Question

if each side of a triangle is doubled,then find the ratio of area of the new triangle thus formed and the given triangle​

Answer 1

Let a,b,c be the sides of the triangle.

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 doubled: 2a, 2b, 2c

Let s' be the semi-perimeter.

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

s' = a + b + c

or s' = 2s

Area of the triangle, A' = √s′(s′−2a)(s′−2b)(s′−2c)

A' = √(2s)(2s−2a)(2s−2b)(2s−2c)

A' = √24s(s−a)(s−b)(s−c)

A' = 4√s(s−a)(s−b)(s−c)

A' = 4A

A':A = 4:1

Ratio of area of the new triangle and old triangle is 4:1

Answer 2

Answer:

the answer is 4:1

Step-by-step explanation:

Let a,b,c be the sides of the triangle.

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 doubled: 2a, 2b, 2c

Let s' be the semi-perimeter.

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

s' = a + b + c

or s' = 2s

Area of the triangle, A2 = √s′(s′−2a)(s′−2b)(s′−2c)

A2 = √(2s)(2s−2a)(2s−2b)(2s−2c)

A2 = √24s(s−a)(s−b)(s−c)

A2 = 4√s(s−a)(s−b)(s−c)

A2 = 4A

A2:A = 4:1

Ratio of area of the new triangle and old triangle is 4:1