Question

The ratio of the perimeters of two similar triangles is 4:7. What is the area of each of these triangles if the sum of their areas is 65cm2.

Answer 1

Answer:

Hence the area of each of the triangles are:

49 cm^2 and 16 cm^2.

Step-by-step explanation:

If the two triangles are similar; the perimeters of the two triangles are in the same ratio as the sides.

We are given that the ratio of the perimeters of two similar triangles is 4:7.

Let a and b denotes  perimeter of two triangles.

i.e. here a:b=4:7.

Let A and B denote the area of two triangles.

also we know that for two similar triangles;

$\dfrac{A}{B}=\dfrac{a^2}{b^2}$

Hence,

$\dfrac{A}{B}=\dfrac{4^2}{7^2}----(1)$

Also the sum of their areas is 65 cm^2.

i.e. $A+B=65\\\\A=65-B----(2)$

on putting equation (2) in equation (1) we have:

$\dfrac{(65-B)}{B}=\dfrac{4^2}{7^2}=\dfrac{16}{49}$

49(65-B)=16B

49×65-49B=16B

49×65=49B+16B

49×65=65B

B=49 cm^2

Hence by equation (2) we have:

A=65-B

A=65-49=16 cm^2.

Hence the area of each of the triangles are:

49 cm^2 and 16 cm^2.

Answer 2

The area of the two triangles are $\boxed{49{\text{ c}}{{\text{m}}^2}}$ and $\boxed{16{\text{ c}}{{\text{m}}^2}}.$

Further explanation:

The areas of two similar triangles are proportional to the squares of their perimeters.

$\boxed{\frac{{{\text{Area of triangle 1}}}}{{{\text{Area of triangle 2}}}} = \frac{{{{\left( {{\text{Perimeter of triangle 1}}} \right)}^2}}}{{{{\left( {{\text{Perimeter of triangle 1}}} \right)}^2}}}}$

Given:

The sum of areas of triangle is $65{\text{ c}}{{\text{m}}^2}.$

The ratio of the perimeters of two similar triangles is $4:7.$

Explanation:

Consider the area of first triangle as X.

Consider the area of second triangle as Y.

The ratio of the perimeter of the triangle is 4:7.

The ratio of the areas of the triangles can be obtained as follows,

$\begin{aligned}\frac{{{\text{Area of triangle 1}}}}{{{\text{Area of triangle 1}}}} &= \frac{{{{\left( {{\text{Perimeter of triangle 1}}} \right)}^2}}}{{{{\left( {{\text{Perimeter of triangle 2}}}\right)}^2}}}\\\frac{X}{Y} &= \frac{{{4^2}}}{{{7^2}}}\\\frac{X}{Y} &= \frac{{16}}{{49}}\\49X &= 16Y\\\end{aligned}$

The sum of areas of triangle is $65{\text{ c}}{{\text{m}}^2}.$

$\begin{aligned}X + Y &= 65\\X &= 65 - Y \\ \end{aligned}$

Substitute $65 - Y$ for $X$ in equation $49X = 16Y.$

$\begin{aligned}49\times \left( {65 - Y} \right) &= 16Y\\49 \times 65 - 49Y &= 16Y\\49 \times 65 &= 16Y + 49Y\\49 \times 65 = 65Y\\\frac{{49 \times 65}}{{65}}&= Y\\49&= Y\\\end{aligned}$

The area of first triangle can be obtained as follows,

$\begin{aligned} X + 49 &= 65\\X &= 65 - 49\\X&= 16\\\end{aligned}$

The area of the two triangles are $\boxed{49{\text{ c}}{{\text{m}}^2}}$ and $\boxed{16{\text{ c}}{{\text{m}}^2}}.$

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
2. Learn more about equation of circle brainly.com/question/1506955.
3. Learn more about range and domain of the function https://brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Triangles

Keywords: Ratio, perimeter, similar, similarity, triangles, proportional, square, area, area of triangle, two similar triangles, sum, sum of areas, 65 cm2, 4:7, corresponding sides.