Question

∆ABC~∆PQRand ar (∆ABC):ar(∆PQR)=16:25.if QR=15cm,then find BC​

Answer 1

$\bf \huge{ \orange{ \underline{ \underline{ \:Question:}}}}$

∆ABC~∆PQRand ar (∆ABC):ar(∆PQR)=16:25.if QR=15cm,then find BC

$\bf \huge{ \orange{ \underline{ \underline{Answer:}}}}$

$\sf{ \pink{ \underline{Given:-}}}$

• ar ( ∆ ABC )/ ar ( ∆ PQR ) = 16/25
• QR = 15 cm

$\sf { \pink{ \underline{ \: To \: \: Find :-}}}$

• Find BC

$\sf { \pink{ \underline{Soln :-}}}$

• ar ( ∆ ABC ) = 16
• ar ( ∆ PQR ) = 25

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\large \implies \sf \: \frac{16}{25} = \left(\frac{BC}{QR} \right) ^{2} \\ \\ \large \implies \sf \: \frac{16}{25} = \left(\frac{BC}{15} \right) ^{2} \\ \\ \large \implies \sf \: \frac{16}{25} = \frac{ {{BC}}^{2} }{225} \\ \\ \large \implies \sf \: {(BC)}^{2} = \frac{3600}{25} \\ \\ \large \implies \sf \: {(BC)}^{2} = 144 \\ \\ \large \implies \sf \: {BC} = \sqrt{144} \\ \\ \large \implies \sf \: {BC} = 12$

•°• BC is 12 cm....