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Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.
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Solution
Given that two parallel sides of trapezium are AB = 77 and CD = 60 cm
Other sides are BC = 26 m and AD = 25 cm.
Join AE and CF
Now, DE ⊥ AB and CF ⊥ AB
∴ DC = EF = 60 cm
Let AE = x
⇒ BF = 77 – 60 – x = 17 – x
`In ΔADE, DE^2 = AD^2 – AE^2 = 25^2 – x^2` [∵ Pythagoras theorem]
And in ΔBCF, `CF^2= BC^2 – BF^2` [∵ By Pythagoras theorem]
`⇒25=sqrt(26^2-(17-x)^2)`
`⇒25^2-x^2=25^2-(289-x^2-34-x)` [ ∵`(a-b)^2=a^2-2ab+b^2` ]
`⇒265-x^2=676-289-x^2+34x`
`34x=238`
`x=7`
`∴ DE =sqrt(25^2-x^2)=sqrt(625-7^2)=sqrt(516)=24cm`
∴ Area of trapezium = `1/2`(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑠𝑖𝑑𝑒𝑠)×ℎ𝑒𝑖𝑔ℎ𝑡=`1/2`(60×77)×24=`1644cm^2`
Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals
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