Chapter 11.2, Problem 8PSA

a

To determine

To find : The area of the triangle in part a

Answer to Problem 8PSA

The area of the triangle is $A=\frac{56}{\sqrt{3}}$

Explanation of Solution

Given information : The following information has been given

The base of the triangle is $7$

The height of the triangle is $16\tan {30}^0=\frac{16}{\sqrt{3}}$

Formula used : The area of any triangle will be

  $A=\frac{1}{2}(base)(height)$

Calculation : We know that the area of a triangle can be calculated as

  $A=\frac{1}{2}(base)(height)$

Thus, substituting the values in the above formula, we get

  $A=\frac{1}{2}(7)(\frac{16}{\sqrt{3}})=\frac{56}{\sqrt{3}}$

b

To determine

To find : The area of the triangle in part b

Answer to Problem 8PSA

The area of the triangle is $60$

Explanation of Solution

Given information : The following information has been given

The base of the triangle is $10$

The height of the triangle is $12$

Formula used : The area of any triangle will be

  $A=\frac{1}{2}(base)(height)$

Calculation : We know that the area of a triangle can be calculated as

  $A=\frac{1}{2}(base)(height)$

Thus, substituting the values in the above formula, we get

  $A=\frac{1}{2}(10)(12)=60$