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The question is from CAT Geometry - Triangles. It tests our understanding on properties of a triangle. CAT Geometry questions are heavily tested in CAT exam. Make sure you master Geometry problems. In this question it discusses about the properties of area of a triangle.

Question 30: Two circles with centres O1 and O2 touch each other externally at a point R. AB is a tangent to both the circles passing through R. P’Q’ is another tangent to the circles touching them at P and Q respectively and also cutting AB at S. PQ measures 6 cm and the point S is at distance of 5 cms and 4 cms from the centres of the circles. What is the area of the triangle SO1O2?

  1. 9 cm2
  2. 3(4+√7)/2 cm2
  3. 27/2 cm2
  4. (3√41)/2 cm2

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Explanatory Answer

Method of solving this CAT Question from Triangles: To calculate the area of a triangle, determine its base and altitude!!
CAT Question - Geometry - Triangles

From the diagram we see that SP, SR are tangents to circle1 from same point S. Similarly SR, SQ are tangents from same point to circle 2.
SP = SR; SQ = SR implies SP = SQ
Given PQ = 6cm
SP + SQ = 6
Therefor SR = SP = SQ = 3 cm.

SR is the altitude to the triangle SO1O2. We need to find the length of the base O1O2 to determine the area.
O1RS is a right angled triangle with hypotenuse = 5 and one side = 3
Therefore, O1R = √(52 - 32) = 4cm
Similarly, O2RS is a right angled triangle with hypotenuse = 4 and one side = 3
Therefore, O2R = √(42 - 32) = √7cm
O1O2 = O1R + O2R = 4 + √7
Area of the triangle SO1O2 = 1/2 * SR * O1O2 = 1/2 * 3 * (4 + √7) cm2

The question is " What is the area of the triangle SO1O2?"

Hence, the answer is 3(4+√7)/2 cm2.

Choice B is the correct answer.


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