In Delta PQR and Delta XYZ,if frac{PQ}{XY} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$.By the $SSS$ (Side-Side-Side) similarity criterion,$\Delta PQR \sim \Delt

Vedclass · Accepted Answer

$4:25$

Vedclass · Answer

(B) Given that $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$.By the $SSS$ (Side-Side-Side) similarity criterion,$\Delta PQR \sim \Delta XYZ$.According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta PQR)}{	ext{Area}(\Delta XYZ)} = \left( \frac{PQ}{XY} ight)^2$.Substituting the given value: $\left( \frac{2}{5} ight)^2 = \frac{4}{25}$.Thus,the ratio is $4:25$.

In $\Delta PQR$ and $\Delta XYZ$,if $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$,then the ratio of their areas $\text{Area}(\Delta PQR) : \text{Area}(\Delta XYZ) = \ldots$

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

In $\Delta PQR, m \angle Q = 90^{\circ}$. If $PR = 17$ and $PQ = 8$,then $QR = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 2 CM$,prove that $AC = 5 CM$.

With the help of the definition of the similarity of triangles,prove that all equilateral triangles are similar.

In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. $A$ line passing through $G$ (the centroid) intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$. If $3 AN = 2 AC$,prove that $AM = 2 MB$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module