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A-5100-cm-2-trapezium-has-the-perpendicular-distance-between-the-two-parallel-sides-60-m-If-one-of-the-parallel-sides-be-40-m-then-find-the-length-of-the-other-parallel-side-




Question Number 35791 by spherecity@gmail.com last updated on 23/May/18
A 5100 cm^2  trapezium has the perpendicular  distance between the two parallel sides  60 m. If one of the parallel sides be 40 m   then find the length of the other parallel side.
$$\mathrm{A}\:\mathrm{5100}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{trapezium}\:\mathrm{has}\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{sides} \\ $$$$\mathrm{60}\:\mathrm{m}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallel}\:\mathrm{sides}\:\mathrm{be}\:\mathrm{40}\:\mathrm{m}\: \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{parallel}\:\mathrm{side}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18
A=(1/2)(l_1 +l_2 )×d  5100=(1/2)(40+x)×60  40+x=((5100)/(30))  40+x=170  x=130
$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({l}_{\mathrm{1}} +{l}_{\mathrm{2}} \right)×{d} \\ $$$$\mathrm{5100}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{40}+{x}\right)×\mathrm{60} \\ $$$$\mathrm{40}+{x}=\frac{\mathrm{5100}}{\mathrm{30}} \\ $$$$\mathrm{40}+{x}=\mathrm{170} \\ $$$${x}=\mathrm{130} \\ $$

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