Question

In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm^2 , Area of trapezium BCED = 24 cm^2 and DE = 14 cm.

Answer 1

$hey \: mate \: here \: is \: ur \: answer \\ \\ \frac{ar(abc)}{ar(ade)} = \frac{ {bc}^{2} }{ {de}^{2} } \\ \\ \frac{25}{49} = \frac{ {bc}^{2} }{ {14}^{2} } \\ \\ in \: trapezium \: bced \\ ar(bced) = \frac{1}{2} (bc + de) \times h \\ \\ given \: ar(bced) = 24 \: {cm}^{2} \\ bc = 10 \: cm \\ de = 12 \: cm \\ 24 \: {cm}^{2} = \frac{1}{2} (10 + 14) \times h \\ \\ h = \frac{48}{10 + 14} \\ \\ h = \frac{48}{24} \\ h = 2 \: cm \\ ar(bcd) = \frac{1}{2} \times d \times h \\ = \frac{1}{2} \times 10 \times h \\ \\ ar(bcd) = \: 10 {cm}^{2} \\ \\ plzzz \: mark \: as \: brainliest$

Answer 2

Answer:

10 cm² is the answer plz check it ...