CAT Geometry Questions & Solutions

A sample of real CAT Geometry past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 189 Geometry questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0) , then the value of a is

   • $\dfrac{27}{7}$

   • $\dfrac{10}{3}$

   • $\dfrac{13}{4}$

   • $\dfrac{29}{9}$

   Show solution

   Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).

   For parallelogram (PQRS), S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5,4)
   

   Diagonal SQ passes through S(5,4) and Q(1,-5).

   Slope m = $\dfrac{-5-4}{1-5} = \dfrac{9}{4}$
   

   Equation of SQ is   $y - 4 = \dfrac{9}{4}(x - 5)$
   

   At the x-axis, y = 0: $-4 = \dfrac{9}{4}(a - 5)$

   $-16 = 9(a - 5)$

   $9a = 29$

   $a = \dfrac{29}{9}$
   

   So, $a = {\dfrac{29}{9}}$

2. Q2.CAT 2024

   A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is

   • $72(2 + \sqrt{2})$

   • $36(1 + \sqrt{2})$

   • $72(1 + \sqrt{2})$

   • $36(2 + \sqrt{2})$

   Show solution

   This is the figure in the question, 

   

   We are given that each side is 6cm long, 
   To find the side AC, we can use cosine rule, since we know each interior angle of the octagon(which is 135 degrees)
   
   

   $\cos\left(\angle ABC\right)=\dfrac{\left(AB^2+BC^2-AC^2\right)}{2\left(AB\right)\left(AC\right)}$
   
   

   $\cos\left(135\right)=\dfrac{\left(36+36-AC^2\right)}{2\left(36\right)}$
   
   

   $-\dfrac{1}{\sqrt{2}}=\dfrac{\left(72-AC^2\right)}{72}$
   
   

   $AC^2=72+36\sqrt{2}$
   
   

   $AC^2=36\left(2+\sqrt{2}\right)$

   Since AC is the side of the square, and the area of a square is square of the side. 

   Answer is $36\left(2+\sqrt{2}\right)$

3. Q3.CAT 2022

   Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is

   Answer: 9

   Show solution

   

   Area of ABD : Area of BDC = 1:1

   Therefore, area of ABD = 54

   Area of ADE : Area of EDB = 1:1

   Therefore, area of ADE = 27

   O is the centroid and it divides the medians in the ratio of 2:1

   Area of BEO : Area of EOD = 2:1

   Area of EOD = 9

4. Q4.CAT 2020

   The points (2,1) and (-3,-4) are opposite vertices of a parallelogram.If the other two vertices lie on the line $x+9y+c=0$, then c is

   • 12

   • 13

   • 15

   • 14

   Show solution

   The midpoints of two diagonals of a parallelogram are the same

   Hence the midpoint of (2,1) and (-3,-4) lie on  $x+9y+c=0$

   midpoint of (2,1) and (-3,-4) = ( $\frac{2-3}{2},\frac{1-4}{2}$ ) = (-1/2 , -3/2)

   Keeping this cordinates in the above line equation, we get c = 14

5. Q5.CAT 2019

   If the rectangular faces of a brick have their diagonals in the ratio $3 : 2 \surd3 : \surd{15}$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

   • $\sqrt{3} : 2$

   • $1 : \sqrt{3}$

   • $2 : \sqrt{5}$

   • $\sqrt{2} : \sqrt{3}$

   Show solution

   Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $\surd3$ and $\surd{15}$

   Hence, $a^{2\ }+\ b^2$ = $3^2$   ......(1)

   $b^{2\ }+\ c^2$ = $(2\sqrt{3})^2$ ......(2)

   $c^{2\ }+\ a^2$ = $(\sqrt{15})^2$ ......(3)

   Adding the three equations, 2($a^2+b^2+c^2$) = 9+12+15=36

   =>$a^2+b^2+c^2$ = 18......(4)

   Subtracting (1) from (4), we get $c^2$ = 9    =>c=3

   Subtracting (2) from (4), we get $a^2$ = 6    =>a=$\sqrt{6}$

   Subtracting (3) from (4), we get $b^2$ = 3    =>b=$\sqrt{3}$

   The ratio of the length of the shortest edge of the brick to that of its longest edge is = $\ \frac{\ \sqrt{3}}{3}$ = $1 : \sqrt{3}$

6. Q6.CAT 2017

   The area of the closed region bounded by the equation 

   I x I + I y I = 2 in the two-dimensional plane is

   • $4\pi$ sq. units

   • 4 sq. units

   • 8 sq. units

   • $2\pi$ sq. units

   Show solution

   The following equation will form a square of side $2\sqrt{2}$ .

   The area of the square = $(2\sqrt{2})^2$ = 8 units.

   
7. Q7.CAT 2008

   Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the other. What is the area (in sq. cm.) of the intersecting region?

   • $\frac{\pi}{3}-\frac{\sqrt 3}{4}$

   • $\frac{2\pi}{3}+\frac{\sqrt 3}{2}$

   • $\frac{4\pi}{3}-\frac{\sqrt 3}{2}$

   • $\frac{4\pi}{3}+\frac{\sqrt 3}{2}$

   • $\frac{2\pi}{3}-\frac{\sqrt 3}{2}$

   Show solution

   

   The circumferences of the two circle pass through each other's centers. Hence, O1A = O1B=O1O2 = 1cm

   By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P.
   
   O1P = Half of O1O2 = 1/2 cm
   So, the angle AO1P = 60 degrees as cos 60 = 1/2
   By symmetry, BO1P = 60 degrees.
   So, angle AO1B= 120 degrees

   

   In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) - A(triangleO1AB )(red)

   Area of sector = 120°/360° * $\pi * 1^2$ = $\pi/3$

   Area of triangle = 1/2 * b * h = 1/2 * (2* 1 cos 30°) * (1/2) = √3/4

   Hence, required area = $\frac{\pi}{3}-\frac{\sqrt 3}{4}$ . Hence so the required area is 2 times the above value which is $\frac{2\pi}{3}-\frac{\sqrt 3}{2}$

8. Q8.CAT 2005

   P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

   • $2r(1+ \sqrt3)$

   • $2r(2+ \sqrt3)$

   • $r(1+ \sqrt5)$

   • $2r+ \sqrt3$

   Show solution

   

   Let PQR be an equilateral triangle with side equal to x and let the intersection point of PS and QR be M.

   Clearly, the circle is the circumcircle of the triangle PQR.

   QR = x => QM = $\frac{x}{2}$ because a perpendicular from the centre to any chord bisects the chord.

   Angle OQM = 30 degrees and QM is equal to $\frac{x}{2}$ => OQ = $\frac{\frac{x}{2}}{cos(30)}$ = $\frac{x}{\sqrt{3}}$

   Hence the radius of the circumcircle of an equilateral triangle is equal to $\frac{x}{\sqrt{3}}$.

   Angle PQS = 90 degrees as it is an angle in a semicircle. PS bisects angle QPR => angle QPS is 30 degrees. Hence QS subtends an angle of 30 degrees in the major arc => QS subtends an angle of 60 degrees at the centre because angle subtended by a chord at the centre is twice the angle subtended by the chord in the major arc.

   Angle QOS = 60 degrees => Triangle QOS is equilateral and hence QS is equal to radius of the circle => QS = $\frac{x}{\sqrt{3}}$

   Given that radius is r => r = $\frac{x}{\sqrt{3}}$ => x = $r\sqrt{3}$

   => Perimeter of PQRS = PQ+QS+SR+RP= $r\sqrt{3} + r + r + r\sqrt{3}$ = $2r(1+\sqrt{3})$

9. Q9.CAT 2003

   In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP:QC is equal to

   • 1:1

   • 3:2

   • 4:1

   • 3:8

   Show solution

   

   Let BD = x .Semi-perimeter of triangle ABC = 12. Now by herons formula area of ABC is 24. Also Area = 0.5*x*10 . We get x = 24/5 .  AP = 6/5  and CQ = 16/5 . Hence the required ratio is 3:8. 

10. Q10.CAT 2001

    A square, whose side is 2 m, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon, in metres, is

    • $\frac{\sqrt 2}{\sqrt 2 +1}$

    • $\frac{2}{\sqrt 2 + 1}$

    • $\frac{2}{\sqrt 2 - 1}$

    • $\frac{\sqrt 2}{\sqrt 2-1}$

    Show solution

    Let the length of each side of the octagon be x.

    

    So, length of the square will be x+2*(x/√2) = 2

    => x(1+√2) = 2 => x = 2/(1+√2)

11. Q11.CAT 1999

    Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?

    • 495

    • 550

    • 1045

    • 2475

    Show solution

    For a triangle to be formed, we need three points.

    Case 1: Select 2 points on the line that has 10 points and 1 point on the line that ha 11 points.

    This can be done in $^{10}C_2$*$^{11}C_1$ ways = 495 ways.

    Case 2: Select 2 points on the line that has 11 points and 1 point on the line that ha 10 points.

    This can be done in $^{11}C_2$*$^{10}C_1$ ways = 550 ways.

    495 + 550 = 1045 ways.

12. Q12.CAT 1997

    In the adjoining figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of CBE to that of ADE?

    

    • 1 : 4

    • 1 : 2

    • 1 : 3

    • Data insufficient

    Show solution

    As we know angles of same sectors are equal
    Hence angle B and angle D will be equal. Angle BCE and angle EAD will be equal.
    So triangles BCE and EAD will be similar triangles with sides ratio as 12:24 or 1:2.
    Area will be in ratio of 1:4.

CAT Geometry previous year questions with solutions — AthenaPrep