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

• 123 3/11 dm
• 102 8/11 dm

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

`\large\boxed{(1130)/(11)\ dm=102(8)/(11)\ dm\ and\ (1356)/(11)\ dm=123(3)/(11)\ dm}`

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}\ (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=(72)/(50)\\\\k^2=(72:2)/(50:2)\\\\k^2=(36)/(25)\to k=\sqrt{(36)/(25)}\\\\k=(√(36))/(√(25))\\\\\boxed{k=(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}\ (P_1)/(P_2)=k\to P_1=kP_2`

We have:

`P_1+P_2=226\ dm^2`

Substitute:

`k=(6)/(5),\ P_1=kP_2\to P_1=(6)/(5)P_2`

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