Question

The area of a trapezium of height 25 cm is 400 cm2

. If the height of

one parallel sides is 25 cm, find the length of other parallel sides.​

Answer 1

Step-by-step explanation:

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$\huge{\underline{\mathrm{Question}}}$

The area of a trapezium of height 25 cm is 400 cm2. If the height of one parallel sides is 25 cm, find the length of other parallel sides.

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$\huge{\underline{\mathrm{Answer}}}$

∴ CF = DF − CD

⇛(77−60)

⇛ 17cm

∴ For △BCF

Permiter of triangle △BCF = BC + BF + CF

⇛ (25 + 26 + 17) cm

⇛ 68cm

$\:∴ Semiperimeter \: \: of \: \: the \: \: triangle  \: \: (s)= \: \frac{p}{2} = \frac{68}{2} = 34cm$

∴ By \: Heron's \: \: formula [/tex] Area \: of \: triangle  \: △BCF = \sqrt{S(s−BC)(s−BF)(s−CF)}

[/tex] = \sqrt{34(34−25)(34−26)(34−17)} [/tex] = \sqrt{34×9×8×17} [/tex]\sqrt{41616} = 204 {cm}^{2} [/tex] ∴let  \: h \:  cm  \: be \: the \: length \: of \:perpendicular \: to \:  CF \: \\ ∴Area \: of  \: BCF \: = \frac{1}{2} \times CF×BE [/tex] = 204 = \times \frac{1}{2} \times 17 \times h[/tex] = h \times \frac{204 \times 2}{14} = 24cm [/tex] ∴Area \: trapezium \: of \: ABCD \: ⇛\: \frac{1}{2} \: 21[(60+77)×24]= \frac{1}{2} = (137 \times 29) 1⇛ \frac{1}{2} \times 3288 [/tex] [/tex]

= ${\boxed{1644}}$

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