12 root 2

Can you let me know the explanation?

gunjan1208 wrote:12 root 2

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.

Alright to make things easy I have attached a diagram and will try to solve the problem using different appraoches!

Approach A:

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.

i.e. CO/OD = 2/1
4/OD = 2/1 (CO = radius of the circle = 4 units)
OD = 2 units.

Let the length of the side of the equilateral triangle be X, then In right angled triangle CDA,
AC = X, AD = 0.5X(CD is a perpendicular bisector) and CD = 6. Applying Pythagorean Theorem,

X^2 = (0.5X)^2 + 6^2
(3/4)*X^2 = 36.
X = 4 * square root(3)
Perimeter = 3X = 12*Square root(3)

Approach B:

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)

pi * R^2 = pi * (X^2) / 3
4^2 = (X^2) / 3
X = 4 * square root(3)
Perimeter = 3X = 12 * Square root(3)

Approach C:

In triangle OAB, Applying sine rule
OA/Sin(OBA) = AB/Sin(AOB)
4/Sin 30 = AB/Sin 120
4/(1/2) = AB/((Square root(3))/2)
AB = 4 * square root(3)
Perimeter = 3X = 12 * Square root(3)

Approach D:

In Triangle ODA,
Cos 30 = AD/AO = Square root(3)/2 = (AB/2)/AO (AO = 4)
AB = 4 * square root(3)
Perimeter = 3X = 12 * Square root(3)

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.

Gunjan,

See the correct answer below.

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.

Hi Karthik,

Thanks. Actually I realized later that the second tule what I applied for finding the hypotenuse was not right. I am good with this.

Thank you very much again.

GMATGuruNY 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.

Hello

Sorry i dint understand two points here

*I am missing something. Please confirm how do we reach - Each side of the triangle ≈ 6?

*√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.

melguy wrote:

GMATGuruNY 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.

Hello

Sorry i dint understand two points here

*I am missing something. Please confirm how do we reach - Each side of the triangle ≈ 6?

The drawing shows 3 congruent triangles, each with two sides of 4 forming a 120 degree angle.
The third side of each triangle must be less than 8, since the third side of a triangle must be less than the sum of the other 2 sides.
Since the triangles are not equilateral, the third side of each triangle must be greater than 4.
Splitting the difference between 8 and 4, I estimated that the third side of each triangle ≈ 6.

*√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.

An equilateral triangle can easily be divided into 30-60-90 triangles:



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.

Thanks for clarifying my doubts!

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.

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)

What am I missing here?