Question

The parallel sides of a trapezium are 15m and 10m long and it's non-parallel sides are 8m and 7m long.Find the area of the trapezium.

Answer 1

Answer: The answer is 50√3 m².

Step-by-step explanation: As given in the question and described in the attached figure, ABCD is a trapezium, where AB = 8 m, BC = 15 m, CD = 7 m and AD = 10 m.

From vertex A and D, perpendiculars AP and DQ are draw on BC so that APQD will form a rectangle with AD = PQ = 10 m.

Let QC = x m, then BP = 15-10-x = (5 - x) m.

From right-angled triangle ABP and CQD, we have

$AB^2=AP^2+BP^2\\\\\Rightarrow 8^2=AP^2+(5-x)^2\\\\\Rightarrow 64=AP^2+25-10x+x^2\\\\\Rightarrow AP^2=39+10x-x^2,$

and

$CD^2=CQ^2+DQ^2\\\\\Rightarrow 7^2=x^2+DQ^2\\\\\Rightarrow DQ^2=49-x^2.$

Since APQD is a rectangle, so we have

$AP=DQ\\\\\Rightarrow AP^2=DQ^2\\\\\Rightarrow 39+10x-x^2=49-x^2\\\\\Rightarrow 10x=10\\\\\Rightarrow x=1.$

Therefore,

$AP=\sqrt{39+10\times 1-1^2}=\sqrt{48}=4\sqrt 3.$

Hence, the area of trapezium ABCD is given by

$A=\dfrac{1}{2}\times(\textup{sum of the parallel sides})\times \textup{height}\\\\\Rightarrow A=\dfrac{1}{2}\times(AD+BC)\times AP\\\\\Rightarrow A=\dfrac{1}{2}\times (10+15)\times 4\sqrt 3=50\sqrt 3.$

Thus, the required area is 50√3 m².