Chapter 11, Problem 43RP

To determine

To calculate:The areaof a trapezoid withbase sides as $12$ and $40$ .

Answer to Problem 43RP

The area of a trapezoid is $390$ .

Explanation of Solution

Given information:

A trapezoidwith sides $12,17,40$ and $25$ with base sides as $12$ and $40$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of trapezoid: $A=\frac{1}{2}\ast (b_1+b_2)\ast h$

b = base of trapezoid

h = height of trapezoid

Calculation:

  

  $40-12=28$ , so the bases of the right triangles can be labeled as x and $28-x$

Using the Pythagoras Theorem,

  $\begin{array}{l}{\left(x\right)}^2+{\left(h\right)}^2={\left(25\right)}^2----Equation1\\ {\left(28-x\right)}^2+{\left(h\right)}^2={\left(17\right)}^2----Equation2\end{array}$

Solving the above simultaneous equations,

  $\begin{array}{l}{x}^2+h^2=625\\ 784-56x+x^2+h^2=289\\ -784+56x=336\\ 56x=1120\\ x=\frac{1120}{56}=20\end{array}$

Substituting value of x in Equation 1, we get

  $\begin{array}{l}{x}^2+h^2=625\\ {\left(20\right)}^2+h^2=625\\ 400+h^2=625\\ h^2=625-400\\ h^2=225\\ h=\sqrt{225}\\ h=15\end{array}$

Area of trapezoid:

  $\begin{array}{l}A=\frac{1}{2}\ast (b_1+b_2)\ast h\\ A=\frac{1}{2}\ast (12+40)\ast 15\\ A=\frac{15}{2}\ast (52)\\ A=390\end{array}$

Area of trapezoid $=390$