Question

Let ∆ABC~∆DEF and their areas be respectively 64 cm² and 121cm² if EF=15.4cm, Find BC.​

Answer 1

Heres Ur Answer:

∆ABC~∆DEF...,..........given

Now,

By similar triangle theorem we get,

∆ABC2 /∆ DEF2 = BC2 / EF2

64/121 = BC2/(15.4)2

By taking square roots of both sides,

8/11= BC/15.4

By cross multiplication,

(15.4×8)/11 = BC

BC=11.2

therefore value of BC is 11.2

Stay Awesome!

Answer 2

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$\sf\underbrace{Appropriate\:Question: }$

• Let △ABC∼△DEF and their areas be respectively, 64cm² and 121cm², If EF = 15.4cm, find BC.??

$\sf\underbrace{Required\:Answer:}$

$\bf\underline{Given:}$

• ∆ABC~∆DEF ar ∆ABC = 64cm² ar ∆DEF = 121 cm² EF = 15.4 cm.

$\bf\underline{To\:Find:}$

• BC

$\bf\underline{Solution:}$

Since ∆ABC~∆DEF

We know that if two triangle are similar,

Ratio of areas is equal to square of ratio of its corresponding sides.

$\sf\underline{Hence,}$ $\sf\dfrac{ ar \:∆ABC}{ar\:∆DEF}$ = $\bigg(\dfrac{BC}{EF}\bigg)$²

$\bf\underline{putting\:The\: values:}$

ㅤㅤ$\implies$$\sf\dfrac{64}{121}$ = $\bigg(\dfrac{BC}{15.4}\bigg)$²

ㅤㅤ$\implies$$\sf\dfrac{64}{121}$ = $\sf\dfrac{BC²}{(15.4²)}$

ㅤㅤ$\implies$$\sf\dfrac{64}{121}$ $\sf{(15.4)²\:=\:BC²}$

ㅤㅤ$\implies$ $\bigg(\dfrac{8\:×\:8}{ 11\:×\:11}\bigg)$ × $\sf{(15.4)²\:=\:BC²}$

ㅤㅤ$\implies$$\sf\dfrac{8²}{11²}$ × $\sf{(15.4)²\:=\:BC²}$

ㅤㅤ$\implies$ $\bigg(\dfrac{8²}{ 11²}$× $\sf{15.4}\bigg)$²$\sf{=\:BC²}$

ㅤㅤ$\implies$$\sf\dfrac{8}{11}$ $\sf{×\:15.4\:=\:BC}$

ㅤㅤ$\implies$$\sf{BC\:=}$ $\sf\dfrac{8}{11}$$\sf{×\:15.4}$

ㅤㅤ$\implies$$\sf{BC\:=\:8\:×\:1.4}$

ㅤㅤ$\implies$$\sf{BC\:=\:11.2}$

$\bf\underline{Hence,\:BC\:=\:11.2\:cm}$

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