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professor190 [17]
3 years ago
8

X22abab tobThe diagianKn showsrectangle. The shadeFind x.​

Mathematics
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

x = 28 cm

Step-by-step explanation:

Given:

Area of link shaded regions = 84 cm²

Required:

The value of x (diameter of the semicircle/length of the rectangle)

Solution:

Diameter of the semicircle = 2r = x

Length of rectangle (L) = 2r = x

Radius of semicircle (r) = ½x

Width of rectangle (W) = radius of semicircle = ½x

Use 3.14 as π

Area of the link shaded regions = area of rectangle - area of semicircle

Thus:

Area of the link shaded regions = (L*W) - (½*πr²)

Plug in the values

84 = (x*½x) - (½*3.14*(½x)²)

84 = x²/2 - (1.57*x²/4)

84 = x²(½ - 1.57/4)

84 = x²(0.5 - 0.3925)

84 = x²(0.1075)

Divide both sides by 0.1075

84/0.1075 = x²

781.4 = x²

√781.4 = x

27.9535329 = x

x = 28 cm

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Refer to attachment for marking of sides.

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<u>•</u><u> </u><u>In </u><u>∆</u><u>A</u><u>B</u><u>D</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>

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\rule{200}2

<u>•</u><u> </u><u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>D</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>

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\rule{200}2

<u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>B</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>,</u>

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