NCERT Exemplar Solutions Class 10 Maths Chapter 10 Exercise 10.4

Strategic angle. List the six equal-tangent pairs first, then split every side at its contact point. The two target sums collapse to the same six tangent lengths, so the equality is bookkeeping once the labels are set.

1. Six equal pairs, one per vertex.
2. Both sums reduce to the same collection of tangent segments.

• Distractor: The cleanest way to avoid label errors is to walk around the hexagon in order, naming contact points as you go.
• Confirm it: Keeping the cyclic order straight is the only place this proof can go wrong; the algebra itself is just substitution.

Strategic angle. Set up the x,y,z tangent labels, sum to find x+y+z=s, then isolate the one you want. BD=y, and y=s-(x+z)=s-b because x+z is exactly side b.

1. a=y+z, b=z+x, c=x+y, so x+y+z=s.
2. BD=y=s-(x+z)=s-b.

• Method note: This identity BD=s-b is a building block for the incircle and for Heron-style results, and the symmetric labels make the analogous formulas CD=s-c and so on fall out for free.
• Why chosen: Memorising the labelling, not the final formula, is what transfers to new problems.

Strategic angle. Slide the contact point E along the tangent CD and use CA=CE, DB=DE to fold the broken path P→ C→ E→ D→ P back onto the two straight tangents PA and PB.

1. CE=CA, DE=DB.
2. Perimeter =PA+PB=2(10)=20 cm.

• Generalises: The striking feature is independence from E: the answer never uses the position of the tangent CD.
• Mark grabber: That invariance is the heart of the problem, and recognising it tells you the radius and the exact place of E are deliberately withheld because they do not matter.

Strategic angle. Show both angles equal the same expression 90^∘-∠ BAC. One comes from the triangle's angle sum after the semicircle gives 90^∘; the other from the tangent being perpendicular to the diameter.

1. ∠ ACB=90^∘-∠ BAC (semicircle).
2. ∠ BAT=90^∘-∠ BAC (tangent ⊥ diameter).

• Pitfall: This question is a guided derivation of the alternate-segment theorem for the special case where the chord and a diameter share the endpoint A.
• One-line proof: Seeing the theorem emerge from elementary facts makes it far easier to trust and reuse later.

Strategic angle. The mutual-tangent condition is what makes ∠ OPO'=90^∘. With a right triangle in hand, the half-chord is simply the altitude to the hypotenuse, 345, and doubling gives the chord.

1. Right triangle legs 3,4, hypotenuse 5.
2. Half-chord =345=2.4; chord PQ=4.8 cm.

• Trap to avoid: The 3,4,5 triple keeps the arithmetic exact and clean. Recognising the common chord as twice the foot-of-altitude segment is the move that avoids a longer coordinate computation, which is why this set-up rewards knowing the altitude formula.
• Why it works: If you forget the altitude shortcut, you can fall back to areas: the area of right triangle OPO' is 12(3)(4)=6, and also 12(OO')(PM)=12(5)(PM), so PM=12/5=2.4 cm gives the same half-chord.
• Exam habit: Two independent derivations of 2.4 confirm the chord 4.8 cm without doubt.

Strategic angle. Spot that BC is itself a tangent (at B), because the diameter AB is perpendicular to it. Then the point M where the tangent at P crosses BC has two equal tangents MB,MP, and the right angle at P converts that into MB=MC.

1. BC tangent at B; MB=MP (equal tangents from M).
2. ∠ BPC=90^∘ gives MP=MC, so MB=MC.

• Quick check: The hidden tangent BC is the insight; once you see that the hypotenuse-adjacent leg is tangent at B, the equal-tangent machinery applies immediately. Students who miss it try coordinate geometry and take far longer for the same midpoint result.
• Common slip: The reason MP=MC deserves a word: in right triangle BPC, the angle ∠ MPC and the angle ∠ MCP are both complements of the equal angles ∠ MPB=∠ MBP, so they are equal, and equal angles give the equal sides MP=MC.
• Key insight: That little angle-chase is the bridge from ``M is equidistant to the two tangent points'' to ``M is the midpoint of BC''.

Strategic angle. Anchor on the isosceles base angle 75^∘, push it into the alternate segment via the tangent-chord theorem, then use RS∥ PQ to make triangle QRS isosceles with two 75^∘ angles. The apex ∠ RQS is the leftover 30^∘.

1. Base angle =75^∘; alternate segment gives ∠ QSR=75^∘.
2. Parallel chord gives ∠ QRS=75^∘, so ∠ RQS=30^∘.

• Cross-check: Notice the answer equals the original ∠ RPQ=30^∘.
• Watch out: That neat coincidence is a quick sanity flag: the apex angle of the inscribed triangle matches the angle between the tangents, which you can use to confirm you have not dropped a factor along the way.

Strategic angle. Build the central angle ∠ BOC=60^∘ from the isosceles triangle OAC, then use the tangent-radius right angle to read ∠ ODC=30^∘. The tangent-chord angle ∠ BCD=30^∘ closes the loop and makes BCD isosceles.

1. ∠ BOC=60^∘; tangent ⊥ OC gives ∠ ODC=30^∘.
2. ∠ BCD=30^∘ (alternate segment), so BC=BD.

• Time saver: The specific value 30^∘ is chosen so the two base angles of BCD coincide; a different ∠ BAC would not give BC=BD. Recognising that the question's data is tuned for this equality helps you anticipate the isosceles conclusion before the algebra confirms it.
• Deeper reason: A general way to see it: the tangent-chord angle ∠ BCD always equals ∠ BAC, and the angle ∠ BDC comes out as 90^∘-∠ ABC from the right angle at C; these two coincide exactly when ∠ BAC=30^∘, since then both equal 30^∘.
• Sanity test: Spotting which special angle makes a figure isosceles is a recurring board skill.

Strategic angle. Route everything through the radius OR to the arc midpoint. The arc midpoint sits on the chord's perpendicular bisector, so OR⊥ PQ; the tangent is ⊥ OR by definition. Two perpendiculars to OR are parallel.

1. Arc midpoint ⇒ OR⊥ PQ.
2. Tangent ⊥ OR, so tangent ∥ PQ.

• Reusable idea: This is the converse companion to Question 28, where a chord parallel to the tangent forced the contact point to bisect the arc. Here the arc midpoint forces the parallel tangent.
• Spot this: Seeing the two statements as a matched pair makes both easier to recall under exam conditions. The single shared fact behind both is that the diameter through the arc midpoint is the axis of symmetry of the chord and its arc: it is perpendicular to the chord, perpendicular to the tangent at the midpoint, and bisects both the chord and the arc.
• Careful here: Naming that diameter early turns either direction of the proof into one or two lines.

Strategic angle. The key realisation is that AB and CD are common tangents, so they form one and the same angle at E for both circles. Each centre lies on the bisector of that single angle, and a single angle has only one bisector.

1. EO bisects ∠ at E (first circle).
2. EO' bisects the same ∠ at E (second circle); one bisector, so collinear.

• Big picture: The proof never needs the radii or the distance between centres, only that the tangents are shared.
• Takeaway: Identifying that the two angle-bisector claims refer to the identical angle is the whole argument, and it is a clean reminder that ``common tangent'' is a strong shared constraint.

Strategic angle. Find TP=12 via the 5,12,13 triple, locate E at ET=8, then use the shared angle at T to set up the proportion AE/ET=OP/TP. Doubling the half-tangent AE gives AB.

1. TP=12 (5,12,13); ET=13-5=8.
2. AE=8·512=103; AB=203 cm.

• Distractor: The similar-triangle proportion is the engine: it converts the radius-and-tangent of the big triangle into the small tangent segment AE. Keeping the matching of opposite-to-adjacent consistent across the two triangles is the one place to be careful, and it lands the exact fraction 203.
• Confirm it: A quick reasonableness test: AB=203≈ 6.67 cm is comfortably less than the chord of contact PQ of the big tangent pair, which fits the picture of AB sitting closer to the circle than PQ.
• Method note: Carrying the answer as an exact fraction rather than a rounded decimal is also what board schemes expect here.

Strategic angle. Strip the 90^∘ tangent-radius angle out of ∠ PCA to get ∠ OCA=20^∘, bounce it to ∠ CAB=20^∘ by the isosceles radii, then use the semicircle's 90^∘ to finish the triangle.

1. ∠ OCA=110^∘-90^∘=20^∘=∠ CAB.
2. ∠ ACB=90^∘ (semicircle), so ∠ CBA=70^∘.

• Why chosen: A fast cross-check uses the tangent-chord theorem directly: the angle between tangent PC and chord CB equals ∠ CAB=20^∘, and since ∠ PCA=110^∘, the chord CB splits it as 110^∘-20^∘=90^∘ at C, agreeing with the semicircle right angle.
• Generalises: Two independent paths to 70^∘ give confidence.

Strategic angle. The whole problem turns on one unknown, the altitude h=AM, and one shared quantity, the half-base BM. Write BM² two different ways: once from the radius using right triangle OMB with hypotenuse OB=9, and once from the equal side using right triangle AMB with hypotenuse AB=6. Setting the two expressions equal eliminates BM entirely and leaves a single linear equation in h.

1. From the radius: BM²=9²-(h-9)². From the equal side: BM²=6²-h².
2. Equate: 81-(h-9)²=36-h². The h² terms cancel, giving 18h=36, so h=2 cm.
3. Back-substitute: BM²=36-4=32, so BM=42 and base BC=82 cm.
4. Area =12× BC× AM=12(82)(2)=82 cm².

• Mark grabber: A qualitative check seals the work: the altitude h=2 is tiny against the radius 9, so the triangle is short and wide and sits low in the circle, exactly as the diagram shows. That mental picture guards against the most common slip, getting the sign of OM=h-9 wrong.
• Pitfall: Had you written OM=9-h instead, the same cancellation still gives h=2 here because the term is squared, but in problems where the apex lies on the far side of the centre the sign genuinely matters, so always read the figure before committing.
• One-line proof: The clean surd 82 is a reassuring sign that the radius and side lengths were chosen to produce an exact answer.

Strategic angle. Find AP=12 from the 5,12,13 triple, then fold the broken boundary A→ B→ R→ C→ A onto the two straight tangents AP and AQ using BR=BP and CR=CQ.

1. AP=AQ=12 cm (5,12,13).
2. Perimeter =AP+AQ=24 cm.

• Trap to avoid: As in Question 33, the position of the contact point R never enters the answer; the perimeter is locked at twice the tangent length. Spotting this invariance saves time and tells you the ``minor arc'' detail is there to test understanding, not to change the number.
• Why it works: The reason R must lie on the minor arc PQ is simply that only then does the tangent at R cross both AP and AQ between A and the contact points, keeping the small triangle ABC well defined.
• Exam habit: Once that condition holds, the broken path folds onto AP+AQ exactly, and 2× 12=24 cm follows.