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Since the triangle is an equilateral triangle we know that the measure of the interior angle is 60 degrees. Join the centre(O here) to the corners of the triangle ABC. The center of the circle acts as a centroid and the line segment AD median/perpendicular bisector. The centroid divides each median in a ratio of 2:1.Approach A:
If an equilateral triangle of side 'X' is inscribed inside a circle, then area of circle = pi (X^2) / 3. We know that the area of the circle = pi * R^2 (Where R = Radius of the circle)Approach B:
In triangle OAB, Applying sine ruleApproach C:
In Triangle ODA,Approach D:
Triangle OAB is 30-60-90 triangle.karthikpandian19 wrote:If the circle above has a radius of 4, what is the perimeter of the inscribed equilateral triangle?
(other details in the picture attached)
gunjan1208 wrote:Think the centre of the circle and connect it to the vertex of the triangle. This is equal to 4. Thus three line to the vertax of the triangle are equal making the triangle keeping two sides equal. Also, any angle on the diameter of the semicircle equals to 90 degreee. This becomes a right angled triangle. Thus the hypotneuse becomes 4 root 2.
It applies to all 3 Hypotneuse. Sum up and we get 12 root 2.
I know my explanation is little rusty, but logically it should suffice.
HelloGMATGuruNY wrote:
What is the perimeter of the inscribed equilateral triangle?
Answer choices: 6√2, 6√3, 12√2, 12√3, 24.
Each side of the triangle ≈ 6.
Perimeter ≈ 18.
Eliminate A and B, which are way too small, and E, which is too big.
The correct answer must be C or D.
√2 implies a 45-45-90 triangle; √3 implies a 30-60-90 triangle.
Since each angle of the equilateral triangle = 60, √2 makes no sense here.
Eliminate C.
The correct answer is D.
Always look for opportunities to ballpark.
Always look at the answer choices.
The approach above requires little insight and allows us to determine the correct answer in mere seconds.
The drawing shows 3 congruent triangles, each with two sides of 4 forming a 120 degree angle.melguy wrote:HelloGMATGuruNY wrote:
What is the perimeter of the inscribed equilateral triangle?
Answer choices: 6√2, 6√3, 12√2, 12√3, 24.
Each side of the triangle ≈ 6.
Perimeter ≈ 18.
Eliminate A and B, which are way too small, and E, which is too big.
The correct answer must be C or D.
√2 implies a 45-45-90 triangle; √3 implies a 30-60-90 triangle.
Since each angle of the equilateral triangle = 60, √2 makes no sense here.
Eliminate C.
The correct answer is D.
Always look for opportunities to ballpark.
Always look at the answer choices.
The approach above requires little insight and allows us to determine the correct answer in mere seconds.
Sorry i dint understand two points here
*I am missing something. Please confirm how do we reach - Each side of the triangle ≈ 6?
An equilateral triangle can easily be divided into 30-60-90 triangles:*√2 implies a 45-45-90 triangle; √3 implies a 30-60-90 triangle.
Since each angle of the equilateral triangle = 60, √2 makes no sense here.
I believe we are not drawing to scale since visually i cannot see any 30-60-90 in the triangle. It appears more like 45-45-90 triangle after drawing radius.
thanks.
Hi. I dont understand why it is not 12√2? we have three 45:45:90 triangles with two legs of 4. So the hypotenuse is 4√2. The perimeter will be 3(4√2)Equilateral triangles and 30-60-90 triangles are common bedfellows.
Equilateral triangles and 45-45-90 triangles are not.
There is almost no chance that the perimeter of an equilateral triangle on the GMAT will involve √2, which is associated with a 45-45-90 triangle.
Inspired from Mitch's analysis...karthikpandian19 wrote:If the circle above has a radius of 4, what is the perimeter of the inscribed equilateral triangle?
(other details in the picture attached)
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