Zollege is here for to help you!!
Need Counselling
GMAT logo

An Equilateral Triangle that has an Area of 9√3 is Inscribed in a Circle GMAT Problem Solving

Overview es 2Overview en 2RegistrationExam PatternPreparation TipsPractice PaperResultCut offmock testNews

Question: An equilateral triangle that has an area of 9√3 is inscribed in a circle. What is the area of the circle?

  1. 12π
  2. 9π√3
  3. 18π√3

Correct Answer: C
Solution and Explanation:
Approach Solution 1:

The problem statement states that:
Given:

  • An equilateral triangle that has an area of 9√3 is inscribed in a circle.

Find out:

  • The area of the circle.

Area of the equilateral triangle = 9√3

As per the formula of the area of an equilateral triangle, we can say:
Area of equilateral triangle= \(a^2\) * √3/4; where “a” is the length of the side of the triangle.

Now, putting the value of the area that is given we can determine the length of the side of the triangle.

Area of the equilateral triangle = \(a^2\) * √3/4

9√3 = \(a^2\) * √3/4

\(a^2\) = 9√3 * 4/ √3

\(a^2\) = 36

a = 6

So, we know that all sides of an equilateral triangle are equal which means each side of the given triangle has a length of 6 units.

It is given that the equilateral triangle is inscribed within a circle.
As per the formula of the radius of the circumscribed circle, we can say:
Radius of circumscribed circle = a * √3/3

Putting the value of a = 6 in this formula, we get:
Radius of circumscribed circle = 6 * √3/3 = 2√3

Now, we know the radius of the circle and it can be used to determine the area of the circle.
Area of the circle = \(πr^2\); where “r” is the radius of the circumscribed circle

Therefore, the area of the circle = \(π [2\sqrt3]^2\)
Hence the area of the circle = 12π

Approach Solution 2:

The problem statement informs that:
Given:

  • An equilateral triangle that has an area of 9√3 is inscribed in a circle.

Find out:

  • The area of the circle.

The area of the equilateral triangle is 9√3.
Let’s assume that the side of the triangle is “a” then we get:
½ ∗ √3/2 ∗ a = 9√3

Solving this we get

a = 6

Since we know that all sides of an equilateral triangle are equal which means each side of the given triangle has a length of 6 units.

Now, in the figure, we need to consider the shaded triangle. Clearly, we can notice that the area of this triangle is a third of the area of the whole triangle.

image1

So we have ½ ∗ h ∗ a/2 = ⅓ ∗ 9√3, therefore from here we get h=√3
This implies that the radius, as we can see from the picture = 3√3 −√3 =2√3

Hence area of the circle = π∗ (2√3)^2
= 12π

Approach Solution 3:

The problem statement indicates that:
Given:

  • An equilateral triangle that has an area of 9√3 is inscribed in a circle.

Find out:

  • The area of the circle.

We know that the formula of the area of an equilateral triangle = √3/4 (side)²
Therefore we can say, 9√3 = √3/4 (side)²
=> 9 = 1/4 (side)²
=> (side)² = 9(4)
=> (side)² = 36
=> Taking square root, we get: Side = 6 cm
Therefore, the length of each side of an equilateral triangle is 6 cm.

Since we know the formula of the radius of circumscribed circle = side * √3/3
Therefore, we can say, the radius of the circle = 6* √3/3 = 2√3.

As per the formula of the area of the circle, we can say,
The area of the circle = \(π(radius)^2\);
Therefore, the area of the circle =\( π [2\sqrt3]^2\)
Hence the area of the circle = 12π

“An equilateral triangle that has an area of 9√3”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “501 GMAT Questions”. GMAT Problem Solving questions facilitate the candidates to assess every requirement of the question to solve numerical problems. GMAT Quant practice papers assist the candidates to get familiar with several sorts of questions that will enable them to improve their mathematical learning.

Suggested GMAT Problem Solving Samples

*The article might have information for the previous academic years, please refer the official website of the exam.

Ask your question

Subscribe To Our News Letter

Get Latest Notification Of Colleges, Exams and News

© 2024 Zollege Internet Private Limited