CAT Quantitative Aptitude Questions | Geometry Questions for CAT - Triangles

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

Explanatory Answer

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

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.