How to find the area of an equilateral triangle - ISEE Upper Level Mathematics Achievement

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Question

The length of one side of an equilateral triangle is 6 inches. Give the area of the triangle.

Answer

$Area=\frac{1}{2}absinC$ ,

where and are the lengths of two sides of the triangle and is the angle measure.

In an equilateral triangle, all of the sides have the same length, and all three angles are always $60^{\circ}$ .

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 6\times 6\times sin60^{\circ}$

$sin60^{\circ}=\frac{\sqrt{3}}{2}\Rightarrow Area=\frac{1}{2}\times 6\times 6\times \frac{\sqrt{3}}{2}=9\sqrt{3}\ in^2$

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Question

The perimeter of an equilateral triangle is $30\sqrt[4]{3}$ . Give its area.

Answer

An equilateral triangle with perimeter $30\sqrt[4]{3}$ has three congruent sides of length

$s = 30\sqrt[4]{3} \div 3 = 10\sqrt[4]{3}$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{ \left ( 10\sqrt[4]{3} \right )^{2}\sqrt{3}}{4} = \frac{100 \sqrt[4]{9} \cdot \sqrt{3}}{4}$

$\sqrt[4]{9} = \sqrt[4]{3^{2}} = 3^{2/4} = 3^{1/2} = \sqrt{3}$ , so

$A = \frac{100 \sqrt{3} \cdot \sqrt{3}}{4} = \frac{100 \cdot 3 }{4}= 75$

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Question

The perimeter of an equilateral triangle is $18\sqrt{3}$ . Give its area.

Answer

An equilateral triangle with perimeter $18\sqrt{3}$ has three congruent sides of length

$s = 18\sqrt{3} \div 3 = 6\sqrt{3}$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{ \left ( 6\sqrt{3} \right )^{2}\sqrt{3}}{4} = \frac{36 \cdot 3 \cdot \sqrt{3}}{4} = 27\sqrt{3}$

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Question

The perimeter of an equilateral triangle is . Give its area in terms of .

Answer

An equilateral triangle with perimeter has three congruent sides of length $\frac{n}{3}$ . Substitute this for in the following area formula:

$A =s^{2} \cdot \frac{\sqrt{3}}{4}$

$A = \left ( \frac{n}{3} \right )^{2} \cdot \frac{\sqrt{3}}{4}$

$A = \frac{n^{2} }{9} \cdot \frac{\sqrt{3}}{4}$

$A = \frac{n^{2} \sqrt{3}}{36}$

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Question

In the above diagram, $\Delta ABC$ is equilateral. Give its area.

Answer

The interior angles of an equilateral triangle all measure 60 degrees, so, by the 30-60-90 Theorem,

$BM = \frac{AM}{\sqrt{3}} = \frac{66}{\sqrt{3}} = \frac{66\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}} = \frac{66 \sqrt{3}}{3} =22 \sqrt{3}$

Also, is the midpoint of $\overline{BC}$ , so $BC = 2 \cdot BM = 44 \sqrt{3}$ ; this is the base.

The area of this triangle is half the product of the base and the height :

$A= \frac{1}{2} \cdot 66 \cdot 44 \sqrt{3}= 1,452 \sqrt{3}$

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Question

The perimeter of an equilateral triangle is . Give its area.

Answer

An equilateral triangle with perimeter 36 has three congruent sides of length

$s = 36 \div 3 = 12$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{12^{2}\sqrt{3}}{4} = \frac{144\sqrt{3}}{4} = 36\sqrt{3}$

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Question

An equilateral triangle is inscribed inside a circle of radius 8. Give its area.

Answer

The trick is to know that the circumscribed circle, or the circumcircle, has as its center the intersection of the three altitudes of the triangle, and that this center, or circumcenter, divides each altitude into two segments, one twice the length of the other - the longer one being a radius. Because of this, we can construct the following:

Each of the six smaller triangles is a 30-60-90 triangle, and all six are congruent.

We will find the area of $\Delta COY$ , and multiply it by 6.

By the 30-60-90 Theorem, $CY = OY \cdot \sqrt{3} = 4\sqrt{3}$ , so the area of $\Delta COY$ is

$A = \frac{1}{2}\cdot OY \cdot CY = \frac{1}{2} \cdot 4\cdot 4\sqrt{3} = 8 \sqrt{3}$ .

Six times this - $6 \cdot 8 \sqrt{3} = 48 \sqrt{3}$ - is the area of $\Delta ABC$ .

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Question

Refer to the above figure. The shaded region is a semicircle with area $24 \pi$ . Give the area of $\bigtriangleup ABC$ .

Answer

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 24 \pi$ :

$\frac{1}{2} \pi r ^{2} = 24 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 24 \pi$

$r ^{2} = 48$

$r = \sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{3} = 8 \sqrt{3}$ ; this is also the length of $\overline{BC}$ .

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle with sidelength $s = 8 \sqrt{3}$ . Substitute this in the area formula:

$A = \frac{s^{2}\sqrt{3}}{4}$

$= \frac{( 8 \sqrt{3})^{2}\sqrt{3}}{4}$

$= \frac{8^{2} \cdot ( \sqrt{3})^{2}\sqrt{3}}{4}$

$= \frac{64 \cdot 3 \cdot \sqrt{3}}{4}$

$= \frac{192 \sqrt{3}}{4}$

$= 48\sqrt{3}$

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