Two parallel sides of a trapezium are $60\ cm$ and $77\ cm$ and other sides are $25\ cm$ and $26\ cm$. Find the area of the trapezium.

Given:

Two parallel sides of a trapezium are $60\ cm$ and $77\ cm$ and other sides are $25\ cm$ and $26\ cm$.

To do:

We have to find the area of the trapezium.

Solution:

Let in trapezium $ABCD$,

$AB \parallel DC$

$AB = 77\ cm, BC = 26\ cm, CD = 60\ cm, DA = 25\ cm$

Through $C$, draw $CE \| DA$ meeting $AB$ at $E$.

$AE = CD = 60\ cm$ and $EB = 77 - 60 = 17\ cm$.

$CE = DA = 25\ cm$

Area of $\triangle BCE$ with sides $17\ cm, 26\ cm, 25\ cm$

$s=\frac{a+b+c}{2}$

$=\frac{17+26+25}{2}$

$=\frac{68}{2}$

$=34$

Area of $\Delta \mathrm{EBC}=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{34(34-17)(34-26)(34-25)}$

$=\sqrt{34 \times 17 \times 8 \times 9}$

$=\sqrt{2 \times 17 \times 17 \times 2 \times 2 \times 2 \times 3 \times 3}$

$=17 \times 2 \times 2 \times 3$

$=204 \mathrm{~cm}^{2}$

Draw $\mathrm{CL} \perp \mathrm{EB}$

$\mathrm{CL}=\frac{\text { area of } \Delta \mathrm{EBC} \times 2}{\text { base }}$

$=\frac{204 \times 2}{17}$

$=24 \mathrm{~cm}$

Area of trapezium $\mathrm{ABCD}=\frac{1}{2}$ (Sum of parallel sides) $\times$ height

$=\frac{1}{2}(77+60) \times 24$

$=\frac{1}{2} \times 137 \times 24$

$=137 \times 12$

$=1644 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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