+0

0
94
4

Jan 18, 2022

#1
+189
-1

The first triangle has an area of 4sqrt3. We can find this by using the Pythagorean Theorem on the first triangle by dividing it into two equal part and get the height is sqrt(4^2-2^2) or 2sqrt3. This multiplied by 4 then divided by 2 is 4sqrt3. The next triangle we can solve it's side lengths by writing equation(s). In this case the equation(s) would be sqrt(s^2-4^2)=4-x and sqrt(2x^2)=s where s is the side length of the second triangle and x is the length of the long lines until the triangle's vertices on it(not the lines with 4's written next to them). we can simplify the second equation to s=x*sqrt2 and substitute. sqrt(x*sqrt2-16)=4-x. We can square both sides to get x*sqrt2-16=x^2-8x+16, this can be simplified to x^2-(8+sqrt2)*x+32=0. This is how much I will do. I think you can try to solve from here.

Jan 18, 2022
#2
+1635
0

First triangle:

A = a2(√3 / 4)

Second triangle:

Side = 4 / cos15º

Area = a2(√3 / 4)

The rest of this problem is entirely up to you. Good luck!

Jan 19, 2022
edited by jugoslav  Jan 19, 2022
#3
0

thank you, may I just ask how you got to know that the side of the second triangle is 4/cos 15

Guest Jan 19, 2022
#4
0

No explanation is needed.

Guest Jan 22, 2022