In $\Delta PQR$ and $\Delta XYZ$,$\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$. Which correspondence between them is a similarity?

  • A
    $\Delta PQR \sim \Delta XYZ$
  • B
    $\Delta PQR \sim \Delta ZYX$
  • C
    $\Delta PQR \sim \Delta YZX$
  • D
    $\Delta PQR \sim \Delta XZY$

Explore More

Similar Questions

If $\Delta XYZ \sim \Delta DEF$ for the correspondence $XYZ \leftrightarrow DEF$,and $2 m \angle X = 3 m \angle Y$ with $m \angle Z = 30^{\circ}$,find $m \angle E$. (in $^{\circ}$)

In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DM^{2} = DN \cdot MC.$

Difficult
View Solution

In $\Delta ABC$ and $\Delta PQR$,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$. If $AB = 15$,$PQ = 20$ and $BC = 12$,find $QR$.

In the figure,$ABC$ and $DBC$ are two triangles on the same base $BC$. If $AD$ intersects $BC$ at $O$,show that $\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(DBC)} = \frac{AO}{DO}$.

Difficult
View Solution

In the figure,$\Delta ABC$ and $\Delta AMP$ are two right-angled triangles,right-angled at $B$ and $M$ respectively. Prove that:
$(i)$ $\Delta ABC \sim \Delta AMP$
$(ii)$ $\frac{CA}{PA} = \frac{BC}{MP}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo