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Find the area of a shaded region shown below, which was formed by cutting an isosceles trapezoid out of the top half of a rectangle. The width of the rectangle is 32 in, the height of the rectangle in 24 in. The leg of the isosceles trapezoid is 15 in.

Find The Area Of A Shaded Region Shown Below Which Was Formed By Cutting An Isosceles Trapezoid Out Of The Top Half Of A Rectangle The Width Of The Rectangle Is class=

Sagot :

Step 1: Redraw the diagram and label it.

From the figure, the hypotenuse of triangles A and B is 15 in and the height is 12 in. We can apply the Pythagoras theorem to find the base.

Let base of the triangle A and B be the adjacent.

Opposite = 12

Adjacent = ?

Hypotenuse = 15

[tex]\begin{gathered} Next,\text{ apply the Pythagoras theorem to find the adjacent.} \\ \text{Opposite}^2+Adjacent^2=Hypotenuse^2 \\ 12^2+Adj^2=15^2 \\ 144+Adj^2\text{ = 225} \\ \text{Collect like terms.} \\ \text{Adj}^2\text{ = 225 - 144} \\ \text{Adj}^2\text{ = 81} \\ F\text{ ind the square root of both sides.} \\ \sqrt[]{Adj^2\text{ }}=\text{ }\sqrt[]{81} \\ \text{Adj = 9 in} \end{gathered}[/tex]

The area of the shaded region = Area of A + Area of B + Area of C

[tex]\begin{gathered} \text{Area of A = }\frac{Base\text{ x Heigth}}{2} \\ \text{Base = 9} \\ \text{Height = 1}2 \\ \text{Area of A = }\frac{9\text{ x 12}}{2} \\ =\text{ }\frac{108}{2} \\ =54in^2 \\ \text{Area of B = }\frac{Base\text{ x Height}}{2} \\ =\text{ }\frac{9\text{ x 12}}{2} \\ =\text{ }\frac{108}{2} \\ \text{= 54 in}^2 \end{gathered}[/tex][tex]\begin{gathered} \text{Area of rectangle C = Length }\times\text{ Breadth} \\ Lenght\text{ = 32} \\ \text{Breadth = 12} \\ \text{Area of C = 32 x 12} \\ =384in^2 \end{gathered}[/tex]

Therefore,

Area of the shaded region = 54 + 54 + 384 = 492 inches square

Final answer

Area of the shaded region = 492 inches square

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