Answer:
To find the area of the hexagon,we will first find the area of each equilateral triangle then multiply it by 6.
Let s be the side length of the square. The dimensions of the rectangle are three times the side of the square (i.e. 3s), and two less than the side of the square (i.e. s-2).
So, the area of this rectangle is
![3s(s-2)=3s^2-6s](https://tex.z-dn.net/?f=3s%28s-2%29%3D3s%5E2-6s)
The area of the square is
, and we know that the two areas are the same, so we have
![3s^2-6s=s^2 \iff 2s^2-6s=0 \iff 2s(s-3)=0 \iff s=0 \lor s=3](https://tex.z-dn.net/?f=3s%5E2-6s%3Ds%5E2%20%5Ciff%202s%5E2-6s%3D0%20%5Ciff%202s%28s-3%29%3D0%20%5Ciff%20s%3D0%20%5Clor%20s%3D3)
The solution s=0 would lead to the extreme case where the rectangle and the square are actually a point, so we accept the solution s=3.
Answer:
I'm gonna pray for you lol
Step-by-step explanation:
Answer:
61°
Step-by-step explanation:
In order for triangle LMN and PQR to be congruent, angle M and angle Q must equal. thus, we can set 2x-26=x+16
with basic algebraic manipulation, x=42. we then plug it back into the angle M or Q, getting angle M/Q= 58°
since this is a isoceles, the 2 base angle are identical, so 180= 58+2(angle P)
thus, angle P= 61
I think the answer is 7 if i am not mistaking