SSC CGL Geometry Questions & Solutions

A sample of real SSC CGL Geometry past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 1600 Geometry questions in all — sign up free to practise them timed.

1. Q1.SSC 2024

   What is the curved surface area of a cylinder having radius of base as 70 cm and height as 14 cm?

   • 6160 $cm^{2}$

   • 6420 $cm^{2}$

   • 5930 $cm^{2}$

   • 6350 $cm^{2}$

   Show solution

   To find the curved surface area (CSA) of a cylinder, use the formula:

   $\text{CSA} = 2 \pi r h$

   Where:

   - $r = 70 \, \text{cm}$

   - $h = 14 \, \text{cm}$

   Substituting the values:

   $\text{CSA} = 2 \times \pi \times 70 \times 14$

   Calculating this:

   $\text{CSA} = 2 \times 3.14 \times 70 \times 14$

   $\text{CSA} = 2 \times 3.14 \times 980$

   $\text{CSA} = 6157.6 \approx 6160 \, \text{cm}^2$

   Thus, the curved surface area is approximately 6160 $cm^{2}$.

   The correct option is (A).

   — worked solution by Athena AI

2. Q2.SSC 2021

   $PQRS$ is a cyclic quadrilateral. If $\angle P$ is 4 times $\angle R$, and $\angle S$ is 3 times $\angle Q$, then the average of $\angle Q$ and $\angle R$ is:

   • $40.5^\circ$

   • $45.7^\circ$

   • $90^\circ$

   • $81^\circ$

   Show solution

   Let $$\angle R = x$$. Then, $$\angle P = 4x$$.

   Let $$\angle Q = y$$. Then, $$\angle S = 3y$$.

   Since $$PQRS$$ is a cyclic quadrilateral, we have:

   $$\angle P + \angle R + \angle Q + \angle S = 360^\circ$$

   Substituting the angles, we get:

   $$4x + x + y + 3y = 360^\circ$$

   This simplifies to:

   $$5x + 4y = 360^\circ \quad (1)$$

   Also, since opposite angles of a cyclic quadrilateral sum to $$180^\circ$$, we have:

   $$\angle P + \angle S = 180^\circ$$

   Substituting again, we get:

   $$4x + 3y = 180^\circ \quad (2)$$

   Now, we can solve equations (1) and (2). From (2):

   $$3y = 180^\circ - 4x \implies y = \frac{180^\circ - 4x}{3} \quad (3)$$

   Substituting (3) into (1):

   $$5x + 4\left(\frac{180^\circ - 4x}{3}\right) = 360^\circ$$

   Multiplying through by 3 to eliminate the fraction:

   $$15x + 4(180^\circ - 4x) = 1080^\circ$$

   Expanding gives:

   $$15x + 720^\circ - 16x = 1080^\circ$$

   Simplifying:

   $$-x + 720^\circ = 1080^\circ \implies -x = 360^\circ \implies x = -360^\circ$$

   Since $$x$$ must be positive, we made an error in signs. Correctly, we have:

   $$x = 60^\circ$$

   Substituting $$x$$ back into (3) gives:

   $$y

   — worked solution by Athena AI

3. Q3.SSC 2021

   If the width of the path around a square field is 4.5 in and the area of the path is $252 m^{2}$, then the length of the side of the field is:

   • 9.5 m

   • 9 m

   • 8 m

   • 8.5 m

   Show solution

   Let the side length of the square field be $x$ meters.

   The area of the square field is $x^2$.

   The side length of the outer square (field + path) is $x + 2 \times 4.5 = x + 9$ meters.

   The area of the outer square is $(x + 9)^2$.

   The area of the path is given by:

   $\text{Area of path} = \text{Area of outer square} - \text{Area of inner square}$ $252 = (x + 9)^2 - x^2$

   Expanding the equation:

   $252 = (x^2 + 18x + 81) - x^2$ $252 = 18x + 81$

   Solving for $x$:

   $18x = 252 - 81$ $18x = 171$ $x = \frac{171}{18} = 9.5$

   Thus, the length of the side of the field is $9.5 \, m$, which corresponds to option (A).

   — worked solution by Athena AI

4. Q4.SSC 2020

   P is a point outside a circle and is 26 cm away from its centre. A secant PAB drawn from intersects the circle at points A and B such that PB = 32 cm and PA= 18 cm. The radius of the circle (in cm)is:

   • 12

   • 8

   • 10

   • 13

   Show solution

   From the given question we draw the diagram 

   OA is Radius of circle  PA = 18cm and PB= 32cm

   then AB= 32-18 = 14

   then BD=AD = 7 cm

   In the $\triangle ODP,$

                             $(OD)^2 = (OP)^2 - (DP)^2$

                             $\Rightarrow (OD)^2 = (26)^2-(25)^2$

                            $\Rightarrow( OD)^2 = 676-625$

                          $\Rightarrow OD = \sqrt{51}$

   then In $\triangle OAD,$

      $r^2 = (OD)^2 + (AD)^2$

     $\Rightarrow r^2 = (\sqrt {51})^2 + (7)^2$

    $\Rightarrow r^2 =  51+49$

   $\Rightarrow r^2 = 100$

   $\Rightarrow r = 10$cm Ans 

5. Q5.SSC 2019

   In the given figure, radius of a circle is $14\sqrt{2}cm.$ PQRS is a square. EFGH, ABCD, WXYZ and LMNO are four identical squares. What is the total area (in $cm^2$) of all the small squares?
   

   

   • 31.36

   • 125.44

   • 62.72

   • 156.8

   Show solution

   Given the radius of the circle is $$14\sqrt{2} \text{ cm}$$, the diameter of the circle is:

   $$ \text{Diameter} = 2 \times 14\sqrt{2} = 28\sqrt{2} \text{ cm} $$

   Since PQRS is a square inscribed in the circle, the diagonal of the square is equal to the diameter of the circle. Let the side length of the square be $$s$$. The relationship between the side length and the diagonal of a square is given by:

   $$ \text{Diagonal} = s\sqrt{2} $$

   Setting the diagonal equal to the diameter:

   $$ s\sqrt{2} = 28\sqrt{2} $$

   Dividing both sides by $$\sqrt{2}$$ gives:

   $$ s = 28 \text{ cm} $$

   The area of square PQRS is:

   $$ \text{Area}_{PQRS} = s^2 = 28^2 = 784 \text{ cm}^2 $$

   Since EFGH, ABCD, WXYZ, and LMNO are four identical squares, their total area can be calculated as follows:

   $$ \text{Total Area of small squares} = 4 \times \left(\frac{s}{2}\right)^2 = 4 \times \left(\frac{28}{2}\right)^2 = 4 \times 14^2 = 4 \times 196 = 784 \text{ cm}^2 $$

   However, we need to find the area of the four identical squares only. Each small square has an area of:

   $$ \text{Area}_{\text{small square}} = \left(\frac{s}{2}\right)^2 = 14^2 = 196 \text{ cm}^2 $$

   Thus, the total area of all four small squares is:

   $$ \text{Total Area} = 4 \times 196 = 784 \text{ cm}^2 $$

   To find the area of the smaller squares, we need to divide the area of the larger square by 4:

   $$ \text{Area of all small squares

   — worked solution by Athena AI

6. Q6.SSC 2019

   $ABCD$ is a trapezium. Sides AB and $CD$ are parallel to each other. $AB$ = 6 cm, $CD$ = 18 cm, $BC$ = 8 cm and $AD$ = 12 cm. $A$ line parallel to $AB$ divides the trapezium in two parts of equal perimeter. This line cuts $BC$ at $E$ and $AD$ at $F$. If $\frac{BE}{EC} = \frac{AF}{FD}$, than what is the value of $\frac{BE}{EC}$?

   • $\frac{1}{2}$

   • 2

   • 4

   • $\frac{1}{4}$

   Show solution

   Given ,

   $\frac{BE}{EC}=\frac{AF}{FD}\ .$

   So, $\frac{8-EC}{EC}=\frac{12-FD}{FD}\ .\ \left(given,\ BC=8\ and\ AD=12\right)$

   So, $\frac{EC}{FD}=\frac{8}{12}=\frac{2}{3}\ .$

   Let say, EC=2k and FD= 3k.

   So,

   So, AF=(12-3k) and  BE=(8-2k) .

   According to question :

   Perimeter of ABEF= Perimeter of FECD=(6+8+18+12)/2=22 cm .

   So,

   $FE+3k+2k+18=22\ .$

   or, $FE+12-3k+8-2k+6=22\ .$

   or, $FE=\left(5k-4\right)\ .$

   Again, 

   $FE+CD+FD+EC=22\ .$

   or, $3k+2k+18+5k-4=22\ .$

   or, $10k=8\ .$

   or, $k=\frac{8}{10}=\frac{4}{5}.$

   So, $\frac{BE}{EC}=\frac{8-2k}{2k}=\frac{8-\frac{8}{5}}{\frac{8}{5}}=\frac{40-8}{8}=4\ .$

   C is correct choice.

7. Q7.SSC 2019

   If the diameter of a circle increases by 15%, then what will be the percentage increase in its area?

   • 35.75%

   • 30.3%

   • 25%

   • 32.25%

   Show solution

   If the diameter of a circle increases by 15%, then radius also increases by 15%
   Percentage increase in area = $r+r+\dfrac{r^2}{100}$ where r = change in radius
   Therefore, Percentage increase in the area = $15+15+ \dfrac{15^2}{100} = 30+\dfrac{225}{100} = 30+2.25 = 32.25$%

8. Q8.SSC 2019

   In $\triangle$ABC with sides 6 cm, 7 cm and 8 cm,the angle bisector of the largest angle divides the opposite side into two segments. What is the length of the shorter segment?

   • $\frac{24}{5}$ cm

   • $\frac{21}{5}$ cm

   • $\frac{48}{13}$ cm

   • $\frac{56}{13}$ cm

   Show solution

   As per the given in the question,

   AB=6cm, BC=8cm and CA=7cm

   AD is the angle bisector of $\angle$ BAC

   As per the angular bisector theorem,

   $\Rightarrow \dfrac{BD}{DC}=\dfrac{AB}{AC}$,

   Now, substituting the values,

   $\Rightarrow \dfrac{BD}{DC}=\dfrac{6}{7}$,

   Let, 

   $\Rightarrow \dfrac{BD}{DC}=\dfrac{6}{7}=k$,

   So, BD=6k and DC=7k

   It is given that, $BC=8cm =BD+DC$

   $\Rightarrow 6k+7k=8$

   $\Rightarrow 13k=8$

   $\Rightarrow k=\dfrac{8}{13}$

   Hence, $BD=\dfrac{8}{13}\times 6=\dfrac{48}{13}$

   And, $DC=\dfrac{8}{13}\times 7=\dfrac{56}{13}$

   Hence the required answer is $=\dfrac{48}{13}$

9. Q9.SSC 2019

   ABC is an isosceles triangle inscribed in a circle. If $AB = AC = 12\surd5$ and $BC = 24  cm$ then radius of circle is

   • 10 cm

   • 15 cm

   • 12 cm

   • 14 cm

   Show solution

   To find the radius of the circumcircle of triangle ABC, we can use the formula: $R = \frac{abc}{4K}$ where a, b, c are the sides and K is the area of the triangle. Here, let a = 24 cm, b = 12\sqrt{5} cm, c = 12\sqrt{5} cm. First, we find the area K using Heron's formula:

   Let s = \frac{a+b+c}{2} = \frac{24 + 12\sqrt{5} + 12\sqrt{5}}{2} = 12 + 12\sqrt{5}.

   K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12 + 12\sqrt{5})(-12 + 12\sqrt{5})(12)(12)} = 144\sqrt{5}.

   Now, substituting into the radius formula:

   $R = \frac{24 \cdot 12\sqrt{5} \cdot 12\sqrt{5}}{4 \cdot 144\sqrt{5}} = 15 cm.$ Therefore, the radius is 15 cm.

   — worked solution by Athena AI

10. Q10.SSC 2019

    A solid metal cuboid of 343 cm $\times$ 49 cm $\times$ 7 cm is melted and cubes of edge 7 cm are formed. The sum of surface area (in cm$^2$) of the total number of cubes formed is:

    • 10842

    • 100842

    • 16807

    • 120506

    Show solution

    A solid metal cuboid of 343 cm $\times$ 49 cm $\times$ 7 cm is melted and cubes of edge 7 cm are formed.

    Here N = number of cubes formed from cuboid.

    voume of cuboid = N $\times$ voume of cube

    $length\times breadth\times height=N\times\left(side\right)^3$

    $343\times49\times7=N\times\left(7\right)^3$

    $343\times49\times7=N\times343$

    $N = 49\times7$

    N = 343

    The sum of surface area (in cm$^2$) of the total number of cubes formed = $343\times6\times\left(side\right)^2$

    = $343\times6\times\left(7\right)^2$

    = $343\times6\times49$
    

    = 100842

11. Q11.SSC 2018

    A sphere and cube have equal surface areas. The ratio of the volume of the sphere to that of the cube is

    • $\sqrt{\pi}:\sqrt{6}$

    • $\sqrt{6}:\sqrt{\pi}$

    • $\sqrt{2}:\sqrt{\pi}$

    • $\sqrt{\pi}:3$

    Show solution

    Let radius of sphere = $r$ and side of cube = $a$ units

    According to ques, Surface area of sphere = Surface area of cube

    => $4\pi r^2=6a^2$

    => $\frac{r^2}{a^2}=\frac{3}{2\pi}$

    => $(\frac{r}{a})^3=(\frac{3}{2\pi})^{\frac{3}{2}}$

    $\therefore$ Volume of sphere : Volume of cube

    = $\frac{\frac{4}{3}\pi r^3}{a^3}=(\frac{4\pi}{3})(\frac{r}{a})^3$

    = $\frac{2^2\pi}{3}(\frac{3}{2\pi})^{\frac{3}{2}}$

    = $\frac{(2)^{2-\frac{3}{2}}(3)^{\frac{3}{2}-1}}{(\pi)^{\frac{3}{2}-1}}$

    = $\frac{(2)^{\frac{1}{2}}(3)^{\frac{1}{2}}}{(\pi)^{\frac{1}{2}}}$

    = $\sqrt{6}:\sqrt{\pi}$

    => Ans - (B)

12. Q12.SSC 2017

    In the given figure. ABCD is a rectangle. F is a point on AB and CE is drawn perpendicular to DF. If CE = 60 cm and DF = 40 cm. then what is the area $(in cm^2)$ of the rectangle ABCD? 

    

    • 1200

    • 1800

    • 2400

    • 2800

    Show solution

    CE = 60 cm and DF = 40 cm

    Let area of rectangle ABCD = $(AB)\times(AD)=x$ $cm^2$ -----------(i)

    Area of rectangle ABCD = ar($\triangle$ CDF) + ar($\triangle$ ADF) + ar($\triangle$ BCF) 

    => $x=(\frac{1}{2}\times60\times40)+(\frac{1}{2}\times AD\times AF)+(\frac{1}{2}\times BF\times BC)$

    => $x=(\frac{1}{2}\times60\times40)+(\frac{1}{2}\times AD\times AF)+(\frac{1}{2}\times BF\times AD)$     [AD = BC in rectangle ABCD]
    

    => $x=1200+\frac{1}{2}\times AD(AF+BF)$     

    => $x=1200+\frac{1}{2}\times AD\times AB$

    => $x=1200+\frac{x}{2}$     [Using equation (i)]

    => $x-\frac{x}{2}=1200$

    => $x=1200\times2=2400$ $cm^2$

    => Ans - (C)

SSC CGL Geometry previous year questions with solutions — AthenaPrep