Question

If the ratio of areas of two similar triangles are 25:49 then what will be the ratio of the perimeter

Answer 1

Step-by-step explanation:

Let the triangles be

ABC andDEF

We know that ar(ABC) /ar(DEF) =(AB/DE) *2

25/49=(AB/DE)*2

So AB/DE=5/8 (by doing underroot)

So AB:DE=5:7

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Answer 2

Step-by-step explanation:

Let the radius of two semicircles be $r_1$ and $r_2$

→ Given :-

▶ The ratio of areas of two semicircles = 49:25 .

$\begin{lgathered}= > \frac{a_1}{a_2} = \frac{49}{25} . \\ \\ = > \frac{ \frac{ \cancel\pi {r_1}^{2} }{ \cancel2} }{ \frac{ \cancel\pi {r_1}^{2} }{ \cancel2} } = \frac{49}{25} . \\ \\ = > {( \frac{r_1}{r_2}) }^{2} = \frac{49}{25} . \\ \\ = > \frac{r_1}{r_2} = \sqrt{ \frac{49}{25} } . \\ \\ = > \frac{r_1}{r_2} = \frac{7}{5} .\end{lgathered}$

→ To find :-

▶ The ratio of their circumference.

$\begin{lgathered}\therefore \frac{c_1}{c_2} \\ \\ = \frac{ \cancel\pi r_1}{ \cancel\pi r_2} . \\ \\ = \frac{r_1}{r_2} . \\ \\ = \boxed{ \green{ \frac{7}{5} .}}\end{lgathered}$

Hence, ratio of their circumference is 7 : 5 .