Geometry ( Math 2 ) Practice paper 2026

In this post students will find 40 marks practice paper for SSC students Maharashtra Board. Students must solve this prectice paper and score 90 percentage in SSC exam.

Answer:

Distance of point (-3, 4) from the origin

Step 1: Apply the distance formula The distance

The distance $d$ of a point $(x,y)$from the origin $(0,0)$ by the distance formula:

the distance formula:

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{x^2+y^2}$

$d=\sqrt{(-3{)}^2+4^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d=5$

Answer:

(C) 5

2 MCQ. Answer: ( B )

3rd MCQ. Answer: (A) 2

4th MCQ : Answer:(B) 0.

Q.1. (B)

i. Find AB ?

$\cos {30}^{\circ }=\frac{AB}{14}$

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}=\frac{AB}{14}$

$AB=\frac{14\sqrt{3}}{2}$

$AB=7\sqrt{3}\text{cm}$

Answer:

$AB=\mathbf{7}\sqrt{\mathbf{3}}\text{cm}$.

ctor.

ii. Area of corresponding major sector

Area of Circle=Area of Minor Sector+Area of Major Sector

$314{\text{cm}}^2=100{\text{cm}}^2+\text{Area of Major Sector}$314 cm2=100 cm2+Area of Major Sector

$\text{Area of Major Sector}=314{\text{cm}}^2-100{\text{cm}}^2$

$\text{Area of Major Sector}=214{\text{cm}}^2$

Answer:

The area of the corresponding major sector is

$214{\text{cm}}^{\mathbf{2}}$.

---

iii. 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.

The value of

$\frac{A\left(\Delta DEF\right)}{A\left(\Delta MNK\right)}$

$\frac{\mathbf{4}}{25}$

iv. Value of acute angle 𝜃 :

Given

$\sec \theta =\frac{2}{\sqrt{3}}$

Answer:

The value of the acute angle 𝜃 =

${30}^{\mathbf{\circ }}$

Q.2. (A)

i. Capacity of a frustum-shaped bucket

formula

The formula for the volume of a frustum is

$V=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$

$=\frac{1}{3}\times \mathbf{\pi }\times 30({14}^2+7^2+14\times 7)$

$=\frac{1}{3}\times \frac{22}{7}\times 30(196+49+98)$

$=\frac{220}{7}\left(343\right)$

$=220\times 49$

$=10780{\text{cm}}^3$

$\text{litres}=\frac{10780}{1000}$

𝟏𝟎.𝟕𝟖 litres

ii. Circle problems

$m\left(\text{arc}AXB\right)={360}^{\circ }-\angle AOB$

= 360 – 120 = 240

Answers:

$m\left(\text{arc}AXB\right)={240}^{\mathbf{\circ }}$

$m\left(\text{arc}CAB\right)={175}^{\mathbf{\circ }}$

$\angle COB={175}^{\mathbf{\circ }}$

iii. Finding AB and BC

In the right-angled triangle

$ABC$, the sum of angles is 180°

$\angle ACB={90}^{\circ }$and

$AB=BC$ the triangle is an isosceles right-angled triangle, meaning

$\angle BAC=\angle BCA$ = 45 °

Q.2. (B) Solve the following sub-questions (Any four):

i) Answer:

Yes, the triangles in the given figure are similar by the 𝐒𝐒𝐒 (Side-Side-Side) test of similarity.

ii. Area of a sector.

The area of a sector of a circle is given by the formula

$A=\frac{\theta }{360}\times \pi r^2$

Answer:

The area of the sector associated with the arc is 𝟒𝟕.𝟏 cm².

iii. Constructing a tangent

Answer:The problem requires a geometric construction which cannot be represented in a text response.

iv. Find cos ?

${\cos }^2\theta =1+{\cot }^2\theta$

$\frac{41}{\mathbf{9}}$ = Cos

5th. Volume of a sphere

volume of a sphere is

$V=\frac{4}{3}\pi r^3$

Answer:

The volume of the sphere is

$113.04{\text{cm}}^{\mathbf{3}}$

Q.3. (A) Complete the following activity and rewrite it (Any one):

ii. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram. Complete the following activity.

We know that, slope of line

$\frac{{\mathbf{y}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}}{{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}}$

Slope of side AB = 3

Slope of side BC =

$\frac{\mathbf{5}\mathbf{-}\mathbf{2}}{\mathbf{8}\mathbf{-}\mathbf{\left(}-1\mathbf{\right)}}=\frac{\mathbf{3}}{\mathbf{9}}=\frac{\mathbf{1}}{\mathbf{3}}$

Slope of side CD =

$\frac{-4\mathbf{-}\mathbf{5}}{\mathbf{5}\mathbf{-}\mathbf{8}}=\frac{-9}{-3}=\mathbf{3}$

Slope of side AD =

$\frac{\mathbf{1}}{\mathbf{3}}$

Slope of side AB =𝐒𝐥𝐨𝐩𝐞 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐂𝐃,

Slope of side BC =𝐒𝐥𝐨𝐩𝐞 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐀𝐃

both the pairs of opposite sides of ABCD are parallel.
Points A, B, C and D are the vertices of a 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥𝐨𝐠𝐫𝐚𝐦.

B. Solve the following sub questions (any two):

i. Find co-ordinates of point P, if P divides the line segment joining the points A(-1,7) and B(4,-3) in the ratio 2: 3

ii. Draw a circle with centre O of radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at point M and N to the circle.

iii. A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60∘ with the horizontal. Find the height of the tree.

iv. Prove that, ‘In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides’.

4. Solve any two of the following sub-questions:

1. 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 iron ball?

2. 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

$\angle APB={70}^{\circ }$

3. Prove that cot𝜃+tan𝜃=cosec𝜃×sec𝜃.

Q.5. Solve the following sub question (anyone). (03)

1. Show that the points (2, 0), (-2, 0) and (0, 2) are the vertices of a triangle. Also state with reason the type of the triangle.

2. prove that :

$C{A}^2=CB\times CD$

Students can download this practice paper for free !

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