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:
The areas of the triangles are

Answer 1

Answer:

The area of triangles are 16 cm^2 and 49 cm^2

Step-by-step explanation:

we know that

If two figures are similar, the ratio of its perimeters is equal to the scale factor and the ratio of its areas is equal to the scale factor squared

Let

z ----> the scale factor

x ----> the area of the smaller triangle in square centimeters

y ----> the area of the larger triangle in square centimeters

we know that

$z=\frac{4}{7}$

$\frac{x}{y}=z^2$

so

$\frac{x}{y}=(\frac{4}{7})^2$      

$\frac{x}{y}=\frac{16}{49}$

$x=\frac{16}{49}y$ -----> equation A

$x+y=65$ ----> equation B

solve the system by substitution

substitute equation A in equation B

$\frac{16}{49}y+y=65$

solve for y

$\frac{65}{49}y=65$

$y=49\ cm^2$

Find the value of x

$x=\frac{16}{49}(49)$

$x=16\ cm^2$

therefore

The area of triangles are 16 cm^2 and 49 cm^2

Answer 2

Answer:

49 sq. cm amd 16 sq. cm

Step-by-step explanation:

Use formula for the area of the triangle:

$A=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}$

The ratio of the perimeters of two similar triangles is 4:7, so

• if the larger base is x units, the smaller base is $\frac{4}{7}x$ units;
• if the larger height is h units, then the smaller height is  $\frac{4}{7}h$ units.

So, the sum of the area is

$\dfrac{1}{2}xh+\dfrac{1}{2}\cdot \dfrac{4}{7}x\cdot \dfrac{4}{7}h=65\\ \\\dfrac{1}{2}xh\left(1+\dfrac{16}{49}\right)=65\\ \\\dfrac{1}{2}xh\cdot \dfrac{65}{49}=65\\ \\A_{larger}=\dfrac{1}{2}xh=49\ cm^2\\ \\A_{smaller}=65-49=16\ cm^2$