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Question

$$ABCD$$  is a trapezium  $$\overline { A B } \| \overline { C D }$$  If  $$A B = 20 \mathrm { cm }\mathrm { BC } = 8 \mathrm { cm } , \mathrm { CD } = 10 \mathrm { cm }$$  and  $$\mathrm { AD } = 6 \mathrm { cm } .$$ Find the area of  $$\mathrm { ABCD } .$$
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Solution

R.E.F. Image.
ABCD is trapezium. Draw CM.|| AD.
In $$ \Delta $$ CMB ; 6, 8, 10 cm.
$$ 8 = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12 cm$$ a = 6 b = 8 c = 10
$$ \Delta  = \frac{1}{2}\times b\times h = \sqrt{S(S-a)(S-b)(S-c)}$$
$$ = \frac{1}{2}\times 8\times h = \sqrt{12(12-6)(12-8)(12-10)}$$
$$ = 4h = \sqrt{12.6.4.2} = \sqrt{576} = 24$$
$$ x = h = 6 $$
Area of trapezium = Area of AMCD + Area of $$ \Delta $$ MBC 
$$ = 10\times 6+24 = 84 cm^{2}$$

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Mathematics

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