Chapter 11.2, Problem 38WE

To determine

To Find: The area of shaded region.

Answer to Problem 38WE

  $A=3(36-x^2)$

Explanation of Solution

Given: Two squares each with side $12cm$ are placed so that the vertex of one lies at the center of the other.

  

  

Area of the quadrilateral is the sum of area of $A_{1}and{A}_2$

Here $A_1$ is the triangle and $A_2$ is the trapezium.

Thus

  $\begin{array}{l}AP=BC=x\\ PB=\frac{DC}{2}=\frac{12}{2}=6cm\\ AB=6-x\\ OB=BC=\frac{1}{2}(side)=\frac{12}{2}=6cm\\ now\\ A_1=\frac{1}{2}bh=\frac{1}{2}(OB)(AB)=\frac{1}{2}(6)(6-x)=3(6-x)\\ A_2=\frac{1}{2}h(b_1+b_2)=\frac{1}{2}(BC)(DC+OB)\\ =\frac{1}{2}(6)(6-x)=3(6+x)\end{array}$

Now sum is

  $\begin{array}{l}A=A_1+A_2\\ A=3(6-x)+3(6+x)\\ A=3(6^2-x^2)\\ A=3(36-x^2)\end{array}$

Hence, area is $A=3(36-x^2)$