It is given that $\triangle FED \sim \triangle STU$. Is it true to say that $\frac{DE}{ST} = \frac{EF}{TU}$? Why?
(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.
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.
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The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.
Medium
View SolutionWhich of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.
| Part $I$ | Part $II$ |
| $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ | $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. |
| $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ | $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. |
| $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ | $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. |
| $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ | $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity. |
Difficult
View SolutionIn the given figure,$OB$ is the perpendicular bisector of the line segment $DE$,$FA \perp OB$,and $FE$ intersects $OB$ at the point $C$. Prove that $\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}$.
Difficult
View SolutionIn $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $T$ is the midpoint of $\overline{PR}$. If $PQ = 6$ and $QR = 8$,then $QT = \ldots$
Medium
View SolutionFor the correspondence $ABC \leftrightarrow XYZ$ being a similarity, which of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ $a. SSS$ condition $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ $b. SAS$ condition $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ $c. AAA$ condition - $d. \text{None of the conditions apply}$
| Part $I$ | Part $II$ |
|---|---|
| $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ | $a. SSS$ condition |
| $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ | $b. SAS$ condition |
| $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ | $c. AAA$ condition |
| - | $d. \text{None of the conditions apply}$ |
Medium
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