Chapter 13, Problem 4NT

To determine

(a)

To find:

The area of rectangle of Figure $24$ (d).

Answer to Problem 4NT

Solution:

Area of rectangle be $\left(\frac{bh}{4}\right){\text{unit}}^{\text{2}}$.

Explanation of Solution

Given:

A rectangle with dimension $\frac{b}{2}\times \frac{h}{2}$.

Approach:

$\text{Area of rectangle}=\text{length}\times \text{breadth}$.

Calculation:

$\begin{array}{rll}\text{Length of rectangle}& =& \frac{b}{2}\text{units}.\\ \text{breadth of rectangle}& =& \frac{h}{2}\text{units}.\\ \text{Area of rectangle}& =& \text{length}\times \text{breadth}\\ & =& \frac{b}{2}\times \frac{h}{2}\\ & =& \left(\frac{bh}{4}\right){\text{unit}}^{\text{2}}\end{array}$

To determine

(b)

To find:

The area of triangle from rectangle of figure 24(d).

Answer to Problem 4NT

Solution:

$\text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}$.

Explanation of Solution

Given:

A rectangle of dimension $\frac{b}{2}\times \frac{h}{2}$ made from two triangles of dimension $b\times h$.

Approach:

$\text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}$.

Calculation:

Here rectangle, Figure $24$ (d) of dimension $\frac{b}{2}\times \frac{h}{2}$ made from two triangles, Figure $24$ (a) of dimension $b\times h$.

So,

$\begin{array}{rll}2\text{area of rectangle}& =& \text{ area of triangle}.\\ 2\left(\frac{bh}{4}\right)& =& \text{area of triangle}\\ \text{area of triangle}& =& \frac{1}{2}bh\end{array}$

Thus $\text{area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}$.