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# What is the area of an equilateral triangle that has an altitude of le

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What is the area of an equilateral triangle that has an altitude of le  [#permalink]

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26 Apr 2018, 12:31
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What is the area of an equilateral triangle that has an altitude of length 24?

(A) $$24 \sqrt{3}$$

(B) $$48 \sqrt{3}$$

(C) $$96 \sqrt{3}$$

(D) $$192 \sqrt{3}$$

(E) $$384 \sqrt{3}$$

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What is the area of an equilateral triangle that has an altitude of le  [#permalink]

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26 Apr 2018, 12:58
Bunuel wrote:
What is the area of an equilateral triangle that has an altitude of length 24?

(A) $$24 \sqrt{3}$$

(B) $$48 \sqrt{3}$$

(C) $$96 \sqrt{3}$$

(D) $$192 \sqrt{3}$$

(E) $$384 \sqrt{3}$$

An equilateral triangle of side 2 will have an altitude $$\sqrt{3}$$
(because the altitude of the equilateral triangle will form a 30-60-90 triangle with the base)

Since, the altitude of the equilateral triangle is 24, the side is $$\frac{48}{\sqrt{3}}$$

Formula used: Area(equilateral triangle) = $$\frac{\sqrt{3}}{4} * side^2$$

Therefore, the area of the equilateral triangle for side $$\frac{48}{\sqrt{3}}$$ is $$\frac{\sqrt{3}}{4}*48*\frac{48}{3} = 192 \sqrt{3}$$ (Option D)
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What is the area of an equilateral triangle that has an altitude of le  [#permalink]

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26 Apr 2018, 19:05
Bunuel wrote:
What is the area of an equilateral triangle that has an altitude of length 24?

(A) $$24 \sqrt{3}$$

(B) $$48 \sqrt{3}$$

(C) $$96 \sqrt{3}$$

(D) $$192 \sqrt{3}$$

(E) $$384 \sqrt{3}$$

Attachment:

equi4.26.18.png [ 62.33 KiB | Viewed 616 times ]

• An equilateral triangle's altitude always creates
creates two 30-60-90 triangles
Each vertex = 60°
Vertex B is bisected: two 30° angles
Vertex C = 60°
Point D = two 90° angles

A 30-60-90 triangle has corresponding sides opposite those angles in ratio
$$x : x\sqrt{3} : 2x$$

We need to find $$x$$ = $$\frac{1}{2}$$ of base

Divide the altitude by $$\sqrt{3}$$
Altitude BD, opposite the 60° angle, corresponds with $$x\sqrt{3}$$
$$x\sqrt{3}=24$$
$$x=\frac{24}{\sqrt{3}}=( \frac{24}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})=\frac{24\sqrt{3}}{3}=8\sqrt{3}$$

$$x=8\sqrt{3}=\frac{1}{2}b,$$ base*

Area of triangle, $$A=\frac{1}{2}*b*h$$
$$A= (8\sqrt{3}* 24)=192\sqrt{3}$$

*No need to find the whole base. Area is divided by 2. But IF you wanted to:
Base = $$2x =(2*8\sqrt{3})=16\sqrt{3}$$
Area, $$A=\frac{b*h}{2}$$
$$A=\frac{16\sqrt{3}*24}{2}=(8\sqrt{3}*24)=192\sqrt{3}$$

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What is the area of an equilateral triangle that has an altitude of le  [#permalink]

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26 Apr 2018, 23:29
Bunuel wrote:
What is the area of an equilateral triangle that has an altitude of length 24?

(A) $$24 \sqrt{3}$$

(B) $$48 \sqrt{3}$$

(C) $$96 \sqrt{3}$$

(D) $$192 \sqrt{3}$$

(E) $$384 \sqrt{3}$$

Ans: E

In equilateral triangle of side a altitude is given by 3^(1/2) (a/2) so we can get the a from here.
and area is 1/2(altitude* side of the triangle) = which here is $$192 \sqrt{3}$$
Ans E
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Re: What is the area of an equilateral triangle that has an altitude of le  [#permalink]

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30 Apr 2018, 15:29
Bunuel wrote:
What is the area of an equilateral triangle that has an altitude of length 24?

(A) $$24 \sqrt{3}$$

(B) $$48 \sqrt{3}$$

(C) $$96 \sqrt{3}$$

(D) $$192 \sqrt{3}$$

(E) $$384 \sqrt{3}$$

When we drop an altitude in an equilateral triangle we create two 30-60-90 right triangles in which the altitude is opposite the 60 degree angle; thus, the altitude = side√3/2, thus:

side√3/2 = 24

side = 48/√3

We may recall that the area formula for an equilateral triangle is (side^2 * √3)/4, thus:

area = [(48/√3)^2 * √3]/4

area = 768√3/4 = 192√3

Alternate Solution:

When we drop a perpendicular (altitude) in an equilateral triangle, we create two 30-60-90 triangles, each with a ratio of sides of x : 2x : x√3. The altitude is x√3, and so we have:

x√3 = 24

x = 24/√3

x = (24√3)/3

x = 8√3

We see that 8√3 is the base of one of the 30-60-90 triangles, so the base of the entire equilateral triangle is 16√3.

Using the formula for the area of a triangle: A = 1/2 b x h, we have:

A = (1/2) x 16√3 x 24 = 192√3

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Re: What is the area of an equilateral triangle that has an altitude of le &nbs [#permalink] 30 Apr 2018, 15:29
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