Chapter 7.4, Problem 25P

The area of a triangle is $80x^5{y}^3.$

The height of the triangle is $x^{4}y.$

What is the length of the base of the triangle?

To determine

To find the length of the base of the triangle.

Answer to Problem 25P

The length of base of triangle is $160x{y}^2$ .

Explanation of Solution

Given information:

The height of the triangle is $x^{4}y$ and the area of the triangle is $80x^5{y}^3$ .

Formula Used: Area of a triangle $\left(A\right)=\frac{1}{2}\times b\times h$ , where b is length of base of the triangle and h is height of the triangle.

Calculations: We have given height of the triangle $h=x^{4}y$ and Area $A=80x^5{y}^3$ .

Plugging $h=x^{4}y$ and $A=80x^5{y}^3$ in above formula $A=\frac{1}{2}\times b\times h$

$80x^5{y}^3=\frac{1}{2}\times b\times x^{4}y$

Switch sides

$\frac{1}{2}\times b\times x^{4}y=80x^5{y}^3$

Multiply by 2 on both sides

$2\times \frac{1}{2}\times b\times x^{4}y=2\times 80x^5{y}^3$

Simplify

$b{x}^{4}y=160x^5{y}^3$

Now divide by $x^{4}y$ on both sides

$\frac{b\times x^{4}y}{x^{4}y}=\frac{160x^5{y}^3}{x^{4}y}$

Apply exponential rule:

  $\frac{a^m}{a^n}=a^{m-n}$

$b\times x^{4-4}{y}^{1-1}=160x^{5-4}{y}^{3-1}$

$b{x}^0{y}^0=160x^{5-4}{y}^{3-1}$

Simplify

$b=160x{y}^2$

Therefore, the length of base of triangle is $160x{y}^2$ .