The areas of two similar triangles are 25 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.Let the areas be $A_1 = 25$ an

Vedclass · Accepted Answer

$5: 4$

Vedclass · Answer

(B) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.Let the areas be $A_1 = 25$ and $A_2 = 16$.Let the corresponding sides be $s_1$ and $s_2$.Then,$\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.$\frac{25}{16} = (\frac{s_1}{s_2})^2$.Taking the square root on both sides,$\frac{s_1}{s_2} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.Therefore,the ratio of their perimeters is $5: 4$.

The areas of two similar triangles are $25$ and $16$. Then,the ratio of their perimeters is $\ldots \ldots \ldots \ldots$

Similar Questions

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AB = 10.8$,$MB = 4.5$ and $AC = 4.8$,find $AN$.

In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

In $\Delta ABC$,$\angle B$ is a right angle. Then,according to Pythagoras' theorem,$\ldots \ldots \ldots$ holds good.

$\Delta PQR \sim \Delta DEF$ for the correspondence $PQR \leftrightarrow DEF$. If $3 PQ = 2 DE$,$EF = 6$,and $PR = 8$,find $QR$ and $DF$.

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module