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The area of an isosceles trapezium is 176 sq.cm. and the height is $\frac{2}{11}$ th of the sum of its parallel sides.If the ratio of the length of the parallel sides is 4:7,then the length of a diagonal (in cm)is1). 282). $\sqrt{137}$3). $2\sqrt{137}$4). 24 |
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Answer» right answer is $\SQRT{137}$ AREA =$ \LARGE \frac{1}{2} $(sum of parallel sides)$\times$ DISTANCE between them $ \Large \frac{1}{2}(7x+4x) \times 2x=176 $ $ \Large 11x^{2}=176\Rightarrow x^{2}16 $ $ \Large \Rightarrow x=4 $ $ \Large AB=7 \times 4=28cm $ $ \Large CD=4 \times 4=16cm $ $ \Large CM=2 \times 4=8cm $ $ \Large AM=AN+NM $ $ \Large \Rightarrow AN+16 $ $ \Large \Rightarrow 6+16=22 $ $ \Large (AN=BM=\frac{12}{2}=6) $ $ \Large AC^{2}=CM^{2}+AM^{2} $ $ \Large AC^{2}=8^{2}+22^{2} $ $ \Large AC=\sqrt{64+484}\Rightarrow \sqrt{548}\Rightarrow 2\sqrt{137} $ |
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