Question

The areas of two similar triangles are 20m2 and 180m2. The length of one of the sides of the second triangle is 12m. What is the corresponding side of the first triangle?

Answer 1

Answer:

The corresponding side of the first triangle is 4 m.

Step-by-step explanation:

Area of the first triangle: A1=20 m^2

Area of the second triangle: A2=180 m^2

Lenght of one of the sides of the second triangle: s2=12 m

Corresponding side of the first triangle: s1=?

A1/A2=(s1/s2)^2

Replacing the known values:

(20 m^2)/(180 m^2)=[s1/(12 m)]^2

Simplifying:

2/18=[s1/(12 m)]^2

Simplifying the fraction on the left side of the equation dividing the numerator and denominator by 2:

(2/2)/(18/2)=[s1/(12 m)]^2

1/9=[s1/(12 m)]^2

Solving for s1: Square root both sides of the equation:

sqrt(1/9)=sqrt{[s1/(12 m)]^2}

sqrt(1)/sqrt(9)=s1/(12 m)

1/3=s1/(12 m)

Multiplying both sides of the equation by 12 m:

(12 m)*(1/3)=(12 m)*s1/(12 m)

Simplifying:

(12 m)/3=s1

4 m=s1

s1=4 m

Answer 2

Answer: 4 meters.

Step-by-step explanation:

1. First, let's calculate the ratio of the sides of the triangles as following:

$ratio=\sqrt{\frac{Area_1}{Area_2}}\\ratio=\sqrt{\frac{20m^{2}}{180m^{2}}}\\ratio=\frac{1}{3}$

2. You know that the length of one of the sides of the second triangle is 12 meters, therefore, you must multiply the ratio calculated by that side, as following:

$side_1=\frac{1}{3}*12m\\side_1=4m$

3. Therefore,  the lenght of the corresponding side of the first triangle is 4 meters.