Chapter 6.5, Problem 37WE

a.

To determine

Show that each leg of isosceles triangle PQR has length $\sqrt{25+x^2}$ if $QR=10\text{ cm}$ and altitude $\left(PN\right)$ of triangle is $x\text{ cm}$ .

Explanation of Solution

Given information: It is given that the base isosceles triangle is $10\text{ cm}$ and altitude $\left(PN\right)$ of triangle is $x\text{ cm}$ as shown,

  

Formula used: The altitude of an isosceles triangle bisects perpendicular to its base.

Calculation: By Pythagoras Theorem in triangle PNQ,

  $\begin{array}{c}PQ=\sqrt{{\left(QN\right)}^2+{\left(PN\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(x\right)}^2}\\ =\sqrt{25+x^2}\mathrm{....}\left(1\right)\\ PR=\sqrt{{\left(NR\right)}^2+{\left(PN\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(x\right)}^2}\\ =\sqrt{25+x^2}\mathrm{....}\left(2\right)\end{array}$

From (1) and (2), each leg of triangle has equal length. Hence proved.

b.

To determine

The value of $\left(x\right)$ if perimeter of triangle is 3 times the length of $\left(PN\right)$ .

Answer to Problem 37WE

The value of $\left(x\right)$ is equal to $12$ .

Explanation of Solution

Given information: It is given that the base isosceles triangle is $10\text{ cm}$ , perimeter of triangle PQR is equal to $3$ times the length of $\left(PN\right)$ and altitude $\left(PN\right)$ of triangle is $x\text{ cm}$ as shown,

  

Formula used: The altitude of an isosceles triangle is equal to sum of its three sides

Calculation: According to question,

  $\begin{array}{c}3x=PQ+QR+PR\\ 3x=\sqrt{25+x^2}+10+\sqrt{25+x^2}\mathrm{....}\left(\begin{array}{l}\text{From }\left(1\right)\text{ and }\left(2\right)\\ PQ=PR=\sqrt{25+x^2}\end{array}\right)\\ 3x=2\sqrt{25+x^2}+10\\ \left(3x-10\right)=2\sqrt{25+x^2}\mathrm{....}\left(3\right)\end{array}$

Squaring (3) and simplify,

  $\begin{array}{c}9x^2-60x+100=4\left(25+x^2\right)\\ \text{or, }5x\left(x-12\right)=\mathrm{0....}\left(\text{Simplified}\right)\\ x=0\text{or}x=12\\ x=\mathrm{12....}\left(PN\ne 0\right)\end{array}$