Maharashtra Board Class 10 Mathematics (Geometry) Question Paper 2024 with Answer Key

Step 1: A Pythagorean triplet satisfies the condition \( a^2 + b^2 = c^2 \). Let us check the given options:

- For \( 3, 5, 17 \):
\[ 3^2 + 5^2 = 9 + 25 = 34 \neq 17^2. \]
This is incorrect.

- For \( 4, 9, 15 \):
\[ 4^2 + 9^2 = 16 + 81 = 97 \neq 15^2. \]
This is incorrect.

Thus, none of the options form a Pythagorean triplet. Quick Tip: For a Pythagorean triplet, always verify \( a^2 + b^2 = c^2 \). Ensure the largest number is \( c \), the hypotenuse.

Step 1: Using the trigonometric identity: \[ \sin x \times \csc x = 1 \quad for all x \neq 0. \]
Substituting \( x = 0 \), we get: \[ \sin 0 \times \csc 0 = 1. \] Quick Tip: The reciprocal identity \( \sin x \times \csc x = 1 \) is always valid for non-zero angles.

Step 1: The X-axis is a horizontal line. The slope of a horizontal line is given by: \[ slope = \frac{\Delta y}{\Delta x} = 0. \] Quick Tip: For horizontal lines (e.g., X-axis), the slope is always \( 0 \). For vertical lines (e.g., Y-axis), the slope is undefined.

Step 1: The largest chord of a circle is its diameter. The diameter of a circle is given by: \[ Diameter = 2 \times Radius. \]

Step 2: Substituting the given radius \( 3 \, cm \), we find: \[ Diameter = 2 \times 3 = 6 \, cm. \]

Hence, the largest chord is \( 6 \, cm \). Quick Tip: The diameter of a circle is the longest chord, always equal to \( 2 \times Radius \).

Step 1: For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: \[ \frac{A(\triangle ABC)}{A(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2. \]

Step 2: Substituting \( AB : PQ = 2 : 3 \): \[ \frac{A(\triangle ABC)}{A(\triangle PQR)} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}. \] Quick Tip: For similar triangles, the area ratio is the square of the ratio of corresponding sides.

Step 1: If two circles touch externally, the distance between their centres is equal to the sum of their radii: \[ Distance = R_1 + R_2. \]

Step 2: Substituting \( R_1 = 5 \, cm \) and \( R_2 = 3 \, cm \): \[ Distance = 5 + 3 = 8 \, cm. \] Quick Tip: For externally touching circles, the distance between their centres equals the sum of their radii.

Step 1: The diagonal \( d \) of a square relates to its side \( a \) by: \[ d = a\sqrt{2}. \]

Step 2: Substituting \( d = 10\sqrt{2} \, cm \): \[ 10\sqrt{2} = a\sqrt{2}. \]

Step 3: Dividing both sides by \( \sqrt{2} \): \[ a = 10 \, cm. \] Quick Tip: The diagonal of a square is always \( \sqrt{2} \) times the length of its side.

Step 1: The slope of a line is given by: \[ Slope = \tan \theta, \]
where \( \theta \) is the angle made with the positive X-axis.

Step 2: Substituting \( \theta = 45^\circ \): \[ Slope = \tan 45^\circ = 1. \] Quick Tip: The slope of a line at \( 45^\circ \) with the X-axis is always \( 1 \), as \( \tan 45^\circ = 1 \).

Step 1: Using the property of inscribed angles, the measure of an inscribed angle is half the measure of the arc it subtends: \[ \angle ABC = \frac{1}{2} m(arc AXC). \]

Substituting \( \angle ABC = 60^\circ \): \[ 60^\circ = \frac{1}{2} m(arc AXC). \]

Step 2: Solving for \( m(arc AXC) \): \[ m(arc AXC) = 2 \times 60^\circ = 120^\circ. \]

Step 3: Using the property of central angles, the measure of a central angle is equal to the measure of the arc it subtends: \[ m\angle AOC = m(arc AXC) = 120^\circ. \]

Hence, \( m\angle AOC = 120^\circ \). Quick Tip: For inscribed angles, remember that \( Inscribed Angle = \frac{1}{2} \times Intercepted Arc \). For central angles, \( Central Angle = Intercepted Arc \).

Step 1: In \( \triangle ABC \), \( \angle ABC = 90^\circ \) and \( \angle C = \theta \). From the Pythagoras theorem: \[ AB^2 + BC^2 = AC^2. \]

Step 2: Dividing throughout by \( AC^2 \): \[ \frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}. \]

Step 3: Simplifying, we get: \[ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1. \]

Step 4: Substituting trigonometric ratios: \[ \sin \theta = \frac{BC}{AC}, \quad \cos \theta = \frac{AB}{AC}. \]

Thus: \[ \sin^2 \theta + \cos^2 \theta = 1. \] Quick Tip: The Pythagoras theorem is the foundation of the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Always check for right-angle triangles when applying this formula.

Step 1: The area of the square is given by: \[ Area of square = (side)^2. \]

Substituting \( side = 14 \, cm \): \[ Area of square = 14^2 = 196 \, cm^2. \]

Step 2: The area of the circle is given by: \[ Area of circle = \pi r^2. \]

Here, the diameter of the circle is equal to the side of the square, so the radius \( r = \frac{14}{2} = 7 \, cm \). Substituting \( r = 7 \): \[ Area of circle = \frac{22}{7} \times 7 \times 7 = 154 \, cm^2. \]

Step 3: The area of the shaded region is the difference between the area of the square and the area of the circle: \[ Area of shaded region = Area of square - Area of circle. \]

Substituting the values: \[ Area of shaded region = 196 - 154 = 42 \, cm^2. \] Quick Tip: When a circle is inscribed in a square, the diameter of the circle equals the side length of the square. Use this relationship to compute the radius of the circle.

Step 1: The area of the sector is given by: \[ Area of sector = \frac{1}{2} \times r \times arc length. \]

Step 2: Substituting \( r = 3.5 \, cm \) and \( arc length = 2.2 \, cm \): \[ Area of sector = \frac{1}{2} \times 3.5 \times 2.2 = 3.85 \, cm^2. \] Quick Tip: To calculate the area of a sector, use \( Area = \frac{1}{2} \times r \times arc length \) directly if the arc length is given.

Step 1: Using the Pythagoras theorem: \[ Hypotenuse^2 = Side_1^2 + Side_2^2. \]

Step 2: Substituting \( Side_1 = 9 \, cm \) and \( Side_2 = 12 \, cm \): \[ Hypotenuse^2 = 9^2 + 12^2 = 81 + 144 = 225. \]

Step 3: Taking the square root: \[ Hypotenuse = \sqrt{225} = 15 \, cm. \] Quick Tip: For right triangles, always apply the Pythagoras theorem: \( c^2 = a^2 + b^2 \).

Step 1: The measure of the angle \( \angle NMS \) is equal to half the difference of the measures of the intercepted arcs: \[ \angle NMS = \frac{1}{2} \left[ m(arc NS) - m(arc EF) \right]. \]

Step 2: Substituting \( m(arc NS) = 125^\circ \) and \( m(arc EF) = 37^\circ \): \[ \angle NMS = \frac{1}{2} \left( 125^\circ - 37^\circ \right) = \frac{1}{2} \times 88^\circ = 44^\circ. \] Quick Tip: For angles formed by two secants, use the formula: \( Angle = \frac{1}{2} (larger arc - smaller arc) \).

Step 1: The slope of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1}. \]

Step 2: Substituting \( (x_1, y_1) = (2, 3) \) and \( (x_2, y_2) = (4, 7) \): \[ m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2. \] Quick Tip: For slope, remember the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) and ensure you correctly subtract coordinates.

Step 1: The surface area of a sphere is given by: \[ Surface area = 4\pi r^2. \]

Step 2: Substituting \( r = 7 \, cm \): \[ Surface area = 4 \times \frac{22}{7} \times 7^2 = 4 \times \frac{22}{7} \times 49 = 616 \, cm^2. \] Quick Tip: For spheres, the surface area formula is \( 4\pi r^2 \). Substitute \( \pi = \frac{22}{7} \) for approximate calculations.

N/A Quick Tip: For proving ratios in triangles with parallel lines, use the basic proportionality theorem or the angle bisector theorem effectively.

N/A Quick Tip: For intersecting chords in a circle, use the inscribed angle theorem and properties of similar triangles to establish proportionality and cross-multiplication.

Step 1: To determine collinearity, check if the slope of \( AB \) is equal to the slope of \( BC \).

Step 2: Slope of \( AB \): \[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - (-3)}{2 - 1} = \frac{-5 + 3}{1} = -2. \]

Step 3: Slope of \( BC \): \[ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - (-5)}{-4 - 2} = \frac{7 + 5}{-6} = \frac{12}{-6} = -2. \]

Step 4: Since \( m_{AB} = m_{BC} \), the points are collinear. Quick Tip: To check for collinearity, verify if the slopes of the line segments between the points are equal.

Step 1: Construct \( \triangle ABC \):

Draw a base \( BC = 6 \, cm \).
At point \( B \), construct \( \angle ABC \) using a protractor such that \( AB = 5.5 \, cm \).
At point \( C \), construct \( \angle ACB \) using a protractor such that \( CA = 4.5 \, cm \).
Mark the point of intersection of the two arcs as \( A \), and join \( AB \) and \( AC \) to complete \( \triangle ABC \).


Step 2: Construct \( \triangle LMN \) similar to \( \triangle ABC \) with a scale factor of \( \frac{4}{5} \):

Draw a base \( MN \) such that \( MN = \frac{4}{5} \times BC = \frac{4}{5} \times 6 = 4.8 \, cm \).
At point \( M \), construct an angle equal to \( \angle ABC \).
At point \( N \), construct an angle equal to \( \angle ACB \).
Mark the point of intersection of the two arcs as \( L \), and join \( LM \) and \( LN \) to complete \( \triangle LMN \).


Step 3: Verify similarity: \[ \frac{AB}{LM} = \frac{BC}{MN} = \frac{CA}{LN} = \frac{5}{4}. \] Quick Tip: For constructing similar triangles, ensure that the scale factor is applied consistently to all sides, and corresponding angles are preserved.

Step 1: Using the Apollonius theorem for medians in a triangle: \[ PQ^2 + PR^2 = 2PM^2 + \frac{1}{2}QR^2. \]

Step 2: Substituting the given values \( PQ^2 + PR^2 = 290 \), \( PM = 9 \): \[ 290 = 2(9^2) + \frac{1}{2}QR^2. \]

Step 3: Simplify: \[ 290 = 2(81) + \frac{1}{2}QR^2. \]
\[ 290 = 162 + \frac{1}{2}QR^2. \]

Step 4: Rearrange and solve for \( QR^2 \): \[ 290 - 162 = \frac{1}{2}QR^2. \]
\[ 128 = \frac{1}{2}QR^2 \quad \Rightarrow \quad QR^2 = 256. \]

Step 5: Taking the square root: \[ QR = \sqrt{256} = 16. \] Quick Tip: The Apollonius theorem is useful for finding side lengths in triangles when medians are involved.

Step 1: Consider \( \triangle ABC \) with a line \( DE \parallel BC \) intersecting \( AB \) at \( D \) and \( AC \) at \( E \).

Step 2: In \( \triangle ADE \) and \( \triangle ABC \), since \( DE \parallel BC \), corresponding angles are equal: \[ \angle ADE = \angle ABC, \quad \angle DEA = \angle BCA. \]

Step 3: By AA similarity criterion: \[ \triangle ADE \sim \triangle ABC. \]

Step 4: From the similarity property: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Step 5: This proves that the line \( DE \) divides the sides \( AB \) and \( AC \) in the same proportion. Quick Tip: The basic proportionality theorem (Thales' theorem) is a fundamental result for parallel lines in triangles.

Step 1: Rewrite the given expression using trigonometric identities: \[ \frac{1}{\sin^2 \theta} = \csc^2 \theta, \quad \frac{1}{\cos^2 \theta} = \sec^2 \theta, \quad \frac{1}{\tan^2 \theta} = \cot^2 \theta. \]

Substituting these identities: \[ \csc^2 \theta - \sec^2 \theta - \cot^2 \theta - \tan^2 \theta - \sec^2 \theta - \csc^2 \theta = -3. \]

Step 2: Simplify: \[ -\sec^2 \theta - \sec^2 \theta - \cot^2 \theta - \tan^2 \theta = -3. \]

Step 3: Use the Pythagorean identities: \[ \sec^2 \theta = \tan^2 \theta + 1, \quad \csc^2 \theta = \cot^2 \theta + 1. \]

Solving the equation, \( \theta = 45^\circ \). Quick Tip: For such problems, use reciprocal and Pythagorean identities effectively to simplify the expressions.

(a) Drawing the figure:

The figure is shown above, where \( AB \parallel CD \), and the diagonals \( AC \) and \( BD \) intersect at \( P \).

(b) Alternate and opposite angles:

One pair of alternate angles: \[ \angle APD = \angle CPB. \]

One pair of opposite angles: \[ \angle APB = \angle CPD. \]

(c) Names of similar triangles:

The triangles \( \triangle APB \) and \( \triangle CPD \) are similar by the AA similarity test: \[ \angle APB = \angle CPD \quad and \quad \angle PAB = \angle PCD. \] Quick Tip: In a trapezium, diagonals often form similar triangles by the AA similarity criterion due to parallel sides and corresponding angles.

(a) Drawing the figure:

The figure is shown above, where \( AOC \) is the diameter, \( AB \) is a chord, and \( AT \) is a tangent.

(b) Measures of \( \angle CAT \) and \( \angle ABC \):
\[ \angle CAT = 90^\circ \quad (Tangent-radius property: tangent is perpendicular to the radius at the point of contact). \]
\[ \angle ABC = 90^\circ \quad (Angle subtended by the diameter in a semicircle is a right angle). \]

(c) Congruence of \( \angle CAT \) and \( \angle ABC \):

Yes, \( \angle CAT \) and \( \angle ABC \) are congruent as they are both equal to \( 90^\circ \) by different properties (tangent-radius property and semicircle theorem). Quick Tip: For tangents and diameters in circles, remember that the tangent at a point of contact is perpendicular to the radius, and the angle in a semicircle is always \( 90^\circ \).