CAT Mensuration Questions & Solutions

A sample of real CAT Mensuration past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 17 Mensuration questions in all — sign up free to practise them timed.

1. Q1.CAT 2024

   The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

   • 1125 $\pi$

   • 750 $\pi$

   • 1125 $\pi \sqrt{2}$

   • 750 $\pi \sqrt{2}$

   Show solution

   Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm
   So, $2(lb+bh+hl)=846$ . 

   And $4(l+b+h)=144$
   $(l+b+h)=36$
   $\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$
   $1296=\left(l^2+b^2+h^2\right)+846$
   $450=l^2+b^2+h^2$

   We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

   

   This is nothing but,  $\sqrt{l^2+b^2+h^2}=2R$
   $l^2+b^2+h^2=4R^2$
   $450=4R^2$
   $R^2=\frac{225}{2}$
   $R=\frac{15}{\sqrt{2}}$

   Volume of sphere will be  $\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$

   $\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$

   $\pi\ \times\ 1125\sqrt{\ 2}$
   

2. Q2.CAT 2023

   A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is

   • 2 : 1

   • 1 : 1

   • $\sqrt{5} : 1$

   • $\sqrt{2} : 1$

   Show solution

   Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

   

   The radius, l/2 and b in the above diagram form a right-angled triangle.
   

   => $\left(\frac{l}{2}\right)^2+b^2=2^2$
   
   We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $\left(\frac{l}{2}\right)^2$ and $b^2$ .
   
   Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied
   
   => $\left(\frac{l}{2}\right)^2=b^2$ => l = 2b.
   => l/b = 2/1.

3. Q3.CAT 2023

   In a regular polygon, any interior angle exceeds the exterior angle by 120 degrees. Then, the number of diagonals of this polygon is

   Answer: 54

   Show solution

   The sum of the interior angles of a polygon of 'n' sides is given by $\left(2n-4\right)\times\ 90$ , and the sum of the exterior angles of a polygon is 360 degrees.

   So, the difference between them will be 120 * n

   => $\left(2n-4\right)90-360=120n$

   => 60n = 720 => n = 12.

   We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

4. Q4.CAT 2023

   In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is

   • 1:2:4

   • 1:2:1

   • 2:4:1

   • 1:1:2

   Show solution

   

   It is given that AB = 9 cm, BC = 6 cm.

   It is also known that the areas of the figures ABP, APQ, and AQCD are in geometric progression.

   Hence, the area of the ABP, APQ, and AQCD are k, 2k, and 4k respectively.

   The ratio of BP, PQ, QC will be the ratio of the respective triangles. Hence, we can draw a line from point A to point C.

   

   Let the area of triangle AQC be x, which implies the area of triangle ADC = ADQC - AQC = 4k -x, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.

   Therefore, 4k-x = 3k+x 

   => x = k/2

   Hence the ratio of BP: PQ: CQ = k:2k: k/2 =2:4:1

5. Q5.CAT 2022

   A trapezium $ABCD$ has side $AD$ parallel to $BC, \angle BAD = 90^\circ, BC = 3$ cm and $AD= 8$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is

   Answer: 66

   Show solution

   

   CD =  $\sqrt{\ y^2+25}$

   $11+y+\sqrt{y^2+25}=36$

   $\sqrt{y^2+25}=25-y$

   $y^2+25=25^2+y^2-50y$

   2y = 24

   y = 12

   Area of trapezium =  $3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$

6. Q6.CAT 2021

   If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

   • $4\sqrt{6}$

   • $6\sqrt{6}$

   • $\sqrt{6}$

   • $2\sqrt{6}$

   Show solution

   Area of a regular hexagon =  $\frac{3\sqrt{3}}{2}x^2$

   Area of an equilateral triangle =  $\frac{\sqrt{3}}{4}\left(a\right)^2$ ; where a = side of the triangle

   Since the area of the two figures are equal, we can equate them as folllows:  $\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$

   On simplifying:  x^2=24\

   $\therefore\ x=2\sqrt{6}$

7. Q7.CAT 2021

   If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

   • $\sqrt{37}+\sqrt{13}$

   • $\sqrt{13}+\sqrt{12}$

   • $\frac{\sqrt{37}+\sqrt{13}}{2}$

   • $\frac{\sqrt{13}+\sqrt{12}}{2}$

   Show solution

   All the sides of the rhombus are equal.

   The area of a rhombus is  $12\ cm^2$

   Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

   The area of a rhombus is  $\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$

   d1*d2 = 24.

   The length of the side of a rhombus is given by  $\frac{\sqrt{\ d1^2+d2^2}}{2}$ . This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

   $\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$

   Hence  $\sqrt{\ d1^2+d2^2}\ =\ 10$

   $d1^2+d2^2\ =\ 100$

   Using d1*d2 = 24, 2*d1*d2 = 48.

   $d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$

   $d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$

   $d1+d2\ =\ \sqrt{\ 148}$   (1)

   d1-d2 =  $\sqrt{52}$    (2)

   (1) + (2)= 2*(d1) = 2*( $\sqrt{\ 37}+\sqrt{\ 13}$ )

   d1 =  $\sqrt{\ 37}+\sqrt{\ 13}$

   or 

   In a rhombus the area of a Rhombus is given by :

   

   The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

   The area of a Rhombus is :  $\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$

   ab =6.

   The length of each side is :  $\sqrt{\ a^2+b^2}$ = 5,  a^2+b^{2\ }=\ 25,\

   $\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$

   ( $\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$

   $2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$ ,  $2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$ .

   2a is longer diagonal which is equal to  $\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$

8. Q8.CAT 2021

   The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

   • 120000

   • 90000

   • 100000

   • 160000

   Show solution

   Let us draw the rectangle.

   

   Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

   In this question, we are going to assume that the L is greater than B.

   Hence, the one side painted at Rs 200/ft should be B to minimise costs.

   Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

   Now, L x B = 60000

   B = 60000/L

   Hence, total cost = 300B + 200L = 18000000/L + 200L

   To minimise this cost, we can use AM>=GM,

   $\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$

   $\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$

   $\frac{18000000}{L}+200L\ge2\times\ 60000$

   Hence, minimum cost = Rs 120000.

9. Q9.CAT 2021

   Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10cm and 20cm, respectively. If the angle $\angle ADC$ is equal to $30^{0}$ then the area of the parallelogram, in sq.cm is

   • $\frac{25(\sqrt{5}+\sqrt{15})}{2}$

   • $25(\sqrt{3}+\sqrt{15})$

   • $\frac{25(\sqrt{3}+\sqrt{15})}{2}$

   • ${25(\sqrt{5}+\sqrt{15})}$

   Show solution

   

   Applying cosine rule in triangle ACD, 

   $100+X^2-2\times\ 10\times\ X\cos30=400$

   $X^2-10X\sqrt{\ 3}-300=0$

   Solving, we get X =  $\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$

   Hence, area = 10Xsin 30 = $\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$

   = $25(\sqrt{3}+\sqrt{15})$

10. Q10.CAT 2020

    A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

    • 243

    • 232

    • 256

    • 264

    Show solution

    

    Let the base radius be 3r.

    Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

    Volume of frustum= $\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$

    Volume of upper cone =  $\frac{\pi}{3}.r^2.9$

    Difference=  $\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$ =>  $\frac{\pi}{3}\cdot r^2=1$

    Volume of larger cone =  $\frac{\pi}{3}\cdot9r^2\cdot27=243$

11. Q11.CAT 2020

    A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

    • $\frac{6\pi}{25}$

    • $\frac{5\pi}{18}$

    • $\frac{3\pi}{25}$

    • $\frac{2\pi}{15}$

    Show solution

    

    Let the length of radius be 'r'.

    From the above diagram,

    x^2+r^2=6^2\ ....(i)

    \left(10-x\right)^2+r^2=8^2\ ----(ii)

    Subtracting (i) from (ii), we get: 

    x=3.6 =>  $r^2=36-\left(3.6\right)^2$ ==>  $r^2=36-\left(3.6\right)^2\ =23.04$ .

    Area of circle =  \pi\ r^2=23.04\pi\

    Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

    .'. Ratio of areas = 23.04 \pi\ /96= $\frac{6\pi}{25}$
    

12. Q12.CAT 2020

    In a trapezium $ABCD$, $AB$ is parallel to $DC$, $BC$ is perpendicular to $DC$ and $\angle BAD=45^{0}$. If $DC$ = 5cm, $BC$ = 4 cm,the area of the trapezium in sq cm is

    Answer: 28

    Show solution

    

    Given, BC = DE = 4

    CD = BE = 5

    In triangle ADE, $\angle EAD=45^{0}$

    $\tan\ 45\ =\ \frac{DE}{AE}$ => AE = 4

    Area of trapezium = Area of rectangle BCDE + Area of triangle AED

    = 20 + 8 = 28

CAT Mensuration previous year questions with solutions — AthenaPrep