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If the area of the quadrilateral ABCD in the above figure is 180 and
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14 Jan 2025, 17:01
14 Jan 2025, 17:01
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If the area of the quadrilateral ABCD in the above figure is 180 and TD, the shortest distance of the side AB from vertex D, is parallel to BC, what is the value of X?
A. 10
B. 15
C. 20
D. 25
E. 28
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Re: If the area of the quadrilateral ABCD in the above figure is 180 and
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23 Feb 2025, 01:09
23 Feb 2025, 01:09
Expert Reply
Area of the quadrilateral ABCD is same as the sum of the area of trapezium BTDC and the area of right triangle
$\(\mathrm{ATD}=\frac{1}{2} \times(\mathrm{TD}+\mathrm{BC}) \times \mathrm{TB}+\frac{1}{2} \times \mathrm{AT} \times \mathrm{TD}=\frac{1}{2} \times(10+5) \times \mathrm{x}+\frac{1}{2} \times 6 \times 10\)$ which is same as $\(7.5 x+30=180\)$, so we get $\(x=20\)$
$\(\mathrm{ATD}=\frac{1}{2} \times(\mathrm{TD}+\mathrm{BC}) \times \mathrm{TB}+\frac{1}{2} \times \mathrm{AT} \times \mathrm{TD}=\frac{1}{2} \times(10+5) \times \mathrm{x}+\frac{1}{2} \times 6 \times 10\)$ which is same as $\(7.5 x+30=180\)$, so we get $\(x=20\)$
Re: If the area of the quadrilateral ABCD in the above figure is 180 and
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06 Aug 2025, 10:11
06 Aug 2025, 10:11
Expert Reply
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