Maharashtra Board Class 10 MATHEMATICS-ALGEBRA-71-N-927 Question Paper 2023 with Solution PDF

Step 1: Recall the Pythagoras Theorem.

According to the Pythagoras theorem, for a right-angled triangle having sides \(a\), \(b\) (legs), and \(c\) (hypotenuse), the relation is given by:
\[ a^2 + b^2 = c^2 \]


Step 2: Compare with the given condition.

The question states that \(a^2 + b^2 = c^2\). Hence, this satisfies the condition of the Pythagoras theorem.


Step 3: Conclusion.

Therefore, the triangle is a Right-angled triangle.
Quick Tip: If the sum of squares of two sides equals the square of the third side, the triangle is right-angled.

Step 1: Apply the property of intersecting chords.

When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other chord. That is:
\[ AE \times EB = CE \times ED \]


Step 2: Substitute the given values.
\[ 4 \times 10 = 8 \times ED \]


Step 3: Simplify to find ED.
\[ 40 = 8 \times ED \] \[ ED = \frac{40}{8} = 5 \]

Step 4: Verify.

Thus, \(ED = 5\).



Correct Answer: (B) 5
Quick Tip: For intersecting chords in a circle, remember that the products of the two segments of each chord are equal.

Step 1: Understanding the coordinate system.

In the Cartesian coordinate system, every point is represented as an ordered pair \((x, y)\), where \(x\) is the horizontal distance and \(y\) is the vertical distance from the origin.


Step 2: Definition of the origin.

The origin is the point where the X-axis and Y-axis intersect. At this point, both coordinates are zero. Hence, the coordinates of the origin are \((0, 0)\).


Step 3: Conclusion.

Therefore, the coordinates of the origin are (0, 0).
Quick Tip: Always remember — the origin in the coordinate plane is denoted by \((0, 0)\) where both axes meet.

Step 1: Recall the formula for the slant height of a cone.

For a cone with radius \(r\) and height \(h\), the slant height \(l\) is given by the Pythagoras theorem as:
\[ l = \sqrt{r^2 + h^2} \]


Step 2: Substitute the given values.

Given \(r = 7\) cm and \(h = 24\) cm, substitute into the formula:
\[ l = \sqrt{7^2 + 24^2} \] \[ l = \sqrt{49 + 576} = \sqrt{625} \]


Step 3: Simplify.
\[ l = 25 \, cm \]


Step 4: Conclusion.

The slant height of the cone is 25 cm.
Quick Tip: In right circular cones, always use the Pythagoras theorem: \(l = \sqrt{r^2 + h^2}\) to find the slant height.

Step 1: Recall the property of similar triangles.

If two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides. That is,
\[ \dfrac{A(\triangle ABC)}{A(\triangle PQR)} = \left(\dfrac{AB}{PQ}\right)^2 \]


Step 2: Substitute the given values.
\[ \dfrac{16}{25} = \left(\dfrac{AB}{PQ}\right)^2 \]


Step 3: Take the square root on both sides.
\[ \dfrac{AB}{PQ} = \dfrac{4}{5} \]


Step 4: Conclusion.

Hence, the ratio of the corresponding sides is:
\[ AB : PQ = 4 : 5 \]



Correct Answer: \(AB : PQ = 4 : 5\)
Quick Tip: For similar triangles, remember that the ratio of their areas equals the square of the ratio of their corresponding sides.

Step 1: Understanding the triangle.

In \(\triangle RST\), \(\angle S = 90^\circ\) and \(\angle T = 30^\circ\). Therefore, it is a 30°–60°–90° right triangle.


Step 2: Property of 30°–60°–90° triangle.

In a 30°–60°–90° triangle, the ratio of sides is given by:
\[ 1 : \sqrt{3} : 2 \]
where the side opposite 30° is the smallest, and the hypotenuse is twice that side.


Step 3: Identify the sides.

Given that \(RT = 12\) cm (hypotenuse), so the side opposite the 30° angle (\(RS\)) is half the hypotenuse.
\[ RS = \dfrac{1}{2} \times RT = \dfrac{1}{2} \times 12 = 6 \, cm \]


Step 4: Conclusion.

Hence, \(RS = 6\) cm.



Correct Answer: \(RS = 6\) cm
Quick Tip: In a 30°–60°–90° triangle, the hypotenuse is twice the side opposite the 30° angle.

Step 1: Recall the property of chords.

The longest chord of a circle passes through its center and is equal to the diameter of the circle.


Step 2: Use the relationship between radius and diameter.
\[ Diameter = 2 \times radius \] \[ Diameter = 2 \times 5 = 10 \, cm \]


Step 3: Conclusion.

Hence, the length of the longest chord = 10 cm.



Correct Answer: 10 cm
Quick Tip: The diameter is always the longest chord in a circle.

Step 1: Recall the distance formula.

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]


Step 2: Substitute the given values.

Here, \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (3, 4)\)
\[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} \]


Step 3: Simplify.
\[ d = 5 \]


Step 4: Conclusion.

Hence, the distance between the two points is 5 units.



Correct Answer: 5 units
Quick Tip: Always use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) for coordinate geometry problems.

(i)
\[ \angle L = \dfrac{1}{2} \, m(arc MN) \quad (By the Inscribed Angle Theorem) \]

\[ 35^\circ = \dfrac{1}{2} \, m(arc MN) \]

\[ 2 \times 35^\circ = m(arc MN) \]

\[ \boxed{m(arc MN) = 70^\circ} \]




(ii)
\[ m(arc MLN) = 360^\circ - m(arc MN) \quad (By definition of measure of arc) \]

\[ m(arc MLN) = 360^\circ - 70^\circ \]

\[ \boxed{m(arc MLN) = 290^\circ} \]



Final Answers:

(i) \(m(arc MN) = 70^\circ\)

(ii) \(m(arc MLN) = 290^\circ\)
Quick Tip: In a circle, the measure of an inscribed angle is half the measure of its intercepted arc. The sum of the measures of the major and minor arcs is always \(360^\circ\).

L.H.S. \(= \cot \theta + \tan \theta\)
\[ = \dfrac{\cos \theta}{\sin \theta} + \dfrac{\sin \theta}{\cos \theta} \]
\[ = \dfrac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \times \cos \theta} \]
\[ = \dfrac{1}{\sin \theta \times \cos \theta} \quad (Since \sin^2 \theta + \cos^2 \theta = 1) \]
\[ = \dfrac{1}{\sin \theta} \times \dfrac{1}{\cos \theta} \]
\[ = \csc \theta \times \sec \theta \]

Hence, \[ L.H.S. = R.H.S. \]
\[ \boxed{\therefore \cot \theta + \tan \theta = \csc \theta \times \sec \theta} \] Quick Tip: Always use the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify trigonometric expressions.

Step 1: Formula for surface area of a sphere.

The formula for the surface area of a sphere is: \[ Surface area = 4\pi r^2 \]


Step 2: Substitute the given values.

Given that \(r = 7\) cm and \(\pi = \dfrac{22}{7}\), we have: \[ Surface area = 4 \times \dfrac{22}{7} \times (7)^2 \]


Step 3: Simplify the expression.
\[ Surface area = 4 \times \dfrac{22}{7} \times 49 \]
\[ Surface area = 4 \times 22 \times 7 \]
\[ Surface area = 616 \, sq. cm. \]


Step 4: Conclusion.

Therefore, the surface area of the sphere is: \[ \boxed{616 \, sq. cm.} \]
Quick Tip: The surface area of a sphere is always given by \(4\pi r^2\). Don’t forget to square the radius and multiply by 4.

Step 1: Understanding the given figure.

In trapezium \(ABCD\), the sides \(AB\), \(PQ\), and \(DC\) are parallel. The lines \(AD\) and \(BC\) are transversals. Hence, the ratios of the corresponding segments on these transversals are equal.

\[ \frac{AP}{PD} = \frac{BQ}{QC} \]


Step 2: Substitute the given values.
\[ \frac{3}{12} = \frac{BQ}{14} \]


Step 3: Simplify the equation.
\[ \frac{1}{4} = \frac{BQ}{14} \]
\[ BQ = \frac{14}{4} = 3.5 \]


Step 4: Conclusion.

Hence, \(BQ = 3.5\) cm.



Correct Answer: \(BQ = 3.5\) cm
Quick Tip: In trapeziums with parallel sides, the intercept theorem (or basic proportionality theorem) can be used to find missing segment lengths.

Step 1: Recall the formula for the diagonal of a rectangle.

The diagonal \(d\) of a rectangle is given by: \[ d = \sqrt{l^2 + b^2} \]
where \(l\) = length and \(b\) = breadth.


Step 2: Substitute the given values.
\[ d = \sqrt{35^2 + 12^2} \] \[ d = \sqrt{1225 + 144} \] \[ d = \sqrt{1369} \] \[ d = 37 \, cm \]


Step 3: Conclusion.

Hence, the length of the diagonal is 37 cm.



Correct Answer: 37 cm
Quick Tip: The diagonal of a rectangle can always be found using the Pythagoras theorem.

Step 1: Recall the relationship between central angle and its intercepted arc.

The measure of an arc is equal to the measure of its corresponding central angle.


Step 2: For arc DE.

Given that \(\angle ECF = 70^\circ\), \[ m(arc EF) = 70^\circ \]
Also, the total measure of the circle is \(360^\circ\).

Given \(m(arc DGF) = 200^\circ\),
so the remaining part of the circle (arc DEF) is: \[ m(arc DEF) = 360^\circ - 200^\circ = 160^\circ \]

Now, \[ m(arc DEF) = m(arc DE) + m(arc EF) \] \[ 160^\circ = m(arc DE) + 70^\circ \] \[ m(arc DE) = 90^\circ \]


Step 3: Final values.

(i) \(m(arc DE) = 90^\circ\)

(ii) \(m(arc DEF) = 160^\circ\)



Correct Answers:

(i) \(m(arc DE) = 90^\circ\)

(ii) \(m(arc DEF) = 160^\circ\)
Quick Tip: In a circle, the measure of a major arc and minor arc always add up to \(360^\circ\). Use this property to find unknown arcs.

Step 1: Recall the concept of collinearity.

Three points are collinear if the slopes of any two pairs of points are equal. That is, \[ Slope of AB = Slope of BC \]

Step 2: Find the slope of AB.

Using the slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points A(\(-1, -1\)) and B(\(0, 1\)): \[ m_{AB} = \frac{1 - (-1)}{0 - (-1)} = \frac{2}{1} = 2 \]

Step 3: Find the slope of BC.

For points B(\(0, 1\)) and C(\(1, 3\)): \[ m_{BC} = \frac{3 - 1}{1 - 0} = \frac{2}{1} = 2 \]

Step 4: Compare the slopes.
\[ m_{AB} = m_{BC} = 2 \]
Since the slopes are equal, the points A, B, and C are collinear.



Correct Answer: Points A, B, and C are collinear.
Quick Tip: If the slopes of any two pairs of points are equal, the points are collinear.

Step 1: Understand the situation.

Let the height of the temple be \(h\) m and the distance from the person to the temple be 50 m.
The angle of elevation is \(45^\circ\).


Step 2: Apply the tangent trigonometric ratio.
\[ \tan \theta = \frac{opposite side}{adjacent side} \] \[ \tan 45^\circ = \frac{h}{50} \]

Step 3: Simplify.
\[ 1 = \frac{h}{50} \Rightarrow h = 50 \]

Step 4: Conclusion.

Therefore, the height of the temple is 50 m.



Correct Answer: 50 m
Quick Tip: When the angle of elevation is \(45^\circ\), the height and base are equal in right-angled triangle problems.

Step 1: In \(\triangle PMQ\),

Ray \(MX\) is the bisector of \(\angle PMQ\).

By the Angle Bisector Theorem, \[ \frac{MP}{MQ} = \frac{PX}{XQ} \quad .............. (I) \]


Step 2: In \(\triangle PMR\),

Ray \(MY\) is the bisector of \(\angle PMR\).

By the Angle Bisector Theorem, \[ \frac{MP}{MR} = \frac{PY}{YR} \quad .............. (II) \]


Step 3: Since \(M\) is the midpoint of \(QR\),
\[ MQ = MR \]
Hence, \[ \frac{MP}{MQ} = \frac{MP}{MR} \quad .............. (III) \]


Step 4: From (I), (II), and (III),

We get, \[ \frac{PX}{XQ} = \frac{PY}{YR} \]

Step 5: By the Converse of the Basic Proportionality Theorem (Thales Theorem),
\[ XY \parallel QR \]


Hence, it is proved that \(XY \parallel QR\).
Quick Tip: When medians and angle bisectors are involved in triangles, use the Angle Bisector Theorem and the Converse of the Basic Proportionality Theorem to prove parallel lines.

Let \(A(-4, 2) = (x_1, y_1)\) and \(B(6, 2) = (x_2, y_2)\), and let \(P(x, y)\) be the midpoint of AB.


Step 1: According to the midpoint theorem,
\[ x = \frac{x_1 + x_2}{2}, \quad y = \frac{y_1 + y_2}{2} \]


Step 2: Substitute the given values.
\[ x = \frac{-4 + 6}{2} = \frac{2}{2} = 1 \]
\[ y = \frac{2 + 2}{2} = \frac{4}{2} = 2 \]


Step 3: Write the coordinates of midpoint P.
\[ P(x, y) = (1, 2) \]


Step 4: Conclusion.

Therefore, the co-ordinates of midpoint \(P\) are (1, 2).



Correct Answer: \(P(1, 2)\)
Quick Tip: The midpoint of a line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

Step 1: Recall the formula for the length of a median.

If \(AP\) is a median to side \(BC\) in \(\triangle ABC\), then \[ AB^2 + AC^2 = 2(AP^2 + \tfrac{1}{4} BC^2) \]

Step 2: Substitute the given values.
\[ 260 = 2(AP^2 + \tfrac{1}{4} \times 18^2) \]
\[ 260 = 2(AP^2 + \tfrac{1}{4} \times 324) \]
\[ 260 = 2(AP^2 + 81) \]

Step 3: Simplify the equation.
\[ 260 = 2AP^2 + 162 \]
\[ 260 - 162 = 2AP^2 \]
\[ 98 = 2AP^2 \]
\[ AP^2 = 49 \]

Step 4: Find AP.
\[ AP = \sqrt{49} = 7 \]

Step 5: Conclusion.

Therefore, the length of median \(AP\) is 7 cm.



Correct Answer: \(AP = 7\) cm
Quick Tip: For any triangle, the median length can be found using the formula \(AB^2 + AC^2 = 2(AP^2 + \tfrac{1}{4}BC^2)\).

Given: A circle with center \(O\), and points \(A\), \(B\), and \(C\) lying on the circle such that \(\angle ACB\) and \(\angle ADB\) are inscribed in the same arc \(AB\).






To Prove: \(\angle ACB = \angle ADB\)


Proof:

Step 1: Join \(O\) to \(A\), \(B\), \(C\), and \(D\). Then, \(\triangle OAC\), \(\triangle OBC\), \(\triangle OAD\), and \(\triangle OBD\) are formed.


Step 2: \(\angle AOB\) is the central angle subtending arc \(AB\), and \(\angle ACB\), \(\angle ADB\) are inscribed angles subtending the same arc \(AB\).


Step 3: By the property of a circle, the measure of an inscribed angle is half the measure of the central angle subtending the same arc.
\[ \angle ACB = \frac{1}{2}\angle AOB \quad and \quad \angle ADB = \frac{1}{2}\angle AOB \]

Step 4: Therefore, \[ \angle ACB = \angle ADB \]

Hence proved.



Result: Angles inscribed in the same arc are congruent.
Quick Tip: All angles subtending the same arc (or equal arcs) in a circle are equal in measure.

N/A Quick Tip: The tangent to a circle is always perpendicular to the radius at the point of contact.

Step 1: Formula for curved surface area (C.S.A) of a frustum.
\[ C.S.A. = \pi (r_1 + r_2) l \]
where \(r_1 = 14\) cm, \(r_2 = 6\) cm, and \(l\) = slant height.


Step 2: Find slant height using Pythagoras theorem.
\[ l = \sqrt{(r_1 - r_2)^2 + h^2} \] \[ l = \sqrt{(14 - 6)^2 + 6^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, cm \]

Step 3: Substitute in the formula.
\[ C.S.A. = 3.14 \times (14 + 6) \times 10 \] \[ C.S.A. = 3.14 \times 20 \times 10 = 628 \, sq. cm \]

Step 4: Conclusion.

Hence, the curved surface area of the frustum is 628 sq. cm.



Correct Answer: \(628 \, sq. cm\)
Quick Tip: For a frustum, always find the slant height first using \(l = \sqrt{(r_1 - r_2)^2 + h^2}\) before calculating surface area.

Step 1: Given information.

In \(\triangle ABC\), \(DE \parallel BC\). Therefore, \(\triangle ADE \sim \triangle ABC\) by Basic Proportionality Theorem (BPT).


Step 2: Ratio of areas of similar triangles.

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. \[ \frac{A(\triangle ADE)}{A(\triangle ABC)} = \left(\frac{AD}{AB}\right)^2 \]

Step 3: Given condition.

It is given that \(2A(\triangle ADE) = A(\triangle DBCE)\).
Now, \[ A(\triangle ABC) = A(\triangle ADE) + A(\triangle DBCE) \]
Substitute \(A(\triangle DBCE) = 2A(\triangle ADE)\): \[ A(\triangle ABC) = A(\triangle ADE) + 2A(\triangle ADE) = 3A(\triangle ADE) \]

Step 4: Substitute in the area ratio.
\[ \frac{A(\triangle ADE)}{A(\triangle ABC)} = \frac{1}{3} \]
Hence, \[ \left(\frac{AD}{AB}\right)^2 = \frac{1}{3} \] \[ \frac{AD}{AB} = \frac{1}{\sqrt{3}} \] \[ AB : AD = \sqrt{3} : 1 \]

Step 5: Relation between BC and DE.

Since \(\triangle ADE \sim \triangle ABC\), \[ \frac{BC}{DE} = \frac{AB}{AD} = \sqrt{3} \] \[ \therefore BC = \sqrt{3} \, DE \]

Step 6: Conclusion.

Hence, \(AB : AD = \sqrt{3} : 1\) and \(BC = \sqrt{3} \, DE\).



Correct Answer: \(AB : AD = \sqrt{3} : 1\) and \(BC = \sqrt{3} \, DE\)
Quick Tip: In similar triangles, the ratio of areas is equal to the square of the ratio of their corresponding sides. Always use this relation to compare medians, sides, or heights.

N/A Quick Tip: When constructing similar triangles, use the ratio of corresponding sides and draw parallel lines to maintain proportionality.

Step 1: Find the volume of the cylindrical pot.
\[ V_1 = \pi r_1^2 h_1 \]
Given \(r_1 = 12\) cm and \(h_1 = 7\) cm, \[ V_1 = 3.14 \times 12^2 \times 7 = 3.14 \times 144 \times 7 = 3165.12 \, cm^3 \]

Step 2: Find the volume of one ice-cream cone.

Diameter of cone = 4 cm, so radius \(r_2 = 2\) cm, and height \(h_2 = 3.5\) cm.
\[ V_2 = \frac{1}{3} \pi r_2^2 h_2 \] \[ V_2 = \frac{1}{3} \times 3.14 \times 2^2 \times 3.5 = \frac{1}{3} \times 3.14 \times 4 \times 3.5 = 14.66 \, cm^3 \]

Step 3: Find the number of students.
\[ Number of students = \frac{V_1}{V_2} = \frac{3165.12}{14.66} \approx 216 \]

Step 4: Conclusion.

Hence, 216 students can be served one cone each.



Correct Answer: 216 students
Quick Tip: When solids are melted and recast, their volumes remain equal. Always use \(Volume of solid 1 = Volume of solid 2\) for such problems.

Step 1: Given.

A circle touches the sides of \(\triangle ABC\) externally at points \(P\), \(M\), and \(N\) such that the tangents from a single external point to a circle are equal in length.


Step 2: Tangent length properties.

Let the tangents drawn from each vertex be as follows:
\[ \begin{aligned} &From A: \, AM = AN = x,
&From B: \, BP = BM = y,
&From C: \, CP = CN = z. \end{aligned} \]

Step 3: Express the sides of the triangle.
\[ \begin{aligned} AB &= AM + MB = x + y,
BC &= BP + PC = y + z,
CA &= CN + NA = z + x. \end{aligned} \]

Step 4: Find the perimeter of the triangle.
\[ Perimeter of \triangle ABC = AB + BC + CA = (x + y) + (y + z) + (z + x) \] \[ Perimeter = 2(x + y + z) \]

Step 5: Relation of \(AM\).

From the figure, \(AM = x\).

Hence, \[ x + y + z = \frac{1}{2} (Perimeter of \triangle ABC) \]

Therefore, \[ AM = \frac{1}{2} (Perimeter of \triangle ABC) \]

Hence proved.



Result: \(AM = \dfrac{1}{2} (Perimeter of \triangle ABC)\)
Quick Tip: The tangents drawn from an external point to a circle are equal in length. Use this property to relate the perimeter of the triangle with the tangent segment lengths.

Step 1: Given equations.

We have \[ x = r \cos \theta \quad and \quad y = r \sin \theta \]

Step 2: Square both equations and add them.
\[ x^2 = r^2 \cos^2 \theta \quad and \quad y^2 = r^2 \sin^2 \theta \] \[ x^2 + y^2 = r^2 (\cos^2 \theta + \sin^2 \theta) \]

Step 3: Simplify using the trigonometric identity.

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we get \[ x^2 + y^2 = r^2 \]

Step 4: Conclusion.

Thus, the required relation after eliminating \(\theta\) is \[ \boxed{x^2 + y^2 = r^2} \]


Correct Answer: \(x^2 + y^2 = r^2\)
Quick Tip: To eliminate \(\theta\) in polar equations, square both \(x = r \cos \theta\) and \(y = r \sin \theta\) and add them to use \(\sin^2 \theta + \cos^2 \theta = 1\).