Question

The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of these triangles?

Answer 1

Answer:

• 123 3/11 dm
• 102 8/11 dm

Step-by-step explanation:

The ratio of the linear dimensions (perimeter) is the square root of the ratio of area dimensions, so the larger : smaller ratio is ...

  √(72/50) = √1.44 = 1.2

The sum of ratio units is 1.2 + 1 = 2.2, so each ratio unit stands for ...

  (226 dm)/2.2 = 102 8/11 dm . . . . . the perimeter of the smaller triangle

Then the perimeter of the larger triangle is ...

  1.2 × 102 8/11 dm = 123 3/11 dm . . . . the perimeter of the larger triangle

Answer 2

Answer:

$\large\boxed{\dfrac{1130}{11}\ dm=102\dfrac{8}{11}\ dm\ and\ \dfrac{1356}{11}\ dm=123\dfrac{3}{11}\ dm}$

Step-by-step explanation:

We know: The ratio of the areas of similar triangles is equal to the square of the similarity scale.

Threfore:

$\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{A_{\triangle_1}}{A_{\triangle_2}}=k^2$

We have

$A_{\triangle_1}=72\ dm^2\\\\A_{\triangle_2}=50\ dm^2$

Susbtitute:

$k^2=\dfrac{72}{50}\\\\k^2=\dfrac{72:2}{50:2}\\\\k^2=\dfrac{36}{25}\to k=\sqrt{\dfrac{36}{25}}\\\\k=\dfrac{\sqrt{36}}{\sqrt{25}}\\\\\boxed{k=\dfrac{6}{5}}$

We have the similarity scale.

We know: The ratio of the perimeters of similar triangles is equal to the similarity scale.

Therefore:

$\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{P_1}{P_2}=k\to P_1=kP_2$

We have:

$P_1+P_2=226\ dm^2$

Substitute:

$k=\dfrac{6}{5},\ P_1=kP_2\to P_1=\dfrac{6}{5}P_2$

$\dfrac{6}{5}P_2+P_2=226\\\\\dfrac{6}{5}P_2+\dfrac{5}{5}P_2=226\\\\\dfrac{11}{5}P_2=226\qquad\text{multiply both sides by 5}\\\\11P_2=1130\qquad\text{divide both sides by 11}\\\\P_2=\dfrac{1130}{11}\\\\\boxed{P_2=102\dfrac{8}{11}\ dm}\\\\\\P_1=kP_2\to P_1=\dfrac{6}{5}\cdot\dfrac{1130}{11}\\\\P_1=\dfrac{6}{1}\cdot\dfrac{226}{11}\\\\P_1=\dfrac{1356}{11}\\\\\boxed{P_1=123\dfrac{3}{11}\ dm}$