Chapter 9.5, Problem 30PPS

To determine

Find the area of triangle.

Answer to Problem 30PPS

  $216$ square unit

Explanation of Solution

Given Information: Hypotenuse $H=30$ , Perpendicular $P=x$ , Base $B=x+6$

Formula Used : Pythagoras Theorem : $H^2=P^2+B^2$ , $a^2+2ab+b^2={(a+b)}^2$ ,

  $a^2-b^2=(a+b)(a-b)$ and Area of Triangle = $\frac{1}{2}\times Base\times Perpendicular$

Calculation:

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. That’s, Pythagoras Theorem.

Pythagoras Theorem : $H^2=P^2+B^2$

As, Hypotenuse $H=30$ , Perpendicular $P=x$ , Base $B=x+6$

  ${\left(30\right)}^2=x^2+{\left(x+6\right)}^2$

As, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(30\right)}^2=x^2+x^2+12x+36$

Now, simplify

  $2x^2+12x+36-900=0$

So,

  $2x^2+12x-864=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 2.

  $x^2+6x-432=0$

Now, adding & subtracting the square of half the coefficient of x.

  $x^2+6x-432+{\left(\frac{6}{2}\right)}^2-{\left(\frac{6}{2}\right)}^2=0$

Now use formula, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(x+3\right)}^2-432-3^2=0$

So,

  ${\left(x+3\right)}^2-432-9=0$

So,

  ${\left(x+3\right)}^2-441=0$

Now, square root of 441 is 21

  ${\left(x+3\right)}^2-{\left(21\right)}^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x+3+21\right)\left(x+3-21\right)=0$

Now,

  $\left(x+24\right)\left(x-18\right)=0$

Hence,

  $x=-24$ or $x=18$

Perpendicular $P=x=18$ ,

Base $B=x+6=18+6=24$

Area of Triangle = $\frac{1}{2}\times Base\times Perpendicular$

So,

Area of Triangle = $\frac{1}{2}\times 24\times 18$

Area of Triangle = $216$ square unit