Question

Find the area of the shaded regions : ​

Answer 1

$\sf \Large Solution!!$

Firstly, name the triangle and rectangle as shown in the attachment. We have to find the shaded region. So, first we have to find the area of triangle and then find the area of the rectangle. After finding both the areas, we will subtract the area of triangle from the area of rectangle. In triangle, we will find the other side by using Pythagoras Theorem. So, let's do this!

$\sf \large Pythagoras\: Theorem$

$\sf \to (Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}$

$\sf \large So,$

$\sf \to (YZ)^{2}=(ZX)^{2}+(XY)^{2}$

$\sf \to (17\:cm)^{2}=(ZX)^{2}+(8\:cm)^{2}$

$\sf \to 289\:cm^{2}=(ZX)^{2}+64\:cm^{2}$

$\sf \to (ZX)^{2}=289\:cm^{2}-64\:cm^{2}$

$\sf \to (ZX)^{2}=225\:cm^{2}$

$\sf \to ZX=\sqrt{225\:cm^{2}}$

$\sf \to ZX=15\:cm$

$\sf \large Now,$

$\sf \to Let\:the\: Hypotenuse=c$

$\sf \to Let\:the\: Perpendicular=a$

$\sf \to Let\:the\:Base=b$

$\sf \to Semi-perimeter(s)=\dfrac{a+b+c}{2}$

$\sf \to s=\dfrac{15\:cm+8\:cm+17\:cm}{2}$

$\sf \to s=\dfrac{40\:cm}{2}=20\:cm$

$\sf \to Area=\sqrt{s(s-a)(s-b)(s-c)}$

$\sf \to Area=\sqrt{20(20-15)(20-8)(20-17)}cm^{2}$

$\sf \to Area=\sqrt{20\times 5\times 12\times 3}cm^{2}$

$\sf \to Area=\sqrt{3600}cm^{2}$

$\sf \to Area\:of\:triangle=60\:cm^{2}$

$\sf \large Now,$

$\sf \to Area\:of\: rectangle=Length\times Breadth$

$\sf \to Area\:of\:rectangle=16\:cm\times 10\:cm$

$\sf \to Area\:of\: rectangle=160\:cm^{2}$

$\sf \large So,$

$\sf \to Area\:of\:shaded\:portion=Area\:of\: rectangle-Area\:of\:triangle$

$\sf \to Area\:of\:shaded\: portion=160\:cm^{2}-60\:cm^{2}$

$\sf \bigstar \blue{Area\:of\:shaded\:portion=100\:cm^{2}}$