Question

Find the percentage increase in the area of a triangle if each side is increased by 5 times

Answer 1

❏ Question

Find the percentage increase in the area of a triangle if each side is increased by 5 times ?

❏ Solution

Let, In ∆ ABC

• AB = a
• BC = b
• CA = c

And

• First Semiperimeter = S

So,

★ First semiperimeter (S) = (a+b+c)/2

Now, Heron's formula

★ First Area of ∆ABC (A) = √[s(s-a)(s-b)(s-c)]

If all side is increase by 5 times ,

So, New side will be of ∆ABC

• AB = 5a
• BC = 5b
• CA = 5c

And,

• Second semiperimeter = (S1)

So, second semiperimeter (S1) = (5a+5b+5c)/2

Or,

➩ (S1) = 5.[(a+b+c)/2 ]

but, first semiperimeter

• S = (a+b+c)/2

➩ (S1) = 5.[S]

➩ (S1) = 5S. ..........(1)

Again, Heron's formula

★ Second area of ∆ABC(A1) = √[S1(S1-a)(S1-b)(S1-c)]

➩ A1 = √[5S(5S-5a)(5S-5b)(5S-5c)]

➩ A1 = √[5×5×5×5×S(S-a)(S-b)(S-c)]

➩ A1 = 5×5×√[S(S-a)(S-b)(S-c)]

➩A1 = 25×√[S(S-a)(S-b)(S-c)]

But, first area

• A = √[S(S-a)(S-b)(S-c)]

So

➩ A1 = 25.A

Hence, new area of ∆ABC will be 25 times of first area of ∆ABC.

Now, change in area will be

➩ [(A1-A) × 100]/A

= [(25A -A)× 100] / A

= (24A × 100)/A

= 24 × 100

= 2400 %

Hence,

• If we increase 5 times each side of ∆ABC , then new area of ∆ABC also increase by 2400 % percentage of old area of ∆ABC .

Answer 2

Answer:

2400

Step-by-step explanation:

By using (n²-1)10 formula

where n is 5

answer comes out to be 2400