Question

The perimeter of two similar triangles are 35 and 45.Find their area ratio

Answer 1

Answer:

$\frac{49}{81}$

Step-by-step explanation:

Since, if two triangles are similar,

Then their corresponding sides are in same proportion,

Also, the ratio of perimeters of the triangles has the same ratio as corresponding sides,

While the ratio of areas is equal to the square of the ratio of corresponding sides.

That is, If ABC and DEF are two similar triangles,

Then,

$\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle D EF}=(\frac{\text{Perimeter of }\triangle ABC}{\text{Perimeter of }\triangle D EF})^2$

If perimeters of triangles ABC and DEF are 35 and 45 respectively,

Then,

$\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle D EF}=(\frac{35}{45})^2$

$=\frac{1225}{2025}$

$=\frac{49}{81}$

Learn more :

https://brainly.in/question/12586648   ( answered by Zyxwvu )

Answer 2

Answer:

$\red { \frac{Area \:of \:\triangle ABC }{Area \:of \:\triangle PQR} }\green {= \frac{ 49}{81}}$

Step-by-step explanation:

Let ∆ABC ~ ∆PQR .

Perimeter of ∆ABC = 35 ,

Perimeter of ∆PQR = 45 .

$We \:know \:that ,$

$\boxed { \pink { \frac{Area \:of \:\triangle ABC }{Area \:of \:\triangle PQR} = \left( \frac{ Perimeter \:of \:\triangle ABC}{Perimeter \:of \: \triangle PQR }\right)^{2}}}$

$= \left( \frac{35}{45}\right)^{2}\\=\left( \frac{5\times 7}{5\times 9}\right)^{2}\\= \left( \frac{7}{9}\right)^{2}\\= \frac{ 49}{81}$

Therefore.,

$\red { \frac{Area \:of \:\triangle ABC }{Area \:of \:\triangle PQR} }\green {= \frac{ 49}{81}}$

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