Chapter 1.7, Problem 16E

In Exercises 15−20, write the specified quantity as a function of the specified variable. It will help in each case to draw a picture.

16. One leg of a right triangle is twice as long as the other. Write the length of the hypotenuse as a function of the length of the shorter leg.

Solution

Write the length of the hypotenuse as a function of the length of the shorter leg.

  $H=\frac{\sqrt{5}B}{2}$

Given:

One leg of a right triangle is twice as long as the other.

Concept Used:

The Pythagoras theorem, also known as the Pythagorean Theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle. Or, the sum of the squares of the two legs of a right triangle is equal to the square of its hypotenuse.

Calculation:

Let the Hypotenuse, Base & Perpendicular is H, B & P respectively.

Pythagoras theorem-

  $\begin{array}{}$ {(hypotenuse)}^2={(base)}^2+{(perpendicular)}^2$$ {(H)}^2={(B)}^2+{(P)}^2$\end{array}$

  

According to question −

Let length of shorter leg = P

And length of longer leg = B

One leg of a right triangle is twice as long as the other.

  $\begin{array}{l}B=2P\\ P=\frac{B}{2}\end{array}$

  $\begin{array}{c}$ {(H)}^2={(B)}^2+{(P)}^2$$ asP=\frac{B}{2}$$ {(H)}^2={(B)}^2+{(\frac{B}{2})}^2$H^2=B^2+{\frac{B}{4}}^2\\ H^2=\frac{4B^2+B^2}{4}\\ H^2=\frac{5B^2}{4}\\ H=\sqrt{\frac{5B^2}{4}}\\ H=\frac{\sqrt{5}B}{2}\end{array}$

So, the length of the hypotenuse as a function of the length of the shorter leg $H=\frac{\sqrt{5}B}{2}$