Question

The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

Answer 1

Given:
The parallel sides of a trapezium are 25 cm and 13 cm.
Its nonparallel sides are equal in length and each is equal to 10 cm.
A rough sketch for the given trapezium is given below:

In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles.
AD = BC = 10 cm
D  = CN = x cm
\[\angle DMA = \angle CNB = 90^\circ\]
Hence, the third side of both the triangles will also be equal.
\[ \therefore AM=BN\]
Also, MN=13
Since AB = AM+MN+NB:
\[ \therefore 25=AM+13+BN\]
\[AM+BN=25-13=12 cm\]
\[Or, BN+BN=12 cm (\text{  Because AM=BN })\]
\[2 BN=12\]
\[BN=\frac{12}{2}=6 cm\]
∴ AM = BN = 6 cm.
Now, to find the value of x, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10, 6 and x.
\[ {(\text{  Hypotenuse })}^2 {=(\text{  Base })}^2 {+(\text{  Altitude })}^2 \]
\[ {(10)}^2 {=(6)}^2 {+(x)}^2 \]
\[ {100=36+x}^2 \]
\[ x^2 =100-36=64\]
\[x=\sqrt{64}=8 cm\]
∴ Distance between the parallel sides = 8 cm
∴ Area of trapezium\[=\frac{1}{2}\times( \text{  Sum of parallel sides })\times(\text{  Distance between parallel sides })\]
\[=\frac{1}{2}\times(25+13)\times(8)\]
\[ {=152 cm}^2\]