#### Question

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.

#### 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`

Is there an error in this question or solution?

Solution 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. Concept: Application of Heron’S Formula in Finding Areas of Quadrilaterals.