Question

Find the area of a trapezium whose parallel sides are 11m and 25m long, and the nonparallel sides are 15m and 13m long.

Answer 1

From the point C construct CE || DA

We know that ADCE is a parallelogram having AE || DC and AD || EC with AD = 13m and D = 11m

It can be written as

AE = DC = 11m and EC = AD = 13m

So we get

BE = AB - AE

By substituting the values

BE = 25 –-11 = 14m

Consider $$\Delta$$ BCE

We know that BC = 15m, CE = 13m and BE = 14m

Take a = 15m, b = 13m and c = 14m

So we get

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

By substituting the values

$$s=\dfrac{15+13+14}{2}$$

So we get

s=21m

We know that

Area=$$\sqrt{s(s-a)(s-b)(s-c)}$$

By substituting the values

Area=$$ \sqrt{21(21-15)(21-13)(21-14)}$$

So we get

Area=$$ \sqrt{21\times6\times8\times7}$$

It can be written as

Area=$$\sqrt{7\times3\times2\times3\times4\times2\times7}$$

We get

Area=$$7\times3\times2\times2$$

So we get

Area=$$84cm^2$$

We know that

Area of △ BCE = $$\dfrac{1}{2} \times BE \times CL$$

By substituting the values

$$84 =\dfrac{1}{2} \times 14 \times CL$$

On further calculation

$$84 = 7 \times CL$$

By division

CL = 12m

We know that

Area of trapezium ABCD =$$\dfrac{1}{2}\times$$ sum of parallel sides $$ \times$$ height

It can be written as

Area of trapezium ABCD = $$\dfrac{1}{2}\times (AB + CD) \times CL$$

By substituting the values

Area of trapezium ABCD = $$\dfrac{1}{2} \times (11 + 25) \times 12$$

On further calculation

Area of trapezium ABCD = $$36 \times 6$$

By multiplication

Area of trapezium ABCD = $$216 m^2$$

Therefore, the area of trapezium ABCD is $$216 m^2$$.