Answer:
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:
[tex]\large\boxed{\dfrac{1130}{11}\ dm=102\dfrac{8}{11}\ dm\ and\ \dfrac{1356}{11}\ dm=123\dfrac{3}{11}\ dm}[/tex]
Step-by-step explanation:
We know: The ratio of the areas of similar triangles is equal to the square of the similarity scale.
Threfore:
[tex]\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{A_{\triangle_1}}{A_{\triangle_2}}=k^2[/tex]
We have
[tex]A_{\triangle_1}=72\ dm^2\\\\A_{\triangle_2}=50\ dm^2[/tex]
Susbtitute:
[tex]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}}[/tex]
We have the similarity scale.
We know: The ratio of the perimeters of similar triangles is equal to the similarity scale.
Therefore:
[tex]\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{P_1}{P_2}=k\to P_1=kP_2[/tex]
We have:
[tex]P_1+P_2=226\ dm^2[/tex]
Substitute:
[tex]k=\dfrac{6}{5},\ P_1=kP_2\to P_1=\dfrac{6}{5}P_2[/tex]
[tex]\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}[/tex]
