OMTEX AD 2

10th Geometry Board Question Paper March 2024 with Solutions - Maharashtra Board

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Q.1 (A) Choose the correct alternative. (4 Marks)
(1) Out of the dates given below which date constitutes a Pythagorean triplet?
  • (A) 15/8/17
  • (B) 16/8/16
  • (C) 3/5/17
  • (D) 4/9/15
Solution:

Answer: (A)

Reason: In a Pythagorean triplet $(a, b, c)$ where $c$ is the largest number, $a^2 + b^2 = c^2$.
For 15, 8, 17:
$$8^2 + 15^2 = 64 + 225 = 289$$ $$17^2 = 289$$ Since LHS = RHS, it is a Pythagorean triplet.

(2) $\sin \theta \times \text{cosec } \theta = ?$
  • (A) 1
  • (B) 0
  • (C) $\frac{1}{2}$
  • (D) $\sqrt{2}$
Solution:

Answer: (A)

Reason: We know that $\text{cosec } \theta = \frac{1}{\sin \theta}$.
Therefore, $\sin \theta \times \frac{1}{\sin \theta} = 1$.

(3) Slope of X-axis is
  • (A) 1
  • (B) -1
  • (C) 0
  • (D) Cannot be determined
Solution:

Answer: (C)

Reason: The X-axis is a horizontal line. The slope of any horizontal line is 0.

(4) A circle having radius 3 cm, then the length of its largest chord is
  • (A) 1.5 cm
  • (B) 3 cm
  • (C) 6 cm
  • (D) 9 cm
Solution:

Answer: (C)

Reason: The largest chord of a circle is its diameter.
Diameter $= 2 \times \text{radius} = 2 \times 3 = 6 \text{ cm}$.

Q.1 (B) Solve the following sub-questions. (4 Marks)
(1) If $\Delta ABC \sim \Delta PQR$ and $AB : PQ = 2 : 3$, then find the value of $\frac{A(\Delta ABC)}{A(\Delta PQR)}$.
Solution:

By the theorem of areas of similar triangles:

$$\frac{A(\Delta ABC)}{A(\Delta PQR)} = \frac{AB^2}{PQ^2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$$
(2) Two circles of radii 5 cm and 3 cm touch each other externally. Find the distance between their centres.
Solution:

When circles touch externally, the distance between their centres is the sum of their radii.

$$d = r_1 + r_2 = 5 + 3 = 8 \text{ cm}$$
(3) Find the side of a square whose diagonal is $10\sqrt{2}$ cm.
Solution:

Formula: $\text{Diagonal} = \text{Side} \times \sqrt{2}$

$$10\sqrt{2} = \text{Side} \times \sqrt{2}$$ $$\text{Side} = 10 \text{ cm}$$
(4) Angle made by the line with the positive direction of X-axis is $45^{\circ}$. Find the slope of that line.
Solution:

Slope $m = \tan \theta$
Here, $\theta = 45^{\circ}$

$$m = \tan 45^{\circ} = 1$$

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Q.2 (A) Complete any two activities and rewrite it. (4 Marks)
(1) In the figure, $\angle ABC$ is inscribed in arc ABC. If $\angle ABC = 60^{\circ}$, find $m\angle AOC$.
Solution:

$\angle ABC = \frac{1}{2} m(\text{arc } AXC)$ ... (Inscribed angle theorem)

$60^{\circ} = \frac{1}{2} m(\text{arc } AXC)$

120° $= m(\text{arc } AXC)$

But $m\angle AOC = \text{m(arc AXC)}$ ... (Property of central angle)

$\therefore m\angle AOC = $ 120°

(2) Find the value of $\sin^2\theta + \cos^2\theta$.
Solution:

In $\Delta ABC$, $\angle ABC = 90^{\circ}, \angle C = \theta$.

$AB^2 + BC^2 = $ AC² ... (Pythagoras theorem)

Divide both sides by $AC^2$:

$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}$

$(\frac{AB}{AC})^2 + (\frac{BC}{AC})^2 = 1$

But $\frac{AB}{AC} = $ sin θ and $\frac{BC}{AC} = $ cos θ

$\therefore \sin^2\theta + \cos^2\theta = $ 1

(3) ABCD is a square and a circle is inscribed in it. If $AB = 14$ cm, find the area of shaded region.
Solution:

Area of square = Side² (Formula)

$= 14^2 = $ 196 $cm^2$

Area of circle = πr² (Formula)

Since side = 14, radius $r = 7$.

$= \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2$

Area of shaded portion = Area of square - Area of circle

$= 196 - 154 = $ 42 $cm^2$

Q.2 (B) Solve any four of the following sub-questions. (8 Marks)
(1) Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Solution:

Given: $r = 3.5 \text{ cm}, l = 2.2 \text{ cm}$.
Area of sector $A = \frac{l \times r}{2}$

$$A = \frac{2.2 \times 3.5}{2}$$ $$A = 1.1 \times 3.5$$ $$A = 3.85 \text{ cm}^2$$
(2) Find the length of the hypotenuse of a right-angled triangle if remaining sides are 9 cm and 12 cm.
Solution:

Let the sides be $a=9, b=12$. By Pythagoras theorem:

$$\text{Hypotenuse}^2 = 9^2 + 12^2$$ $$\text{Hyp}^2 = 81 + 144 = 225$$ $$\text{Hypotenuse} = \sqrt{225} = 15 \text{ cm}$$
(3) In the figure, $m(\text{arc } NS) = 125^{\circ}, m(\text{arc } EF) = 37^{\circ}$. Find the measure of $\angle NMS$.
Solution:

When two secants intersect outside the circle at point M, the angle formed is half the difference of the intercepted arcs.

$$\angle NMS = \frac{1}{2} [m(\text{arc } NS) - m(\text{arc } EF)]$$ $$\angle NMS = \frac{1}{2} [125^{\circ} - 37^{\circ}]$$ $$\angle NMS = \frac{1}{2} [88^{\circ}]$$ $$\angle NMS = 44^{\circ}$$
(4) Find the slope of the line passing through the points $A(2,3)$ and $B(4,7)$.
Solution:

Let $A(x_1, y_1) = (2,3)$ and $B(x_2, y_2) = (4,7)$.

$$\text{Slope } m = \frac{y_2 - y_1}{x_2 - x_1}$$ $$m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2$$
(5) Find the surface area of a sphere of radius 7 cm.
Solution:

Surface Area $= 4\pi r^2$

$$= 4 \times \frac{22}{7} \times 7 \times 7$$ $$= 4 \times 22 \times 7$$ $$= 88 \times 7 = 616 \text{ cm}^2$$
Q.3 (A) Complete any one activity. (3 Marks)
(1) In $\Delta ABC$, ray BD bisects $\angle ABC$, DE || BC. Show that $\frac{AB}{BC} = \frac{AE}{EB}$.
Solution:

Proof:

In $\Delta ABC$, ray BD bisects $\angle B$.

$\therefore \frac{\text{AB}}{BC} = \frac{AD}{DC}$ ... (I) (Angle Bisector Theorem)

In $\Delta ABC$, DE || BC

$\therefore \frac{\text{AE}}{EB} = \frac{AD}{DC}$ ... (II) (Basic Proportionality Theorem)

From (I) and (II),

$\frac{AB}{\text{BC}} = \frac{\text{AE}}{EB}$

(Filled boxes: AB, AE, BC, AE)

(2) Proof of property of chords intersecting inside the circle: $AE \times EB = CE \times ED$.
Solution:

In $\Delta CAE$ and $\Delta BDE$,

$\angle AEC \cong \angle DEB$ ... (Vertically opposite angles)

$\angle CAE \cong $ ∠BDE ... (Angles inscribed in the same arc)

$\therefore \Delta CAE \sim \Delta BDE$ ... (AA test of similarity)

$\therefore \frac{AE}{DE} = \frac{CE}{\text{EB}}$ ... (Corresponding sides of similar triangles)

$\therefore AE \times EB = CE \times ED$

Q.3 (B) Solve any two sub-questions. (6 Marks)
(1) Determine whether the points are collinear: $A(1,-3), B(2,-5), C(-4,7)$.
Solution:

Points are collinear if the slope of line AB = slope of line BC.

Slope of AB = $\frac{-5 - (-3)}{2 - 1} = \frac{-2}{1} = -2$

Slope of BC = $\frac{7 - (-5)}{-4 - 2} = \frac{12}{-6} = -2$

Since Slope of AB = Slope of BC and point B is common, the points A, B, and C are collinear.

(2) $\Delta ABC \sim \Delta LMN$. In $\Delta ABC$, $AB=5.5, BC=6, CA=4.5$. Construct $\Delta ABC$ and $\Delta LMN$ such that $\frac{BC}{MN} = \frac{5}{4}$.
Solution:

Analysis:
Since $\Delta ABC \sim \Delta LMN$, the corresponding sides are proportional.
$\frac{AB}{LM} = \frac{BC}{MN} = \frac{AC}{LN} = \frac{5}{4}$

Calculation for sides of $\Delta LMN$:

  • $LM = AB \times \frac{4}{5} = 5.5 \times 0.8 = 4.4 \text{ cm}$
  • $MN = BC \times \frac{4}{5} = 6 \times 0.8 = 4.8 \text{ cm}$
  • $LN = AC \times \frac{4}{5} = 4.5 \times 0.8 = 3.6 \text{ cm}$

Construction Steps:
1. Draw $\Delta ABC$ using sides 5.5 cm, 6 cm, and 4.5 cm.
2. Draw $\Delta LMN$ using sides 4.4 cm, 4.8 cm, and 3.6 cm.

(3) Seg PM is a median of $\Delta PQR$. $PM=9$ and $PQ^2 + PR^2 = 290$, then find QR.
Solution:

By Apollonius Theorem:

$$PQ^2 + PR^2 = 2(PM^2 + QM^2)$$ $$290 = 2(9^2 + QM^2)$$ $$145 = 81 + QM^2$$ $$QM^2 = 145 - 81 = 64$$ $$QM = 8$$

Since M is the midpoint (PM is median): $QR = 2 \times QM = 2 \times 8 = 16$.

(4) Prove that: 'If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the side in the same proportion'.
Solution:

Given: In $\Delta ABC$, line $l || \text{ side } BC$ and intersects $AB$ at $P$ and $AC$ at $Q$.
To Prove: $\frac{AP}{PB} = \frac{AQ}{QC}$
Construction: Draw segment $PC$ and segment $BQ$.
Proof:
$\Delta APQ$ and $\Delta BPQ$ have equal height.
$\therefore \frac{A(\Delta APQ)}{A(\Delta BPQ)} = \frac{AP}{PB}$ ... (I) (Areas proportional to bases)
Similarly, $\frac{A(\Delta APQ)}{A(\Delta CPQ)} = \frac{AQ}{QC}$ ... (II)
$\Delta BPQ$ and $\Delta CPQ$ lie between the same parallel lines $PQ$ and $BC$, hence they have equal heights. They also have a common base $PQ$.
$\therefore A(\Delta BPQ) = A(\Delta CPQ)$ ... (III)
From (I), (II), and (III):
$\frac{AP}{PB} = \frac{AQ}{QC}$.

Q.4 Solve any two sub-questions. (8 Marks)
(1) If $\frac{1}{\sin^2\theta} - \frac{1}{\cos^2\theta} - \frac{1}{\tan^2\theta} - \frac{1}{\cot^2\theta} - \frac{1}{\sec^2\theta} - \frac{1}{\text{cosec}^2\theta} = -3$, then find the value of $\theta$.
Solution:

Simplifying the terms:

$\text{cosec}^2\theta - \sec^2\theta - \cot^2\theta - \tan^2\theta - \cos^2\theta - \sin^2\theta = -3$

Rearranging the terms:

$(\text{cosec}^2\theta - \cot^2\theta) - (\sec^2\theta + \tan^2\theta) - (\sin^2\theta + \cos^2\theta) = -3$

Using identities: $\text{cosec}^2 - \cot^2 = 1$ and $\sin^2 + \cos^2 = 1$

$1 - (\sec^2\theta + \tan^2\theta) - 1 = -3$

$-(\sec^2\theta + \tan^2\theta) = -3$

$\sec^2\theta + \tan^2\theta = 3$

Replace $\sec^2\theta$ with $1 + \tan^2\theta$:

$(1 + \tan^2\theta) + \tan^2\theta = 3$

$1 + 2\tan^2\theta = 3$

$2\tan^2\theta = 2 \implies \tan^2\theta = 1$

$\tan \theta = 1$

$\therefore \theta = 45^{\circ}$

(2) A cylinder of radius 12 cm contains water up to the height 20 cm. A spherical iron ball is dropped into the cylinder and thus water level raised by 6.75 cm. What is the radius of the iron ball?
Solution:

The volume of water displaced (rise in level) is equal to the volume of the sphere submerged.

Radius of Cylinder ($R$) = 12 cm
Rise in height ($h$) = 6.75 cm
Let radius of sphere be $r$.

Volume of Sphere = Volume of water raised in Cylinder

$$\frac{4}{3} \pi r^3 = \pi R^2 h$$ $$\frac{4}{3} r^3 = (12)^2 \times 6.75$$ $$r^3 = \frac{144 \times 6.75 \times 3}{4}$$ $$r^3 = 36 \times 20.25$$ $$r^3 = 729$$ $$r = 9 \text{ cm}$$
(3) Draw a circle with centre O having radius 3 cm. Draw tangent segments PA and PB through the point P outside the circle such that $\angle APB = 70^{\circ}$.
Solution:

Steps of Construction:

  1. Draw a circle with center O and radius 3 cm.
  2. We need the angle between tangents to be $70^{\circ}$. The angle subtended by the radii at the center will be supplementary to this angle (Quadrilateral OAPB is cyclic with 90° at A and B).
  3. $\angle AOB = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
  4. Draw a radius OA. Draw another radius OB such that $\angle AOB = 110^{\circ}$.
  5. Draw a line perpendicular to OA at A and a line perpendicular to OB at B.
  6. The point where these two perpendicular lines intersect is point P.
Q.5 Solve any one sub-question. (3 Marks)
(1) ABCD is a trapezium, AB || CD. Diagonals intersects in point P. Write the answers of the following questions:
Solution:

(a) Draw the figure:
(Draw a trapezium ABCD with AB parallel to CD at the bottom/top, and diagonals AC and BD intersecting at P.)

(b) Write any one pair of alternate angles and opposite angles:
Alternate angles: $\angle CDB \cong \angle ABD$ (since AB || CD).
Opposite angles: $\angle APB \cong \angle CPD$ (Vertically opposite angles).

(c) Write the names of similar triangles with test of similarity:
$\Delta APB \sim \Delta CPD$ by AA Test of Similarity (Alternate angles and vertically opposite angles are congruent).

(2) AB is a chord of a circle with centre O. AOC is diameter of circle, AT is a tangent at A. Write answers to the following:
Solution:

(a) Draw the figure:
(Draw a circle with center O. Draw diameter AC. Draw chord AB. Draw line AT tangent at A.)

(b) Measures of $\angle CAT$ and $\angle ABC$:
$\angle CAT = 90^{\circ}$ (Tangent Theorem: Tangent is perpendicular to radius at point of contact).
$\angle ABC = 90^{\circ}$ (Angle inscribed in a semicircle is a right angle).

(c) Are $\angle CAT$ and $\angle ABC$ congruent? Justify:
Yes, they are congruent.
Justification: Both angles measure $90^{\circ}$ as proved above.

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