A Circle Is Inscribed Inside Right Triangle Abc Shown Above GMAT Problem Solving

Question:

A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

1. 4π(√2−1)^2
2. 8π(√2−1)^2
3. 2π
4. 4π
5. 8π(1−√2)

“A circle is inscribed inside right triangle ABC shown above''- is a topic that belongs to the GMAT Quantitative reasoning section of the GMAT exam. GMAT quant section measures the critical thinking and numerical literacy of the students. The students must determine the correct answer choice by precisely calculating the sum mathematically. The students must have advanced knowledge of mathematical calculations to solve GMAT Problem Solving questions. The GMAT Quant topic in the problem-solving part consists of numerous mathematical problems that can be cracked only by better quantitative knowledge.

Solution and Explanation:

Approach Solution 1:

The problem statement informs that:

Given:

• A circle is inscribed inside the right triangle ABC.
• The length of AB of the right triangle ABC is 4.

Find Out:

• The area of the circle.

This is a right isosceles triangle. Since one angle is 45°, and another angle will be 45°.
Therefore, the area of triangle ABC is = ½ x base x height = ½ ∗ 4 ∗ 4 = 8 (given AB = 4 then BC also = 4 since angles are equal sides will also be equal).
Let the centre of the incircle is O and the radius is r.
Therefore, the area of triangle AOB = ½ ∗ r ∗ 4 ( since AB = 4 units and 'r' is the length of height)
Again we can say, the area of triangle BOC = ½ ∗ r∗ 4
And similarly, Area of triangle AOC = ½ ∗ r∗ AC = ½ ∗ r∗ 4√2 = 2√2 ∗ r
Therefore, Area of triangle ABC = A (AOB + BOC + AOC)
                                                     =½ ∗ r∗ 4+½∗ r∗ 4+ 2√2∗r
                                                     = r∗(4+2√2)= 8 units

Therefore, r = 8/(4+2√2)
                      = 4/(2+√2)
                      = 4∗(2−√2)/{(2 +√2)(2−√2)}
                      =2∗ (2−√2)
                      = 2√2(√2−1)

Hence, the area of the circle = π∗ [2√2 (√2−1)]^2
                                                  = π∗ 8 ∗(√2−1)^2
                                                  =8π(√2−1)^2

Correct Answer: (B)
 

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