Find the area of trapezium $A B C D$ , where $A B$ is parallel to $D C$ . $A B = 77 c m, B C = 28 c m, C D = 60 c m, D A = 26 c m$

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Solution

Find the area of trapezium ABCD where AB is parallel
to DC. $A B = 77 c m B C = 28 c m C D = 60 c m D A = 26 c m$
Draw a line BC parallel to AD and draw a
perpendicular BE on CD. It can be observed
that $A B C D$ is a parallelogram
$B C = A D = 28 c m$
$C D = A B = 60 c m$
$C F = 77 c m - C D$
$= 77 - 60 = 17 c m .$
For $△ B C F$
Perimeter of a triangle $25 - (28 + 26 + 17)$
$S = \frac{71}{2} = 35.5$
By Heron's formula $\sqrt{S} (S - a) (S - b) (S - c)$
$△ B C F = \sqrt{35.5 (35.5 - 28) (35.5 - 26) (35.5 - 17)}$
$= 216.3 c m^{2}$
area of $△ B C F = \frac{1}{2} \times B E \times C F$
$216.3 c m^{2} = \frac{1}{2} \times B E \times 17$
$B E = 432.6 / 17 = 25.4 c m$
Area of $A B C D = B E \times C D = 25.4 \times 60$
$= 1524 c m^{2}$
Area of field $= 216.3 + 1524$
$= 1740.3 c m^{2}$

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Find the area of trapezium ABCD, where AB is parallel to DC.AB=77cm,BC=28cm,CD=60cm,DA=26cm

Find the area of trapezium $A B C D$ , where $A B$ is parallel to $D C$ . $A B = 77 c m, B C = 28 c m, C D = 60 c m, D A = 26 c m$