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ABCD is a field in the shape of a trapezium, $A D ‖ B C, ∠ A B C = 90 ° and ∠ A D C = 60 ° .$ Given that AD=55 m, BC=45 m and AB=30 m. Four sectors are formed with centres A, B, C and D as shown in the figure. The radius of each sector is 14 m. Then, the area of the remaining portion is ____.

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Detailed Solution

Given that

A D = 55 m, B C = 45 m and A B = 30 m

.
Observe the given figure:
Question Image

We know that, area of trapezium

= 12 (sum of parallel sides) × height

Area of trapezium = 12 (A D + B C) × A B = 12 (55 + 45) × 30 = 100 × 15 = 1500 m 2

Area of 4 sectors = Area of sector having central angle

90 °

(A) + Area of sector having central angle

90 °

(B) + Area of sector having central angle

120 °

60 °

(D)
=

\frac{90 °}{360 °} π r^{2} + \frac{90 °}{360 °} π r^{2} + \frac{120 °}{360 °} π r^{2} + \frac{60 °}{360 °} π r^{2}

\frac{90 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{120 °}{360 °} π {(14)}^{2} + \frac{60 °}{360 °} π {(14)}^{2}

(\frac{90 °}{360 °} + \frac{90 °}{360 °} + \frac{120 °}{360 °} + \frac{60 °}{360 °}) π {(14)}^{2}

(\frac{360 °}{360 °}) π {(14)}^{2}

= 616

m^{2}

Area of the remaining portion = Area of trapezium ABCD − Area of four sectors

= 1500 − 616 = 884 m 2

Hence, the area of the remaining portion is

884 m 2

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