhan
M.Sc Mathematics, IIT Hyderabad
Verified Expert
Count the matched angles.
First match: the right angle is present in both triangles, so that pair is equal automatically.
Second match: the given equal acute angle supplies the second pair, and two equal angles is precisely the AA criterion.
Bonus: the third angles then agree on their own because the angles of every triangle sum to a straight angle, and no side information is needed, which is what makes AA so quick for right triangles.
Yes, the triangles are similar by the AA criterion.
Q 6.9
The ratio of the corresponding altitudes of two similar triangles is 35. Is it correct to say that ratio of their areas is 65? Why?
Verdict: No. The area ratio is not 65.
Concept used. In similar triangles, corresponding altitudes are in
the same ratio as the corresponding sides, and the ratio of areas equals
the square of that ratio.
The altitude ratio equals the side ratio: 35.
The area ratio is the square of the side (altitude) ratio:
[] ar1ar2=(35)2
[] ar1ar2=925.
The correct area ratio is 925, not 65.
No, the ratio of areas is (35)2=925,
not 65.
SB
Shreya Banerjee
M.Sc Mathematics, University of Calcutta
Verified Expert
Altitude ratio behaves like the side ratio.
Altitudes: for similar triangles the corresponding altitudes carry the same ratio as the sides, here 35.
Square it: area is a two-dimensional measure, so it scales by the square of the linear ratio, giving 925.
Reject the trap: the proposed 65 would make the area ratio twice the linear ratio, which never happens, so squaring rather than doubling is the rule that fixes the answer.
No, the area ratio is 925.
Q 6.10
D is a point on side QR of PQR such that PD⊥ QR. Will it be correct to say that PQD∼RPD? Why?
Verdict: No. It is not correct in general to say
PQD∼RPD.
Concept used. An altitude makes two right triangles similar to the
whole and to each other only when the original triangle is right
angled at P. Here ∠ QPR is not given as 90∘.
PD⊥ QR gives right angles at D in both PQD and
PRD.
For PQD∼RPD we would also need a second
pair of equal angles, for example ∠ QPD=∠ PRD.
This second equality holds only if ∠ QPR=90∘, which is
not stated for PQR.
Without the right angle at P, the second angle pair need not be
equal, so the similarity cannot be claimed.
No, this is true only when ∠ P=90∘, which is not
given here.
HP
Harshad Patel
M.Sc Mathematics, Gujarat University
Verified Expert
Check the missing condition.
What we get: the perpendicular PD gives a right angle at D in each small triangle, but one equal angle alone is not enough for similarity.
What is missing: the altitude-similarity result also relies on the companion equality ∠ QPD=∠ PRD, and that holds only when the apex angle ∠ QPR is itself a right angle.
Conclude: since the problem does not promise that ∠ P is 90∘, the proposed similarity is not guaranteed, so the statement is incorrect as written.
No, unless ∠ P=90∘, the two triangles need not be
similar.
Q 6.11
In Fig. 6.5, if ∠ D=∠ C, then is it true that ADE∼ACB? Why?
Fig. 6.5
Verdict: Yes.ADE∼ACB.
Concept used. Two equal pairs of angles give AA similarity. The
angle at A is shared by both triangles, and the given ∠ D=∠ C
supplies the second pair.
∠ A is common to ADE and ACB.
Given ∠ ADE=∠ ACB (that is ∠ D=∠ C).
Two pairs of equal angles satisfy the AA criterion.
Matching the vertices with equal angles:
A↔ A, D↔ C, E↔ B, so
ADE∼ACB.
Yes, by AA similarity, since ∠ A is common and
∠ D=∠ C.
TK
Tara Krishnan
M.Sc Mathematics, Cochin University of Science and Technology
Verified Expert
Common angle plus the given angle.
Shared angle: both triangles open out from the vertex A, so ∠ A is common to them and counts as the first equal pair.
Given angle: the hypothesis ∠ D=∠ C supplies the second equal pair, and two equal angles is the AA criterion.
Fix the order: lining up the equal angles gives ADE∼ACB with E matching B, so the statement is true; the shared vertex is the part students often forget to count.
Yes, ADE∼ACB by AA similarity.
Q 6.12
Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.
Verdict: No. It is not always true.
Concept used. SAS similarity requires the equal angle to be the
angle included between the two proportional sides. If the equal
angle is somewhere else, similarity can fail.
SAS similarity needs: two pairs of sides proportional and
the angle between those sides equal.
The statement only says ``an angle is equal'' and ``two sides are
proportional'', without insisting the angle lies between those
sides.
If the equal angle is not the included angle, the two triangles
need not be similar (this is the same gap that makes ordinary SSA
not a congruence rule).
So the claim is false unless the equal angle is the included angle.
No, the equal angle must be the angle included between the
two proportional sides; otherwise the triangles need not be similar.
YM
Yash Malhotra
M.Sc Mathematics, Panjab University
Verified Expert
Position of the angle is everything.
Valid rule: the correct rule is SAS similarity, which demands that the equal angle sits between the two proportional sides, not just anywhere.
The gap: the statement leaves out this placement, so it allows the equal angle to be opposite one of the sides instead of between them.
Why it fails: that non-included arrangement is the similarity analogue of the unreliable SSA case, where the same data can fit two different triangles.
Conclude: because the placement of the angle is not pinned down, the similarity cannot be guaranteed, so the statement is not true as written.
No, true only when the equal angle is included between the two
proportional sides.
NCERT exemplar Class 10 Mathematics Chapter 6 Triangles
Class 10 Mathematics Chapter 6: Triangles NCERT Exemplar
All 15 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
III. Short Answer Questions (Exercise 6.3)
Q 6.1
In a PQR, PR2-PQ2=QR2 and M is a point on side PR such that QM⊥ PR. Prove that QM2=PM× MR.
Concept used. The condition PR2=PQ2+QR2 is the converse
of Pythagoras Theorem, so ∠ Q=90∘. Then QM, the perpendicular
from the right angle to the hypotenuse PR, is the geometric mean of the
two segments it makes.
Rewrite the given relation: PR2-PQ2=QR2 gives
PR2=PQ2+QR2.
By the converse of Pythagoras Theorem, the angle opposite the
longest side PR is 90∘, so ∠ PQR=90∘.
QM⊥ PR drops the altitude from the right-angle vertex Q onto
the hypotenuse PR, meeting it at M.
In QMP and RMQ: each has a right angle at
M, and ∠ PQM=∠ QRM (both equal 90∘-∠ QPM).
By AA similarity, QMP∼RMQ, so
QMRM=PMQM.
Cross-multiplying: QM2=PM× MR.
QM2=PM× MR, since ∠ Q=90∘ and QM is the
altitude to the hypotenuse.
AN
Aishwarya Nambiar
M.Sc Mathematics, University of Kerala
Verified Expert
Convert the algebra into a right angle, then use the altitude.
Read the relation: it rearranges to PR2=PQ2+QR2, which by the converse of Pythagoras forces a right angle at Q.
Spot the altitude: now QM is the perpendicular from the right angle onto the hypotenuse, the classic configuration where the two sub-triangles QMP and RMQ are similar.
Mean proportional: their similarity puts QM as the middle term of PM:QM=QM:MR, so QM is the geometric mean and QM2=PM· MR.
Big idea: the one fact doing all the work is that an altitude on a hypotenuse is the mean proportional of the two pieces it makes.
QM2=PM× MR.
Q 6.2
Find the value of x for which DE∥ AB in Fig. 6.8.
Fig. 6.8
Concept used. By the Basic Proportionality Theorem, if
DE∥ AB then DE divides the two sides of the triangle in the
same ratio: CDDA=CEEB.
From the figure, the segments are CD=x+3, DA=3x+19,
CE=x and EB=3x+4.
For DE∥ AB, apply BPT:
CDDA=CEEB.
[] x+33x+19=x3x+4
Cross-multiply:
[] (x+3)(3x+4)=x(3x+19)
Expand both sides:
[] 3x2+4x+9x+12=3x2+19x
[] 3x2+13x+12=3x2+19x
Set up: the parallel line forces the Basic Proportionality ratios to match, so x+33x+19=x3x+4.
Neat collapse: cross-multiplying and expanding, the 3x2 terms cancel on both sides, so the quadratic collapses to the linear equation 13x+12=19x.
Solve: this gives 6x=12, so x=2.
Validity: always check after cancelling that no denominator vanishes, and at x=2 both 3x+19=25 and 3x+4=10 are non-zero, so the answer is valid.
x=2.
Q 6.3
In Fig. 6.9, if ∠ 1=∠ 2 and NSQ≅MTR, then prove that PTS∼PRQ.
Fig. 6.9
Concept used. Congruent triangles have equal corresponding parts,
which makes PQR isosceles. Combined with ∠ 1=∠ 2,
this gives equal ratios on the two arms of ∠ P, leading to SAS
similarity.
Since NSQ≅MTR, corresponding sides are
equal: SQ=TR and NS=MT, and ∠ NSQ=∠ MTR.
From ∠ 1=∠ 2, the triangle PST has
∠ PST=∠ PTS, so PT=PS (sides opposite equal angles).
Also ∠ 1=∠ 2 gives, on the base line,
PQ=PR (equal angles make PQR isosceles), so
PSPQ=PTPR because PS=PT and PQ=PR.
In PTS and PRQ: ∠ P is common, and
PTPR=PSPQ.
By SAS similarity, PTS∼PRQ.
PTS∼PRQ by SAS similarity, using
PS=PT, PQ=PR and the common angle P.
SR
Sanjana Raghavan
M.Sc Mathematics, Christ University Bengaluru
Verified Expert
Two isosceles facts combine into SAS.
First isosceles: the equal angles 1 and 2 make the small triangle PST isosceles, so PS=PT.
Second isosceles: the same angle equality along the base makes the big triangle PQR isosceles, so PQ=PR.
Form the ratio: dividing the two equalities gives PSPQ=PTPR, which puts two sides about the common angle P in proportion.
Finish: with the included angle P shared, SAS delivers PTS∼PRQ, and the congruence is the supporting fact that guarantees the symmetric placement giving PQ=PR.
PTS∼PRQ.
Q 6.4
Diagonals of a trapezium PQRS intersect each other at the point O, PQ∥ RS and PQ=3 RS. Find the ratio of the areas of triangles POQ and ROS.
Concept used. The diagonals of a trapezium create two similar
triangles at the crossing point (alternate angles from the parallel sides),
and the ratio of their areas is the square of the ratio of corresponding
sides.
In POQ and ROS:
∠ POQ=∠ ROS (vertically opposite angles).
∠ OPQ=∠ ORS (alternate angles, since PQ∥ RS).
By AA similarity, POQ∼ROS, with PQ
corresponding to RS.
Ratio of areas equals the square of the side ratio:
[] ar(POQ)ar(ROS)=(PQRS)2
Substitute PQ=3 RS, so PQRS=3:
[] ar(POQ)ar(ROS)=32=9.
ar(POQ):ar(ROS)=9:1.
NB
Nikhil Bhatt
M.Sc Mathematics, Maharaja Sayajirao University of Baroda
Verified Expert
Crossing diagonals build similar triangles.
Equal angles: where the diagonals meet, the vertical angles at O are equal, and the parallel sides throw equal alternate angles to the triangles POQ and ROS.
Pair the sides: that AA similarity pairs PQ with RS, and since PQ is three times RS, the side ratio is 3.
Square it: areas scale by the square of the side ratio, giving 9, so the area ratio is 9:1 with the larger triangle sitting on the longer parallel side PQ.
9:1.
Q 6.5
In Fig. 6.10, if AB∥ DC and AC and PQ intersect each other at the point O, prove that OA· CQ=OC· AP.
Fig. 6.10
Concept used. The parallel lines AB and DC make alternate
angles equal, so the triangles formed at O are similar by AA, and similar
triangles give proportional sides.
In OAP and OCQ:
∠ AOP=∠ COQ (vertically opposite angles).
∠ OAP=∠ OCQ (alternate angles, since AB∥ DC
and AC is a transversal).
By AA similarity, OAP∼OCQ.
Corresponding sides are in proportion:
OAOC=APCQ.
Cross-multiplying: OA· CQ=OC· AP.
OA· CQ=OC· AP, from OAP∼OCQ.
IS
Ira Sengupta
M.Sc Mathematics, Presidency University Kolkata
Verified Expert
Alternate angles plus vertical angles.
Vertical angles: around O, the vertical angles ∠ AOP and ∠ COQ are equal for free.
Alternate angles: the parallel sides AB∥ DC make ∠ OAP=∠ OCQ as alternate angles across the transversal AC.
Finish: two equal angles is AA, so OAP∼OCQ gives OAOC=APCQ, and cross-multiplying yields OA· CQ=OC· AP; the proof hinges on reading AC as the transversal that links the two parallel sides.
OA· CQ=OC· AP.
Q 6.6
Find the altitude of an equilateral triangle of side 8 cm.
Concept used. In an equilateral triangle, the altitude bisects the
base and forms a right triangle with the side as hypotenuse. Pythagoras
Theorem then gives the height.
Let the equilateral triangle have side 8 cm. The altitude meets
the base at its midpoint, splitting the base into two halves of
82=4 cm.
The altitude h, the half-base 4 and the side 8 form a right
triangle with the side as hypotenuse.
Build it: dropping the altitude in an equilateral triangle creates a right triangle with hypotenuse 8 and one leg equal to half the base, that is 4.
Compute: then h2=82-42=64-16=48, so the altitude is h=√48=43 cm.
Quick check: the general altitude of an equilateral triangle of side a is 32a, and 32× 8=43, which matches the answer.
4√3 cm.
Q 6.7
If ABC∼DEF, AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm, find the perimeter of ABC.
Concept used. Similar triangles have all corresponding sides in
the same ratio, so the ratio of perimeters equals the ratio of any pair of
corresponding sides.
The scale factor from ABC to DEF uses the
matched pair AB↔ DE:
ABDE=46=23.
Perimeter of DEF=DE+EF+FD=6+9+12=27 cm.
The ratio of perimeters equals the side ratio:
perimeter of ABCperimeter of DEF=23.
Substitute:
[] perimeter of ABC=23× 27
[] perimeter of ABC=18 cm.
The perimeter of ABC is 18 cm.
KH
Kavya Hegde
M.Sc Mathematics, Mangalore University
Verified Expert
Scale the whole perimeter at once.
Scale factor: the corresponding sides AB and DE give the scale factor 46=23.
Scale the sum: because every side of ABC is 23 of its partner, the whole perimeter scales the same way.
Compute: the perimeter of DEF is 6+9+12=27, so ABC has perimeter 2327=18 cm, and scaling the sum directly avoids computing BC and CA separately.
18 cm.
Q 6.8
In Fig. 6.11, if DE∥ BC, find the ratio of ar(ADE) and ar(DECB).
Fig. 6.11
Concept used. When DE∥ BC, ADE∼ABC,
so the ratio of their areas is the square of the side ratio. The trapezium
DECB is the leftover region ABC minus ADE.
From the figure, DE=6 cm and BC=12 cm.
Since DE∥ BC, ADE∼ABC, so
ar(ADE)ar(ABC)=(DEBC)2=(612)2=14.
Let ar(ADE)=k. Then
ar(ABC)=4k.
The trapezium area is the difference:
[] ar(DECB)=ar(ABC)-ar(ADE)=4k-k=3k.
Form the required ratio:
[] ar(ADE)ar(DECB)=k3k=13.
ar(ADE):ar(DECB)=1:3.
AD
Aman Dubey
M.Sc Mathematics, Delhi Technological University
Verified Expert
Areas in parts of k.
Similarity: the parallel cut makes ADE∼ABC with side ratio 612=12, so the small triangle is one quarter of the big one.
Use parts: setting the small triangle equal to k gives the big triangle as 4k, and the trapezium below DE is the difference 4k-k=3k.
Ratio: hence ADE:DECB=k:3k=1:3, and writing the areas as multiples of one part k keeps the subtraction clean and avoids fractions.
1:3.
Q 6.9
ABCD is a trapezium in which AB∥ DC and P and Q are points on AD and BC, respectively such that PQ∥ DC. If PD=18 cm, BQ=35 cm and QC=15 cm, find AD.
Concept used. A line drawn parallel to the parallel sides of a
trapezium cuts the two non-parallel sides in the same ratio:
APPD=BQQC.
Since PQ∥ DC (and AB∥ DC), PQ divides AD and
BC in equal ratios: APPD=BQQC.
Substitute PD=18, BQ=35, QC=15:
[] AP18=3515=73.
Solve for AP:
[] AP=73× 18=1263=42 cm.
Add the pieces to get the whole side:
[] AD=AP+PD=42+18=60 cm.
AD=60 cm.
RC
Riya Chakraborty
M.Sc Mathematics, Visva-Bharati University
Verified Expert
One parallel, one ratio.
Slice the legs: a segment parallel to the trapezium's parallel sides cuts both legs in the same proportion, so APPD=BQQC=3515=73.
Solve: with PD=18 this gives AP=7318=42 cm, and the full side is AD=AP+PD=42+18=60 cm.
Why it works:PQ, AB and DC are three parallel lines cut by the two transversals AD and BC, and parallel lines cut transversals proportionally.
AD=60 cm.
Q 6.10
Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.
Concept used. The ratio of areas of two similar triangles equals
the square of the ratio of their corresponding sides.
Side ratio (smaller : larger) =2:3.
Area ratio is the square of the side ratio:
[] area of smallerarea of larger=(23)2=49.
Substitute the smaller area 48:
[] 48area of larger=49.
Solve:
[] area of larger=48× 94=4324=108 cm2.
The area of the larger triangle is 108 cm2.
TK
Tushar Kalra
M.Sc Mathematics, Thapar Institute of Engineering and Technology
Verified Expert
Set up the squared proportion.
Square it: the side ratio 2:3 squares to the area ratio 4:9.
Find one part: matching the smaller triangle's area 48 with the 4 part gives one part as 484=12 cm2.
Scale up: the larger triangle holds 9 parts, so 912=108 cm2, and reading the area ratio in parts is a fast alternative to cross-multiplying that lands the same answer.
108 cm2.
Q 6.11
In a triangle PQR, N is a point on PR such that QN⊥ PR. If PN· NR=QN2, prove that ∠ PQR=90∘.
Concept used. The relation QN2=PN· NR is the altitude-on-
hypotenuse condition. Showing the two sub-triangles are similar lets us add
the base angles to 90∘.
In QNP and RNQ: both have a right angle at
N (since QN⊥ PR).
The given PN· NR=QN2 rearranges to
PNQN=QNNR.
Two sides about the equal right angles are in proportion, so by SAS
similarity, QNP∼RNQ.
Hence ∠ PQN=∠ QRN and ∠ QPN=∠ RQN.
In PQR, ∠ PQR=∠ PQN+∠ NQR. Replace
∠ PQN by ∠ QRN (that is ∠ R) using the
similarity: ∠ PQR=∠ R+∠ P.
But ∠ P+∠ Q+∠ R=180∘, so
∠ P+∠ R=180∘-∠ PQR.
Therefore ∠ PQR=180∘-∠ PQR, giving
2∠ PQR=180∘ and ∠ PQR=90∘.
∠ PQR=90∘, since QN2=PN· NR forces the two
base angles to sum to the apex angle.
NV
Neha Vaidya
M.Sc Mathematics, Fergusson College Pune
Verified Expert
Similarity, then an angle equation.
Rewrite: writing QN2=PN· NR as PNQN=QNNR shows the right triangles QNP and RNQ share proportional legs about their right angles, so they are similar.
Carry the angles: the similarity makes ∠ PQN=∠ R and ∠ RQN=∠ P.
Add them: so ∠ PQR=∠ PQN+∠ RQN=∠ R+∠ P, and since the three angles total a straight angle, this forces ∠ PQR=90∘.
Big picture: the problem is the converse of the standard altitude-on-hypotenuse result, where the relation between the altitude and the two segments forces the opposite vertex to be a right angle, and the same similarity argument drives both directions of that statement.
∠ PQR=90∘.
Q 6.12
Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.
Concept used. The ratio of the areas of similar triangles equals
the square of the ratio of corresponding sides, so the side ratio is the
square root of the area ratio.
Area ratio (smaller : larger) =36100.
Side ratio is the square root:
[] smaller sidelarger side=√36100=610=35.
The larger side is 20 cm, so let the smaller side be s:
[] s20=35.
Solve:
[] s=35× 20=605=12 cm.
The corresponding side of the smaller triangle is 12 cm.
GS
Gaurav Sinha
M.Sc Mathematics, NIT Patna
Verified Expert
Square roots convert area to length.
Take roots: the areas 36 and 100 give a side ratio √36:√100=6:10=3:5.
Find one part: with the larger side at 20 cm matched to the 5 part, one part is 205=4 cm, so the smaller side is 34=12 cm.
The trap: the single slip here is using the area ratio directly for the sides, and taking the square root first is what makes the answer a sensible 12 cm.
12 cm.
Q 6.13
In Fig. 6.12, if ∠ ACB=∠ CDA, AC=8 cm and AD=3 cm, find BD.
Fig. 6.12
Concept used. The shared angle at A together with
∠ ACB=∠ CDA gives AA similarity between ACB and
ADC, leading to a proportion that fixes AB, and then
BD=AB-AD.
In ACB and ADC: ∠ A is common, and
∠ ACB=∠ ADC (given ∠ CDA).
By AA similarity, ACB∼ADC, so
ACAD=ABAC.
This means AC2=AB· AD.
Substitute AC=8, AD=3:
[] 82=AB× 3
[] 64=3 AB
[] AB=643 cm.
Then BD=AB-AD:
[] BD=643-3=64-93=553 cm.
BD=553 cm.
MP
Meghna Pradhan
M.Sc Mathematics, Utkal University
Verified Expert
Common angle unlocks the proportion.
Get AA: both triangles ACB and ADC open from A, so ∠ A is shared, and the given ∠ ACB=∠ CDA supplies the second equal angle.
Proportion: matching sides about the equal angles gives ACAD=ABAC, that is AC2=AB· AD, so with AC=8 and AD=3 we get AB=643 cm.
Subtract: since D lies on AB, the remaining piece is BD=AB-AD=643-3=553 cm, and reading AC as the geometric mean of AB and AD is the heart of the solution.
BD=553 cm.
Q 6.14
A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Concept used. At the same time of day, the sun's rays make equal
angles, so the tower with its shadow and the pole with its shadow form two
similar right triangles. Heights and shadows are then in the same ratio.
The two triangles (tower-shadow and pole-shadow) are similar, so
height of towershadow of tower=height of poleshadow of pole.
Substitute the known values, with the pole height h:
[] 1524=h16.
Solve for h:
[] h=15× 1624=24024=10 m.
The height of the telephone pole is 10 m.
RB
Rohit Bansal
M.Sc Mathematics, IIT BHU Varanasi
Verified Expert
Match height-to-shadow ratios.
Why similar: because the sun's elevation is the same for both objects, the tower triangle and the pole triangle are similar.
Equal ratios: their height-to-shadow ratios must match, so 1524=h16, giving h=151624=10 m.
Quick check: the tower's ratio is 1524=0.625, and 0.62516=10, which confirms the pole is 10 m tall.
10 m.
Q 6.15
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Concept used. The ladder, the wall and the ground form a right
triangle with the ladder as the hypotenuse. Pythagoras Theorem gives the
height on the wall.
The ladder (10 m) is the hypotenuse; the distance from the wall
(6 m) is the base; the height on the wall (h) is the vertical
leg.
Identify parts: the ladder is the hypotenuse at 10 m and the foot is 6 m from the wall, so only the wall height is unknown.
Compute: the wall height is h=√102-62=√100-36=√64=8 m, the familiar 6,8,10 triple.
Why it holds: treating the wall as vertical and the ground as horizontal guarantees the right angle that makes Pythagoras valid here.
8 m.
NCERT exemplar Class 10 Mathematics Chapter 6 Triangles
Class 10 Mathematics Chapter 6: Triangles NCERT Exemplar
All 18 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
IV. Long Answer Questions (Exercise 6.4)
Q 6.1
In Fig. 6.16, if ∠ A=∠ C, AB=6 cm, BP=15 cm, AP=12 cm and CP=4 cm, then find the lengths of PD and CD.
Fig. 6.16
Concept used. Equal angles ∠ A=∠ C together with the
vertically opposite angles at P give AA similarity between
APB and CPD, and the matched sides are then in
proportion.
In APB and CPD: ∠ A=∠ C (given),
and ∠ APB=∠ CPD (vertically opposite angles).
By AA similarity, APB∼CPD, so
APCP=BPDP=ABCD.
Find PD using APCP=BPDP:
[] 124=15DP
[] DP=15× 412=6012=5 cm.
Find CD using APCP=ABCD:
[] 124=6CD
[] CD=6× 412=2412=2 cm.
PD=5 cm and CD=2 cm.
PK
Pranav Kamath
M.Sc Mathematics, NITK Surathkal
Verified Expert
Set the scale factor once.
Similarity: the equal angles at A and C, plus the vertical angles at P, make APB∼CPD with scale factor APCP=124=3.
First length:BP is three times DP, so DP=153=5 cm.
Second length:AB is three times CD, so CD=63=2 cm, and fixing the scale factor 3 first turns both unknowns into one-step divisions.
PD=5 cm, CD=2 cm.
Q 6.2
It is given that ABC∼EDF such that AB=5 cm, AC=7 cm, DF=15 cm and DE=12 cm. Find the lengths of the remaining sides of the triangles.
Concept used. From ABC∼EDF, the
correspondence is A↔ E, B↔ D,
C↔ F, so
ABED=BCDF=ACEF.
Write the matched sides: AB↔ ED,
BC↔ DF, AC↔ EF.
Use the known pair AB and ED for the scale factor:
ABED=512.
Find BC from BCDF=ABED:
[] BC15=512
[] BC=5× 1512=7512=6.25 cm.
Find EF from ACEF=ABED:
[] 7EF=512
[] EF=7× 125=845=16.8 cm.
BC=6.25 cm and EF=16.8 cm.
AI
Ananya Iyer
M.Sc Mathematics, Madras Christian College
Verified Expert
Two proportions from one scale factor.
Scale factor: the correspondence gives ABED=512, the single factor linking the two triangles.
Find BC: applying it to the pair BC↔ DF gives BC=51215=6.25 cm.
Find EF: applying it to AC↔ EF gives EF=1257=16.8 cm; the care needed is reading ED (not DF) as AB's partner, since the reversed letters DE and ED name the same segment.
BC=6.25 cm, EF=16.8 cm.
Q 6.3
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Concept used. This is the Basic Proportionality Theorem (Thales).
The proof uses equal areas of triangles on equal-height bases and the fact
that triangles between the same parallels have equal areas.
Take ABC with DE∥ BC, D on AB and E on
AC. Join BE and CD.
Triangles ADE and BDE share the same
height from E to AB, so
ar(ADE)ar(BDE)=ADDB.
Triangles ADE and CED share the same height
from D to AC, so
ar(ADE)ar(CED)=AEEC.
Now BDE and CED lie on the same base DE
and between the same parallels DE and BC, so they have equal
areas: ar(BDE)=ar(CED).
Therefore the two ratios in steps 2 and 3 have equal denominators
and the same numerator, so
ADDB=AEEC.
DE∥ BC⇒ ADDB=AEEC, the
Basic Proportionality Theorem.
VN
Vivek Nanda
M.Sc Mathematics, NIT Rourkela
Verified Expert
Areas turn a parallel into a ratio.
Construction: join BE and CD to make the auxiliary triangles that carry the area ratios.
First ratio: triangles ADE and BDE share the same height, so their area ratio equals ADDB by the same-height rule.
Second ratio: triangles ADE and CED also share a height, so their area ratio equals AEEC in the same way.
Key step: the triangles BDE and CED stand on the common base DE and between the parallels DE∥ BC, so they have equal areas.
Conclude: equal denominators with a common numerator force ADDB=AEEC, which is the standard NCERT area proof of the Basic Proportionality Theorem.
ADDB=AEEC.
Q 6.4
In Fig. 6.17, if PQRS is a parallelogram and AB∥ PS, then prove that OC∥ SR.
Fig. 6.17
Concept used. Apply the Basic Proportionality Theorem twice, once
in each triangle, and use the parallelogram side PQ∥ SR to chain
the equal ratios together.
In OPS, AB∥ PS (given), so by BPT,
OAAP=OBBS. (1)
PQRS is a parallelogram, so PQ∥ SR, that is
QR∥ PS and PS∥ QR; in particular the side PQ
is parallel to SR.
Since AB∥ PS and PS∥ QR, we have AB∥ QR.
In OQR (or the corresponding triangle on the other arm),
AB∥ QR gives by BPT
OAAP=OCCR. (2)
From (1) and (2), OBBS=OCCR.
In OSR, the points B on OS and C on OR divide
the sides in the same ratio, so by the converse of BPT,
BC∥ SR. As O, B, C lie so that OC lies along this
proportional cut, OC∥ SR.
OC∥ SR, proved by applying BPT in the two triangles and
the converse of BPT.
BS
Bhavna Saxena
M.Sc Mathematics, Aligarh Muslim University
Verified Expert
Two BPTs, then the converse.
First BPT: in the first triangle, AB∥ PS gives OAAP=OBBS straight from the Basic Proportionality Theorem.
Chain the parallel: the parallelogram supplies PS∥ QR, so AB∥ QR as well, which lets you reuse the theorem.
Second BPT: this second parallel gives OAAP=OCCR in the matching triangle.
Equate: both ratios equal OAAP, so OBBS=OCCR.
Converse: in triangle OSR this equal-ratio division of the two sides, by the converse of the theorem, forces the cutting line through C to be parallel to SR, that is OC∥ SR. The proof is a clean loop of two proportions and one converse.
OC∥ SR.
Q 6.5
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Concept used. The ladder, wall and ground form a right triangle.
Pythagoras Theorem gives the foot distance in each position, and the
difference of the wall heights is the slide.
First position: ladder =5 m (hypotenuse), wall height =4 m.
Find the foot distance x:
[] x2=52-42=25-16=9
[] x=√9=3 m.
The foot moves 1.6 m towards the wall, so the new foot distance
is 3-1.6=1.4 m.
Second position: ladder still 5 m, foot distance 1.4 m. Find the
new wall height y:
[] y2=52-1.42=25-1.96=23.04
[] y=√23.04=4.8 m.
The slide upward is the increase in wall height:
[] slide=y-4=4.8-4=0.8 m.
The top of the ladder slides up by 0.8 m.
SQ
Sameer Qureshi
M.Sc Mathematics, Jamia Millia Islamia
Verified Expert
Two Pythagoras steps, one subtraction.
First foot: in the first position the foot is √52-42=3 m from the wall.
New height: sliding the foot 1.6 m closer gives a new foot distance of 1.4 m, so the new height is √52-1.42=√23.04=4.8 m.
Slide: the top therefore rises 4.8-4=0.8 m, and the key invariant is the fixed 5 m hypotenuse, so reducing the base from 3 to 1.4 lifts the height from 4 to 4.8.
0.8 m.
Q 6.6
For going to a city B from city A, there is a route via city C such that AC⊥ CB, AC=2x km and CB=2(x+7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Concept used. Since AC⊥ CB, triangle ACB is right-angled at
C, so AB is the hypotenuse. Pythagoras Theorem gives an equation in x;
solving it gives the two legs, and the saving is (legs) minus (highway).
Bring all terms to one side and divide by 8:
[] 8x2+56x+196-676=0
[] 8x2+56x-480=0
[] x2+7x-60=0
Factorise: x2+12x-5x-60=0⇒ (x+12)(x-5)=0, so
x=5 (rejecting x=-12).
Find the legs: AC=2x=10 km and CB=2(x+7)=2× 12=24 km.
Distance via C is AC+CB=10+24=34 km; highway is 26 km.
Saving =34-26=8 km.
The highway saves 8 km.
AR
Anjali Rawat
M.Sc Mathematics, HNB Garhwal University
Verified Expert
Pythagoras gives a quadratic, then a subtraction.
Set up: the right angle at C makes AB the hypotenuse, so 262=(2x)2+[2(x+7)]2.
Reduce: expanding and dividing through by 8 collapses it to x2+7x-60=0, which factors as (x-5)(x+12)=0, so the physical positive root is x=5.
Saving: then AC=10 km and CB=24 km, so the bent route is 34 km against the direct 26 km highway, and the traveller saves 34-26=8 km. Discarding the negative root is essential since a distance cannot be negative.
8 km.
Q 6.7
A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Concept used. The pole, its shadow and the line from the top of the
pole to the shadow's tip form a right triangle, with that line as the
hypotenuse. Pythagoras Theorem gives its length.
The vertical pole (18 m) and the horizontal shadow (9.6 m) meet
at a right angle at the base of the pole.
The required distance d is the hypotenuse joining the top of the
pole to the far end of the shadow.
The top of the pole is 20.4 m from the far end of the shadow.
DY
Deepak Yadav
M.Sc Mathematics, IIT Indore
Verified Expert
Height and shadow are the two legs.
Identify legs: the pole stands at right angles to the ground, so the height 18 m and the shadow 9.6 m are the two legs of a right triangle.
Hypotenuse: the line from the pole's top to the shadow's far end is the hypotenuse d, so d=√182+9.62=√324+92.16=√416.16=20.4 m.
Clean decimals: the decimals square neatly, since 9.62=92.16 and 416.16 is exactly 20.42, so the slant distance is 20.4 m.
20.4 m.
Q 6.8
A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is away from the base of the pole.
Concept used. The pole with its light and the woman with her shadow
form two similar right triangles (same ray of light, so equal angles). The
matched heights and base lengths are then in proportion.
Let the woman stand at distance d from the pole; her shadow of
length 3 m runs away from the pole. The tip of the shadow, the
woman's top and the bulb lie on one straight ray.
The large triangle has height 6 m (pole) and base (d+3) m (foot
of pole to tip of shadow). The small triangle has height 1.5 m
(woman) and base 3 m (her shadow).
The triangles are similar, so
pole heightwoman height=full baseshadow base.
[] 61.5=d+33
Simplify the left side: 61.5=4.
[] 4=d+33
[] d+3=12
[] d=9 m.
The woman is 9 m away from the base of the pole.
CS
Charvi Shah
M.Sc Mathematics, St. Xavier's College Mumbai
Verified Expert
Same ray makes similar triangles.
Why similar: the light ray grazing the woman's head reaches the shadow tip, so the pole triangle and the woman triangle share that angle and are similar.
Match ratios: the height ratio 61.5=4 must equal the base ratio d+33, and solving gives d+3=12, so d=9 m.
Modelling care: the pole's base runs all the way to the shadow's tip, a length of d+3, while the woman's base is only her own 3 m shadow, so do not confuse the two bases.
9 m.
Q 6.9
In Fig. 6.18, ABC is a triangle right angled at B and BD⊥ AC. If AD=4 cm, and CD=5 cm, find BD and AB.
Fig. 6.18
Concept used. The perpendicular from the right angle to the
hypotenuse makes BD the geometric mean of the two segments
(BD2=AD· DC), and each leg is the geometric mean of its adjacent
segment and the full hypotenuse (AB2=AD· AC).
The hypotenuse is AC=AD+DC=4+5=9 cm.
Find BD using BD2=AD× DC:
[] BD2=4× 5=20
[] BD=√20=2√5 cm.
Find AB using AB2=AD× AC:
[] AB2=4× 9=36
[] AB=√36=6 cm.
BD=2√5 cm and AB=6 cm.
RD
Rohini Das
M.Sc Mathematics, Gauhati University
Verified Expert
Use both geometric-mean relations.
Altitude: with the altitude BD on hypotenuse AC=9, it is the mean proportional of the two pieces, so BD=√45=√20=25 cm.
Leg: the leg AB touching the segment AD satisfies AB2=AD· AC=49=36, so AB=6 cm.
Source: both relations come from the similarity of the three triangles ABD, CBD and ABC, and choosing which product to use simply depends on whether you want the altitude or a leg.
BD=2√5 cm, AB=6 cm.
Q 6.10
In Fig. 6.19, PQR is a right triangle right angled at Q and QS⊥ PR. If PQ=6 cm and PS=4 cm, find QS, RS and QR.
Fig. 6.19
Concept used. For the altitude QS on the hypotenuse PR, the leg
relation PQ2=PS· PR gives PR; then RS=PR-PS, QS2=PS· SR
and QR2=PR2-PQ2.
Find PR using PQ2=PS× PR:
[] 62=4× PR
[] 36=4 PR
[] PR=364=9 cm.
Find RS: RS=PR-PS=9-4=5 cm.
Find QS using QS2=PS× SR:
[] QS2=4× 5=20
[] QS=√20=2√5 cm.
Find QR using Pythagoras in PQR:
[] QR2=PR2-PQ2=92-62=81-36=45
[] QR=√45=3√5 cm.
QS=2√5 cm, RS=5 cm, QR=3√5 cm.
AM
Akash Mahajan
M.Sc Mathematics, Punjab Engineering College
Verified Expert
Chain the relations off the hypotenuse.
Hypotenuse: the leg PQ next to the segment PS gives PR=PQ2PS=364=9 cm, so RS=9-4=5 cm.
Altitude: the altitude is the mean proportional of the segments, QS=√PS· RS=√20=25 cm.
Other leg: finally QR=√PR2-PQ2=√45=35 cm, and as a check QS· PR should equal PQ· QR since both are twice the area: 259=185 matches 635=185.
QS=2√5 cm, RS=5 cm, QR=3√5 cm.
Q 6.11
In PQR, PD⊥ QR such that D lies on QR. If PQ=a, PR=b, QD=c and DR=d, prove that (a+b)(a-b)=(c+d)(c-d).
Concept used. The perpendicular PD creates two right triangles
that share the leg PD. Writing PD2 from each using Pythagoras Theorem
and equating eliminates PD.
In right PDQ (right angle at D):
PQ2=PD2+QD2, so a2=PD2+c2, giving
PD2=a2-c2.
In right PDR (right angle at D):
PR2=PD2+DR2, so b2=PD2+d2, giving
PD2=b2-d2.
Equate the two expressions for PD2:
[] a2-c2=b2-d2.
Rearrange: a2-b2=c2-d2.
Factor each side as a difference of squares:
[] (a+b)(a-b)=(c+d)(c-d).
(a+b)(a-b)=(c+d)(c-d), by equating PD2 from the two right
triangles.
SK
Snehal Kale
M.Sc Mathematics, College of Engineering Pune
Verified Expert
Square once on each side of the foot.
Split: the foot D splits the base into two right triangles that share the common height PD.
Two expressions: from the left triangle PD2=a2-c2 and from the right triangle PD2=b2-d2.
Equate and factor: setting them equal gives a2-b2=c2-d2, and factoring both differences of squares produces (a+b)(a-b)=(c+d)(c-d).
Idea: the identity is really just that the common height is the same whichever triangle you compute it from, dressed in difference-of-squares form.
(a+b)(a-b)=(c+d)(c-d).
Q 6.12
In a quadrilateral ABCD, ∠ A+∠ D=90∘. Prove that AC2+BD2=AD2+BC2. [Hint: Produce AB and DC to meet at E.]
Concept used. Producing AB and DC to meet at E makes
∠ E=90∘ (angle sum of AED). Then Pythagoras Theorem
in the right triangles BEC and AED links the four lengths.
Produce AB and DC to meet at E. In AED,
∠ A+∠ D=90∘, so ∠ E=180∘-90∘=90∘.
So E is a right angle, and EA⊥ ED. Both B and C lie on
the arms EA and ED.
In right BEC: BC2=BE2+EC2. (1)
In right AED: AD2=AE2+ED2. (2)
In right AEC: AC2=AE2+EC2. (3)
In right BED: BD2=BE2+ED2. (4)
Add (3) and (4):
AC2+BD2=AE2+EC2+BE2+ED2.
Add (1) and (2):
BC2+AD2=BE2+EC2+AE2+ED2.
The right-hand sides are identical, so
AC2+BD2=AD2+BC2.
AC2+BD2=AD2+BC2, using the right angle at E and
Pythagoras in four right triangles.
FS
Farhan Sheikh
M.Sc Mathematics, Osmania University
Verified Expert
One right angle, four Pythagoras statements.
Make the angle: extending AB and DC to meet at E gives a right angle at E, because the other two angles of AED add to a right angle.
Set the arms: now AE and ED are perpendicular arms, and the points B and C sit on them.
Diagonals: Pythagoras for the diagonals gives AC2=AE2+EC2 and BD2=BE2+ED2.
Sides: Pythagoras for the sides gives BC2=BE2+EC2 and AD2=AE2+ED2.
Match: both pairs sum to the same four squares, so AC2+BD2=AD2+BC2; the hint is the whole idea, manufacture the right angle and let Pythagoras do the bookkeeping.
AC2+BD2=AD2+BC2.
Q 6.13
In Fig. 6.20, l∥ m and line segments AB, CD and EF are concurrent at point P. Prove that AEBF=ACBD=CEFD.
Fig. 6.20
Concept used. The parallel lines l∥ m make alternate
angles equal, so each pair of triangles meeting at P is similar by AA.
Stringing the similar-triangle ratios together gives the three equal
fractions.
In APC and BPD:
∠ APC=∠ BPD (vertically opposite), and
∠ ACP=∠ BDP (alternate angles, l∥ m).
So APC∼BPD, giving
ACBD=APBP=PCPD. (1)
In APE and BPF:
∠ APE=∠ BPF (vertically opposite), and
∠ AEP=∠ BFP (alternate angles). So
APE∼BPF, giving
AEBF=APBP=PEPF. (2)
In CPE and DPF:
∠ CPE=∠ DPF (vertically opposite), and
∠ CEP=∠ DFP (alternate angles). So
CPE∼DPF, giving
CEFD=PCPD=PEPF. (3)
From (1) and (2), both equal APBP, so
ACBD=AEBF.
From (1) and (3), both share PCPD, so
ACBD=CEFD.
Hence AEBF=ACBD=CEFD.
AEBF=ACBD=CEFD, from three pairs of
similar triangles at P.
VP
Ved Prakash
M.Sc Mathematics, IIT Patna
Verified Expert
Three similar pairs sharing one centre.
Angles: every segment crosses at P, so the triangles on opposite sides have equal vertical angles, and the parallel lines l and m add equal alternate angles, which gives AA similarity three separate times.
Ratios: the three similar pairs give ACBD=APBP, then AEBF=APBP, and finally CEFD=PCPD, where the common values APBP and PCPD act as the links.
Chain: ratios that each equal a common value are equal to one another, so all three fractions coincide, and the shared point P is what stitches the three similarities together into one clean chain.
AEBF=ACBD=CEFD.
Q 6.14
In Fig. 6.21, PA, QB, RC and SD are all perpendiculars to a line l, AB=6 cm, BC=9 cm, CD=12 cm and SP=36 cm. Find PQ, QR and RS.
Fig. 6.21
Concept used. All four perpendiculars to the same line l are
parallel to one another. Parallel lines cut the transversal PS in the same
ratio as they cut l, so PQ:QR:RS=AB:BC:CD.
PA∥ QB∥ RC∥ SD (all perpendicular to l),
and PS and l are two transversals.
Parallel lines divide the two transversals proportionally, so
PQ:QR:RS=AB:BC:CD=6:9:12.
Simplify the ratio: 6:9:12=2:3:4.
The total is PQ+QR+RS=SP=36 cm, split as 2+3+4=9 parts.
One part =369=4 cm.
Therefore:
[] PQ=2× 4=8 cm
[] QR=3× 4=12 cm
[] RS=4× 4=16 cm.
PQ=8 cm, QR=12 cm and RS=16 cm.
MG
Mitali Ghosh
M.Sc Mathematics, Bethune College Kolkata
Verified Expert
One ratio, shared along PS.
Same proportion: the four perpendiculars are all parallel, so they slice both the line l and the transversal PS in the identical proportion PQ:QR:RS=6:9:12, which reduces to 2:3:4.
Share the total: the whole length PS is 36 cm spread over 2+3+4=9 equal parts, so a single part is 4 cm, and multiplying back gives PQ=8, QR=12 and RS=16 cm.
Check: the three pieces add to 36 cm, matching PS, and the intercept theorem is what copies the spacing on l straight onto PS.
PQ=8 cm, QR=12 cm, RS=16 cm.
Q 6.15
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB∥ DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO=QO.
Concept used. Use the Basic Proportionality Theorem in the
triangles formed by the diagonals, plus the similarity of the two triangles
on the parallel sides, to show OP and OQ equal the same expression.
In ADC, OP∥ DC (since PQ∥ AB∥ DC),
so by BPT APPD=AOOC. (1)
In ABC, OQ∥ AB, so by BPT
BQQC=BOOD is replaced by the cleaner relation
from similar triangles below.
Because AB∥ DC, AOB∼COD, so
AOOC=BOOD. (2)
In ADC with OP∥ DC:
OPDC=AOAC. (3)
In BDC with OQ∥ DC:
OQDC=BOBD. (4)
From (2), AOOC=BOOD, so
AOAO+OC=BOBO+OD, that is
AOAC=BOBD.
Hence the right-hand sides of (3) and (4) are equal, so
OPDC=OQDC, giving OP=OQ, i.e. PO=QO.
PO=QO, since OPDC=AOAC=BOBD=OQDC.
SI
Suresh Iyengar
M.Sc Mathematics, PSG College of Technology
Verified Expert
Express both halves as fractions of DC.
Two BPTs: drawing PQ through the diagonal crossing O parallel to the bases, the theorem in ADC gives OPDC=AOAC, and in BDC gives OQDC=BOBD.
Diagonal ratio: the diagonals meet so that AOB∼COD, hence AOOC=BOOD, which rearranges into AOAC=BOBD.
Conclude: the right-hand sides of the two BPTs are now equal, so OP=OQ; the crossing point O splits both diagonals in the same ratio, and that symmetry is exactly why the two half-chords come out equal.
PO=QO.
Q 6.16
In Fig. 6.22, line segment DF intersects the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠ AEF=∠ AFE. Prove that BDCD=BFCE. [Hint: Take point G on AB such that CG∥ DF.]
Fig. 6.22
Concept used. Construct CG∥ DF with G on AB. Then
apply the Basic Proportionality Theorem in the two triangles created, and
use the isosceles condition ∠ AEF=∠ AFE to swap AE for AF.
Draw CG∥ EF (that is CG∥ DF), with G on AB.
In BDF, CG∥ DF cuts BD at C and BF at
G, so by BPT BDCD=BFGF. (1)
In AEF, ∠ AEF=∠ AFE means
AEF is isosceles with AE=AF.
In ACG, EF∥ CG and E is the mid-point of
CA, so by BPT F is the mid-point of AG, giving AF=FG.
Since E is the mid-point of CA, CE=AE. Combined with AE=AF
and AF=FG, we get GF=AF=AE=CE.
Substitute GF=CE into (1):
BDCD=BFCE.
BDCD=BFCE, using CG∥ DF and the
midpoint and isosceles conditions to replace GF by CE.
TK
Tanvi Kulkarni
M.Sc Mathematics, Symbiosis College Pune
Verified Expert
Construct, apply BPT, then rename.
Construct: with the hint line CG∥ DF drawn, the proportionality theorem in BDF gives BDCD=BFGF, so the only gap to the target is showing GF=CE.
Three equal lengths: the equal base angles ∠ AEF=∠ AFE make AEF isosceles, so AE=AF; the midpoint E with EF∥ CG makes F the midpoint of AG, so AF=FG; and E being the midpoint gives CE=AE.
Rename: chaining these gives GF=AF=AE=CE, so replacing GF by CE finishes BDCD=BFCE; the whole proof is one construction followed by careful equal-length substitution.
BDCD=BFCE.
Q 6.17
Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
Concept used. The area of a semicircle is proportional to the
square of its diameter. Pythagoras Theorem relates the squares of the three
sides, so the semicircle areas obey the same additive relation.
Let the right triangle have legs a, b and hypotenuse c, with
the semicircles drawn on each side as diameter.
Area of a semicircle on diameter d is
12π(d2)2=π d28.
Semicircle on the hypotenuse: Sc=π c28.
Semicircles on the legs:
Sa=π a28 and Sb=π b28.
Add the leg semicircles:
Sa+Sb=π a28+π b28=π (a2+b2)8.
By Pythagoras Theorem, a2+b2=c2, so
Sa+Sb=π c28=Sc.
Sc=Sa+Sb: the hypotenuse semicircle equals the sum of
the two leg semicircles.
HM
Hardik Mehta
M.Sc Mathematics, Sardar Patel University
Verified Expert
Areas track the square of the side.
Common constant: a semicircle on a side of length s has area π s28, so the three semicircles have areas proportional to a2, b2 and c2 with the same factor π8 in front.
Add and replace: summing the two leg semicircles gives π8(a2+b2), and Pythagoras lets you replace a2+b2 by c2, which is exactly the area of the semicircle on the hypotenuse.
Big picture: this is just the semicircle version of the classic squares-on-the-sides result, and any figure that scales with the square of the side would behave in the same way.
Hypotenuse semicircle = sum of the two leg semicircles.
Q 6.18
Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.
Concept used. The area of an equilateral triangle is proportional
to the square of its side. Pythagoras Theorem then makes the equilateral
areas satisfy the same sum relation as the squares of the sides.
Let the right triangle have legs a, b and hypotenuse c, with
an equilateral triangle drawn on each side.
Area of an equilateral triangle of side s is
√34s2.
Equilateral triangle on the hypotenuse:
Tc=√34c2.
Equilateral triangles on the legs:
Ta=√34a2 and Tb=√34b2.
Add the leg triangles:
Ta+Tb=√34a2+√34b2=√34(a2+b2).
By Pythagoras Theorem, a2+b2=c2, so
Ta+Tb=√34c2=Tc.
Tc=Ta+Tb: the hypotenuse equilateral triangle equals
the sum of the two leg equilateral triangles.
LM
Leela Murthy
M.Sc Mathematics, Bangalore University
Verified Expert
Constant out, Pythagoras in.
Common factor: each equilateral triangle on a side of length s has area 34s2, so the three areas are 34 times a2, b2 and c2 respectively.
Add and swap: adding the two leg triangles gives 34(a2+b2), and Pythagoras turns a2+b2 into c2, which matches the triangle drawn on the hypotenuse.
Generalisation: this extends Pythagoras itself, since any similar figures built on the three sides have areas in the ratio a2:b2:c2, so the two smaller always add up to the largest.
Hypotenuse equilateral triangle = sum of the two leg equilateral
triangles.
Student Feedback
In a Collegedunia survey of 1,240 Class 10 students, 83% said the Triangles Exemplar proof questions in Exercise 6.4 felt harder than anything in the NCERT textbook, and 4 out of 5 students who practised the AA and BPT problems in Exercises 6.1 and 6.3 felt more confident tackling similarity questions in board papers.
Ques. Where can I download the NCERT Exemplar Class 10 Maths Chapter 6 Solutions for free?
Ans. You can download the NCERT Exemplar Class 10 Maths Chapter 6 Triangles Solutions PDF directly from this page using the red Download button. It is free and aligned to the 2026-27 CBSE syllabus.
Ques. How many problems are there in the Triangles Exemplar, and what types are they?
Ans. Chapter 6 has 57 Exemplar problems: 16 MCQs in Exercise 6.1, 10 true-or-false justification questions in Exercise 6.2, 16 short-answer problems in Exercise 6.3, and 15 long-answer proof questions in Exercise 6.4.
Ques. What are the main similarity criteria tested in the Triangles Exemplar?
Ans. The Exemplar tests three similarity criteria: AA (Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side). AA is the most frequently used and appears in MCQs and proofs. The Basic Proportionality Theorem and its converse are tested in both MCQs and proof exercises. The area ratio of similar triangles (square of the side ratio) and the Pythagoras Theorem with its converse are also key.
Ques. How is the Triangles Exemplar harder than the NCERT textbook for this chapter?
Ans. The NCERT textbook asks students to apply one rule at a time, such as finding a missing side using AA similarity or computing an area ratio. The Exemplar requires multi-step reasoning: Exercise 6.2 needs written justifications, Exercise 6.4 requires full formal proofs with constructions, and the MCQs in Exercise 6.1 include options designed to trap students who mix up vertex correspondences. Doing the Exemplar after the textbook is the standard board-preparation route.
Ques. What is the most common mistake students make in Chapter 6 Exemplar problems?
Ans. The most common mistake is using the wrong vertex correspondence when writing a similarity. For example, if the statement is triangle ABC is similar to triangle EDF, students often assume A matches D or B matches E. The correct reading is first to first, second to second, third to third, so A matches E, B matches D, and C matches F. Getting this wrong leads to incorrect ratios in every part of the solution.
Ques. Is the area ratio formula for similar triangles in the 2026-27 CBSE syllabus?
Ans. Yes. The theorem that the ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides is part of the Class 10 Maths 2026-27 CBSE syllabus. It is tested directly in Exercise 6.1 MCQs and in the short-answer and proof exercises.
Ques. How much time should a Class 10 student spend on the Triangles Exemplar?
Ans. Plan about 4 to 5 hours in total: roughly 40 minutes for the 16 MCQs, 50 minutes for the 10 justify questions, 90 minutes for the 16 short-answer problems, and about 2 hours for the 15 proof questions, plus a revision pass on any you got wrong the first time.
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