Your friend claims that a triangle with side lengths of a, b, and c will have half the area of a triangle with side lengths of 2a, 2b, and 2c. Is your friend correct?

My friend is wrong since, the area of the triangle with sides a, b, c is one-quarter the area of triangle with sides 2a, 2b, 2c

Step-by-step explanation:

Using herons formula, the area of the first triangle with sides a, b, c is A = √[s(s -a)(s - b)(s -c)] where s = (a + b + c)/2.

So, A = √[(a + b + c)/2((a + b + c)/2 - a)((a + b + c)/2 - b)((a + b + c)/2 - c)]

= √[(a + b + c)/2((b + c - a)/2)((a - b + c)/2)((a + b - c)/2)]

= 1/4√[(a + b + c)(b + c - a)(a - b + c)(a + b - c)].

For the triangle with sides 2a, 2b, 2c, the area is A' = √[s'(s' - 2a)(s' - 2b)(s' -c)] where s' = (2a + 2b + 2c)/2 = a + b + c.

So, A' = √[(a + b + c)((a + b + c) - 2a)((a + b + c) - 2b)((a + b + c) - 2c)]

= √[(a + b + c)((b + c - a)(a - b + c)(a + b - c)]

= √[(a + b + c)(b + c - a)(a - b + c)(a + b - c)].

Since A = 1/4√[(a + b + c)(b + c - a)(a - b + c)(a + b - c)]. and A' = √[(a + b + c)(b + c - a)(a - b + c)(a + b - c)] ⇒ A = A'/4

So, the area of the triangle with sides a, b, c is one-quarter the area of triangle with sides 2a, 2b, 2c. So, my friend is wrong.

21/58.

hope to be marked brainliest

first one because it can't have a fraction in it

step-by-step explanation: