CAT 2023 QA Questions and Solutions
Question 17: Geometry – Triangles
A triangle is drawn with its vertices on the circle \( C \) such that one of its sides is a diameter of \( C \) and the other two sides have their lengths in the ratio \( a : b \). If the radius of the circle is \( r \), then the area of the triangle is:
A. \( \frac{2 a b r^2}{a^2 + b^2} \)
B. \( \frac{a^2 + b^2}{4 a b r^2} \)
C. \( \frac{a b r^2}{a^2 + b^2} \)
D. \( \frac{a b r^2}{2 (a^2 + b^2)} \)
Answer: A. \( \frac{2 a b r^2}{a^2 + b^2} \)
Detailed Solution:
Using the property that the diameter of a circle forms a right angle with the chord, and the formula for area, we get:
\[ \text{Area} = \frac{2 a b r^2}{a^2 + b^2} \].
Question 18: Geometry – Quadrilaterals & Polygons
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:
A. 2 : 4 : 1
B. 1 : 2 : 4
C. 1 : 1 : 2
D. 1 : 2 : 1
Answer: A. 2 : 4 : 1
Detailed Solution:
Given that the areas are in a geometric progression, we assume the areas as \( A, Ar, \) and \( Ar^2 \).
If \( AQCD = 4 \times \text{Area of } ABP \), solving for \( r \), we get \( r = 2 \).
Thus, the ratio of \( BP : PQ : QC \) is 2 : 4 : 1.
