Question

how much area of triangle will increase in percentage if each side of the triangle is doubled​

Answer 1

Answer:

300%

Step-by-step explanation:

(i)

Initial sides be a,b,c.

∴ s = (a + b + c)/2

∴ A = √s(s - a)(s - b)(s - c)

(ii)

Sides of the triangle is increased.

New sides are 2a,2b,2c.

∴ s' = (2a + 2b + 2c)/2

     = a + b + c

     = 2s

∴ A' = √2s(2s - 2a)(2s - 2b)(2s - 2c)

= √16s(s - a)(s - b)(s - c)

= 4√s(s - a)(s - b)(s - c)

= 4A

Increase in Area = 4A - A = 3A

Percentage Increase = (3A/A) * 100

                                   = 300%

∴ % increase = 300%

Hope it helps!

Answer 2

Let a,b,c be the sides of the original ∆ & s be its semi perimeter.

S= (a+b+c)/2

2s= a+b+c.................(1)

The sides of a new ∆ are 2a,2b,2c

[ given: Side is doubled]

Let s' be the new semi perimeter.

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

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

s'= a+b+c

S'= 2s. ( From eq 1)......(2)

Let ∆= area of original triangle

∆= √s(s-a)(s-b)(s-c).........(3)

&

∆'= area of new Triangle

∆' = √s'(s'-2a)(s'-2b)(s'-2c)

∆'= √ 2s(2s-2a)(2s-2b)(2s-2c)

[From eq. 2]

∆'= √ 2s×2(s-a)×2(s-b)×2(s-c)

= √16s(s-a)(s-b)(s-c)

∆'= 4 √s(s-a)(s-b)(s-c)

∆'= 4∆. (From eq (3))

Increase in the area of the triangle= ∆'- ∆= 4∆ - 1∆= 3∆

%increase in area= (increase in the area of the triangle/ original area of the triangle)× 100

% increase in area= (3∆/∆)×100