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.

the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. hope this ! mark brainliest! you v much! : )

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.