Delta ABC sim Delta PQR for the correspondenc | vedclass

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(D) Given that $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$.For similar triangles,the ratio of their perimeters is eq

Vedclass · Accepted Answer

$5\sqrt{10}$

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(D) Given that $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$.For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.$	herefore \frac{	ext{Perimeter of } \Delta ABC}{	ext{Perimeter of } \Delta PQR} = \frac{AC}{PR}$Given,Perimeter of $\Delta ABC = 35$,Perimeter of $\Delta PQR = 28$,and $PR = 4\sqrt{10}$.Substituting the values:$\frac{35}{28} = \frac{AC}{4\sqrt{10}}$Simplifying the fraction $\frac{35}{28}$ by dividing by $7$,we get $\frac{5}{4}$.$\frac{5}{4} = \frac{AC}{4\sqrt{10}}$$AC = \frac{5}{4} 	imes 4\sqrt{10}$$AC = 5\sqrt{10}$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. The perimeter of $\Delta ABC$ is $35$ and the perimeter of $\Delta PQR$ is $28$. If $PR = 4\sqrt{10}$,then $AC = \ldots$

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