$\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow EFD$. $AB : BC : CA = 4 : 3 : 5$. If the perimeter of $\Delta DEF$ is $36$,find the measures of all the sides of $\Delta DEF$.
(D) Given the similarity $\Delta ABC \sim \Delta DEF$ with the correspondence $ABC \leftrightarrow EFD$,the ratio of corresponding sides is equal.
Thus,$\frac{AB}{EF} = \frac{BC}{FD} = \frac{CA}{DE} = k$.
Given $AB : BC : CA = 4 : 3 : 5$,let $AB = 4x, BC = 3x, CA = 5x$.
From the correspondence $ABC \leftrightarrow EFD$,we have $EF = BC = 3x$,$FD = CA = 5x$,and $DE = AB = 4x$.
The perimeter of $\Delta DEF = EF + FD + DE = 3x + 5x + 4x = 12x$.
Given the perimeter is $36$,we have $12x = 36$,which implies $x = 3$.
Therefore,the sides are $EF = 3(3) = 9$,$FD = 5(3) = 15$,and $DE = 4(3) = 12$.
Thus,the sides of $\Delta DEF$ are $12, 9, 15$.
Thus,$\frac{AB}{EF} = \frac{BC}{FD} = \frac{CA}{DE} = k$.
Given $AB : BC : CA = 4 : 3 : 5$,let $AB = 4x, BC = 3x, CA = 5x$.
From the correspondence $ABC \leftrightarrow EFD$,we have $EF = BC = 3x$,$FD = CA = 5x$,and $DE = AB = 4x$.
The perimeter of $\Delta DEF = EF + FD + DE = 3x + 5x + 4x = 12x$.
Given the perimeter is $36$,we have $12x = 36$,which implies $x = 3$.
Therefore,the sides are $EF = 3(3) = 9$,$FD = 5(3) = 15$,and $DE = 4(3) = 12$.
Thus,the sides of $\Delta DEF$ are $12, 9, 15$.
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In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$. If $PA = 10$,$PC = 15$ and $PD = 12$,then $BD = \ldots$
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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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Medium
View Solution$PQR$ is a triangle right-angled at $P$ and $M$ is a point on $QR$ such that $PM \perp QR$. Show that $PM^{2} = QM \cdot MR$.
Difficult
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