The diagram below shows a logo formed by removing a semicircle with diameter AB and a
rhombus DEFG from an isosceles trapezium ABCE where AE = BC.
Find the area of the logo.

The diagram below shows a logo formed by removing a semicircle with diameter AB and a rhombus DEFG from an isosceles trapezium ABCE where AE BC Find the area of class=

Respuesta :

The area of the logo is 136. 14 cm²

Determining the area

First, let's find the area of the semicircle

Area of semicircle = 1/2(πr²)

radius = 8/2 = 4cm

Substitute into the formula

Area of semicircle = [tex]\frac{1}{2} * 3. 142 * 4 * 4[/tex] = 25. 135 cm²

Now, let's find area of trapezium

Area of trapezium = [tex]\frac{a + b}{2} * h[/tex]

Area =  [tex]\frac{12 + 10. 2}{2} * 10[/tex]

Area = [tex]11. 1 * 10[/tex]

Area = 111 cm²

Area of logo = Area of trapezium + area of semicircle

Area of logo = 111 +  25. 135

Area of logo = 136. 14 cm²

Thus, the area of the logo is 136. 14 cm²

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