Question

If each side of a triangle is tripled, then find the ratio of areas of the new
triangle thus formed and the given triangle ​

Answer 1

Answer:

9:1

Step-by-step explanation:

Let the sides be a,b,c respectively

Perimeter 2s = a + b + c

Semi-perimeter, s = (a+b+c)/2

Using Heron's formula:

Area of the triangle A = √s(s−a)(s−b)(s−c)

Now, if the sides are tripled: 3a, 3b, 3c

Let s' be the semi-perimeter.

2s' = 3a + 3b + 3c

s' = 3(a + b + c)/2

or s' = 3s

Area of the triangle, A' = √s′(s′−3a)(s′−3b)(s′−3c)

A' = √(3s)(3s−3a)(3s−3b)(3s−3c)

A' = √$3^{4}$s(s−a)(s−b)(s−c)

A' = 9√s(s−a)(s−b)(s−c)

A' = 9A

A':A = 9:1

Ratio of area of the new triangle and old triangle is 9:1

Answer 2

Answer:

9:1

Step-by-step explanation:

Let the sides be a,b,c respectively

Perimeter 2s = a + b + c

Semi-perimeter, s = (a+b+c)/2

Using Heron's formula:

Area of the triangle A = √s(s−a)(s−b)(s−c)

Now, if the sides are tripled: 3a, 3b, 3c

Let s' be the semi-perimeter.

2s' = 3a + 3b + 3c

s' = 3(a + b + c)/2

or s' = 3s

Area of the triangle, A' = √s′(s′−3a)(s′−3b)(s′−3c)

A' = √(3s)(3s−3a)(3s−3b)(3s−3c)

A' = √3^{4}3

4

s(s−a)(s−b)(s−c)

A' = 9√s(s−a)(s−b)(s−c)

A' = 9A

A':A = 9:1

Ratio of area of the new triangle and old triangle is 9:1

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