Question

the line parallel to parallel sides of trapezium passing through mid point of slant sides divide it in ratio of 5:2.then the ratio of parallel sides.

Answer 1

Given: A trapzm ABCD in which AB║PQ║CD.Also, $\frac{Area(trap.ABPQ)}{Area(PQCD)}=\frac{5}{2}$

To Find: $\frac{AB}{CD}$

Solution: ∠1=∠2

∠3=∠4

∠5=∠6

∠7=∠8

As AB║CD║PQ , So corresponding angles are equal.

$\frac{AP}{PD}=\frac{BQ}{QC}=1$

Trapzm ABQP ~ Trapzm CDQP

$\frac{Area(trpzmABPQ)}{Area(trapzmPQCD)}=[\frac{AB}{CD}]^{2}$  [ When trpzm are similar their areas is equal to square of corresponding sides.]

$\frac{Area(trpzmABPQ)}{Area(trapzmPQCD)}=\frac{5}{2}$

$[\frac{AB}{CD}]^{2}=\frac{5}{2}$

$\frac{AB}{CD}=\sqrt{\frac{5}{2}}$

Answer 2

Answer:

Step-by-step explanation: