Delta DEF and Delta PQR are similar triangles | vedclass

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Question

(D) For two triangles $\Delta DEF$ and $\Delta PQR$ to be similar,the ratio of their corresponding sides must be equal,and the included angles between

Vedclass · Accepted Answer

$\frac{DE}{PR} = \frac{DF}{RQ}$

Vedclass · Answer

(D) For two triangles $\Delta DEF$ and $\Delta PQR$ to be similar,the ratio of their corresponding sides must be equal,and the included angles between those sides must be equal.Given that $m \angle D = m \angle R$,the sides forming $\angle D$ are $DE$ and $DF$,and the sides forming $\angle R$ are $RQ$ and $RP$.According to the $SAS$ (Side-Angle-Side) similarity criterion,the triangles are similar if $\frac{DE}{RQ} = \frac{DF}{RP}$.Rearranging this,we get $\frac{DE}{RP} = \frac{DF}{RQ}$,which matches option $D$.

$\Delta DEF$ and $\Delta PQR$ are similar triangles if $m \angle D = m \angle R$ and $\ldots \ldots$

Similar Questions

State the necessary conditions,as per definition,for two triangles to be similar.

In the figure,$OA \cdot OB = OC \cdot OD$. Show that $\angle A = \angle C$ and $\angle B = \angle D$.

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,and given $AB + BC = 12$,$PQ + QR = 15$,and $AC = 8$,find the length of $PR$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 6$,then $BM = \dots$

In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively. Then the correspondence $DEF \leftrightarrow \dots$ is a similarity.

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