Question

Triangle ABC ~ triangle PQR and (triangle ABC ) = 4ar (triangle PQR ) if BC = 12cm, find QR.
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Answer 1

Answer:

(triangle ABC ) = 4ar (triangle PQR )

both are similar triangle so height of both triangle is same

$\frac{1}{2} \times hight \times bc = 4 + \times \frac{1}{2} \times hight \: qr$

$\frac{1}{2} \times 12 =4 \times \frac{1}{2} \times qr$

$qr \: = 3cm$

Answer 2

$\sf \large \orange {\underline \orange{Given - }}$

• ∆ABC ~ ∆PQR
• (∆ABC) = 4AR (∆PQR)
• BC = 12cm

$\sf \large \orange {\underline \orange{To \: find - }}$

• Length of side QR

$\sf \large \orange {\underline \orange{S olution \: - }}$

Here,

(∆ABC) = 4AR (∆PQR)

$\sf \therefore \frac{ \triangle ABC }{ \triangle PQR } = \frac{4}{1}$

∆ABC ~ ∆PQR

$\sf \therefore \frac{ \triangle ABC }{ \triangle PQR } = \frac{BC ^{2} }{ {QR}^{2} }$

$\sf\longrightarrow \: \frac{BC ^{2} }{QR ^{2} } = \frac{ {4} }{ {1} }$

The length of side BC is 12cm.

$\sf \longrightarrow \: \frac{12 ^{2} }{QR ^{2} } = \frac{ {4}}{ {1}}$

$\sf \longrightarrow \: QR ^{2} = \frac{ {12}^{2} }{ {4} }$

$\sf\longrightarrow \: QR ^{2} = 36$

By taking square roots,

$\longrightarrow \red {\boxed{ \sf QR = 6}}$

Ans :- The length of QR is 6 cm.