Chapter 7.1, Problem 66E

To determine

Tocalculate: the dimensions of a right triangle with a two inch hypotenuse and an area of 1 square inch.

Answer to Problem 66E

  $\begin{array}{l}a=\sqrt{2}inches\\ b=\sqrt{2}inches\\ c=2inches\end{array}$

Explanation of Solution

Given information:

A right triangle whose hypotenuse is 2 inch and area is 1 sq in.

Formula used:

Area of triangle is:

  $A=\frac{1}{2}bh$

Where,

b is base of the triangle.

h is the height of the triangle.

For right triangle, Pythagoras theorem

  $c^2=a^2+b^2$

Where,

a is opposite side (height)

b base

c is the hypotenuse

Calculation:

Firstly we will use the Pythagoras theorem

  $\begin{array}{l}{c}^2=a^2+b^2\\ 2^2=h^2+b^2\\ h^2+b^2=4---(1)\end{array}$

Now we will use the area formula,

  $\begin{array}{l}A=\frac{1}{2}bh\\ 1=\frac{1}{2}bh\\ bh=2---(2)\end{array}$

Solve for b

  $b=\frac{2}{h}$

Let’s substitute the value of b into the equation (1)

  $\begin{array}{l}{h}^2+{\left(\frac{2}{h}\right)}^2=4\\ h^2+\frac{4}{h^2}=4\\ h^4+4=4h^2\\ h^4-4h^2+4=0\\ $ {(h^2-2)}^2=0$h^2-2=0\\ h^2=2\\ h=\sqrt{2}inches\end{array}$

Put the value of h

  $\begin{array}{l}b=\frac{2}{h}\\ b=\frac{2}{\sqrt{2}}\\ b=\sqrt{2}inches\end{array}$