Find the percentage increase in the area of a triangle if its each side is doubled?

Question

Answer

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)..... And

∆' = 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

% increase in area = 3×100=300 %

Hence, the percentage increase in the area of a triangle is 300%...

Answer

Percentage Increase in Area of a Triangle When its Each Side is Doubled

Explanation:

Let's assume that we have a triangle with sides a, b, and c. Its area can be calculated using Heron's formula:

A = √(s(s-a)(s-b)(s-c))

where s = (a+b+c)/2 is the semi-perimeter of the triangle.

Now, if we double each side of the triangle to get sides 2a, 2b, and 2c, the new semi-perimeter will be:

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

Therefore, the new area of the triangle can be calculated using Heron's formula as:

A' = √(s'(s'-2a)(s'-2b)(s'-2c))

Expanding the terms inside the square root:

A' = √(s(s-a)(s-b)(s-c)) × √((s+a)(s+b-c)(s+c-b)(s+a-b))

Dividing A' by A:

A'_/A = √((s+a)(s+b-c)(s+c-b)(s+a-b))/√(s(s-a)(s-b)(s-c))

Now, using the fact that s = (a+b+c)/2 and s' = a+b+c, we can simplify the expression as:

A'_/A = √(2a/2a) × √(2b/2b) × √(2c/2c) × √((a+b+c)(a+b-c)(a-b+c)(-a+b+c))/√((a+b+c)(-a+b+c)(a-b+c)(a+b-c))

Cancelling out the common terms:

A'_/A = √2 × √((a+b+c)(a+b-c)(a-b+c)(-a+b+c))/√((a+b+c)(-a+b+c)(a-b+c)(a+b-c))

A'_/A = √2 × √((a+b-c)(a-b+c)(b+c-a)(a+b+c))/√((a+b-c)(a-b+c)(b+c-a)(a+b+c))

A'_/A = √2

Conclusion:

As we can see, the area of the triangle is doubled when each of its sides is doubled. Therefore, the percentage increase in the area of the triangle is:

Percentage Increase = (A'_/A - 1) × 100%

Percentage Increase = (√2 - 1) × 100%

Percentage Increase ≈ 41.4%

Answer

300

Similar Class 9 Doubts

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