It is given that $\triangle FED \sim \triangle STU$. Is it true to say that $\frac{DE}{ST} = \frac{EF}{TU}$? Why?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) No,it is not true.
Given that $\triangle FED \sim \triangle STU$,the corresponding vertices are $F \leftrightarrow S$,$E \leftrightarrow T$,and $D \leftrightarrow U$.
According to the property of similar triangles,the ratios of their corresponding sides must be equal.
Therefore,the correct ratios are $\frac{FE}{ST} = \frac{ED}{TU} = \frac{FD}{SU}$.
Comparing this with the given expression $\frac{DE}{ST} = \frac{EF}{TU}$,we can see that it does not match the required correspondence of sides.

Explore More

Similar Questions

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.

In $Fig.$,line segment $DF$ intersects the side $AC$ of a triangle $ABC$ at the point $E$ such that $E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$. Prove that
$\frac{BD}{CD} = \frac{BF}{CE}$

Difficult
View Solution

In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$SY = 4.5$ and $XT = 8$,find $XZ$.

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,$2 AB = PQ$,and $BC = 10$,then $QR = \dots$

The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

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