Question

In the figure , DE ∣∣ BC . If ar ( D B C E ) / ar ( △ A B C ) = 16 / 25 x and BC = 8 . 5 cm , find DE ?

Answer 1

Answer:

Given

ΔABC in which DE ∥BC and DE=4cm, BC=8cm. and ar. (ΔADE)=25sq.cm.

In ΔADE and ΔABC ,

∠A=∠A [Common]

∠ADE=∠ABC [Corresponding angles]

∠AED=∠ACB [Corresponding angles]

∴ΔADE∼ΔABC [AAA similarity]

Since ratio of areas of two similar triangles is equal to ratio of squares on the corresponding sides,

∴

ar(ΔABC)

ar(ΔADE)

=

(BC)

2

(DE)

2

⇒

ar(ΔABC)

25

=

(8)

2

(4)

2

=

64

16

=

4

1

Hence, ar(ΔABC)=25×4=100sq.cm

Answer 2

$\sf \huge\red{Answer}$

$\huge \sf \underline \pink{given}$

DE∥BC and BC = 8.5cm.

$\dfrac{Δ FDE}{ΔABC} = \dfrac{16}{25}$

$\huge \sf \underline \pink{to \: find}$

DE

$\huge \sf \underline \pink{solution}$

Since, ΔABC ~ ΔFDE

$\dfrac{Δ FDE}{ΔABC } = \dfrac{DE^{2} }{ BC ^{2} }$

$\dfrac{16}{25 } = \dfrac{DE ^{2} }{8.5^{2} }$

$\dfrac{16}{25} ⬈ \dfrac{DE}{8.5}$

$\dfrac{16 \times 8.5 \times 8.5}{25} = D{E}^{2}$

$\dfrac{4 \times 4 \times 8.5 \times 8.5}{5 \times 5} = {DE}^{2}$$( \dfrac{4 \times 8.5}{5} )^{2} = {DE}^{2}$

$( \dfrac{4 \times \cancel{8.5}}{ \cancel{5}} )^{ \cancel{2}} = {DE}^{ \cancel{2}}$

$4 \times 1.7 = {DE}$$DE = 6.8 \: cm$

$\red {\boxed{i \: hope \: its \: helpful}}$