Question

can anyone answer this question please?need it immediately

Answer 1

Let ABCD be the trapezium such that AB = 13cm, CD = 25 cm, BC=AD=10 cm.

Also, AB is parallel to CD.

Let BF and AE be perpendiculars on CD such that ABFE is a rectangle. Also,    ED = FC since the given trapezium is isosceles.

CD = DE+EF+FC

or, 25 = DE+13+FC

or, 25-13 = 2 DE  (since DE=FC)

0r, 12/2=DE

or, DE=FC=6 cm.

ΔADE and ΔCFB are right angled triangles where AD = BC, DE = CF. Hence, AE=BF.

In ΔADE, AD=10cm, DE= 6 cm

So, by pythagoras theorem, $P^{2} +B^{2} =H^{2}$

$(AE)^{2}+6^{2} = 10^{2}$

$(AE)^{2}+36=100$

$(AE)^{2}=100-36$

$AE^{2}=64$

$AE=\sqrt{64}=8$.

Hence, AE = BF = 8cm = height of trapezium.

Thus, Area of trapezium = $\frac{1}{2}(sum of parallel sides)(height)$

$\frac{1}{2}(13+25)(8)=152$.

Hence, area of trapezium is 152 sq. cm.