If Delta ABC sim Delta PQR under the correspo | vedclass

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(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.According to the theorem of areas of similar triangles,

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta ABC)}{	ext{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}ight)^2$.Substituting the given values: $\frac{36}{64} = \left(\frac{12}{PQ}ight)^2$.Taking the square root on both sides: $\sqrt{\frac{36}{64}} = \frac{12}{PQ}$.$\frac{6}{8} = \frac{12}{PQ}$.Simplifying the fraction: $\frac{3}{4} = \frac{12}{PQ}$.Cross-multiplying: $3 	imes PQ = 12 	imes 4$.$3 	imes PQ = 48$.$PQ = \frac{48}{3} = 16$.

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$. If $AB = 12$,$\text{Area}(\Delta ABC) = 36$,and $\text{Area}(\Delta PQR) = 64$,then $PQ = \ldots$

Similar Questions

$P$ and $Q$ are the points on the sides $DE$ and $DF$ of a triangle $DEF$ such that $DP = 5 \, cm$,$DE = 15 \, cm$,$DQ = 6 \, cm$,and $QF = 18 \, cm$. Is $PQ \parallel EF$? Give reasons for your answer.

In $\Delta ABC$,$M$ lies on $AB$ and $N$ lies on $AC$ such that $\overline{MN} \parallel \overline{BC}$. If $AM = 6$,$MB = 9$,and $AN = 8$,find $AC$.

In $\Delta ABC,$ the midpoints of the sides are $P, Q, R.$ In $\Delta PQR,$ the midpoints of the sides are $X, Y, Z.$ If the area of $\Delta XYZ$ is $20,$ find the area of $\Delta PQR$ and $\Delta ABC.$

In $\Delta PQR$,$m\angle Q = 90^{\circ}$,$PR = a^{2} + b^{2}$ and $QR = a^{2} - b^{2}$. Then,$PQ = \ldots$ $(a, b \in R^{+}; a > b)$

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $M$ is the midpoint of $\overline{XZ}$. If $XY = 3$ and $YZ = 4$,then $YM = \ldots$

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