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Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset 91影视 Explanations$ Math

Student Edition 路 227 pages

ISBN: 9780395977279

Chapter 11: Q33. (page 432)

a. An equilateral triangle has sides of length s. Show that its area is $\frac{s^{2}}{4} \sqrt{3}$
b. Find the area of an equilateral triangle with side 7

Short Answer

Expert verified

a. The area of triangle is $\frac{s^{2}}{4} \sqrt{3}$ .

b. The area of an equilateral triangle with side 7 is. $\frac{49 \sqrt{3}}{4}$

Step by step solution

01

Step 1. Draw the equilateral triangle.

02

Step 2. Observe the triangle.

In $鈻� A B C$ , draw the altitude

Now

Here two $30^{掳} 鈭� 60^{掳} 鈭� 90^{掳}$ triangles are formed.

In $鈻� A D C$

$\begin{array}{l} A C = 2 D C \\ A D = \sqrt{3} D C \end{array}$

Thus

$\begin{array}{l} A D = \sqrt{3} (\frac{A C}{2}) \\ A D = \sqrt{3} (\frac{s}{2}) \end{array}$

03

Step 3. Find the area of equilateral triangle.

Area of $鈻� A B C$ can be found.

$\begin{array}{l} b = B C = s \\ h = A D = \frac{s}{2} \sqrt{3} \\ A = \frac{1}{2} b h \\ A = \frac{1}{2} (s) (\frac{s}{2} \sqrt{3}) \\ A = \frac{s^{2}}{4} \sqrt{3} \end{array}$

Hence, proved the area of equilateral triangle is $A = \frac{s^{2}}{4} \sqrt{3}$ .

04

Step 1. Draw the equilateral triangle.

05

Step 2. Observe the triangle.

In $鈻� A B C$ , draw the altitude

Now

Here two $30^{掳} 鈭� 60^{掳} 鈭� 90^{掳}$ triangles are formed.

In $鈻� A D C$

$\begin{array}{l} A C = 2 D C \\ A D = \sqrt{3} D C \end{array}$

Thus

$\begin{array}{l} A D = \sqrt{3} (\frac{A C}{2}) \\ A D = \sqrt{3} (\frac{s}{2}) \end{array}$

06

Step 3. Find the area of equilateral triangle.

Area of $鈻� A B C$ can be found.

$\begin{array}{l} b = B C = s \\ h = A D = \frac{s}{2} \sqrt{3} \\ A = \frac{1}{2} b h \\ A = \frac{1}{2} (s) (\frac{s}{2} \sqrt{3}) \\ A = \frac{s^{2}}{4} \sqrt{3} \end{array}$

07

Step 4. Find the area of equilateral triangle with side 7.

We get the area of triangle is $A = \frac{s^{2}}{4} \sqrt{3}$

Substitute $s = 7$

$\begin{matrix} A = \frac{s^{2}}{4} \sqrt{3} \\ = \frac{7^{2}}{4} \sqrt{3} \\ = \frac{49 \sqrt{3}}{4} \end{matrix}$

Therefore, the area of equilateral triangle with side7 is $\frac{49 \sqrt{3}}{4}$ .

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Most popular questions from this chapter

Find the area of each figure:

An isosceles triangle with base10and Perimeter 36 .

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