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Geometry: Need help finding angles of a quadrilateral?!?
The angles are EFGH
E= 8x-10
F= x^2+10
G= 2x+30
H= x^2+2x+10
I have the answers, I just don't know how to solve it. Can't seem to remember this at all.
4 Answers
- Anonymous10 years agoFavorite Answer
Add up the angles to 360°.....
8x - 10 + x² + 10 + 2x + 30 + x² + 2x + 10 = 360
Then, solve for x to get...
2x² + 12x + 40 = 360
2x² + 12x - 320 = 0
2(x² + 6x - 160) = 0
2(x - 10)(x + 16) =
x = 10
Hence, the angles are...
E = 70°
F = 110°
G = 50°
H = 130°
I hope this helps!
Source(s): Σ - L. E. GantLv 710 years ago
You also know that the angles add up to 360 degrees....
So, E + F + G + H = 360
So, 8x - 10 + x^2 + 10 + 2x + 30 + x^2 + 2x + 10 = 360
==> 2x^2 + 12x + 40 = 360
==> 2x^2 + 12x - 320 = 0
==> x^2 + 6x - 160 = 0
Will leave the rest to you....
- 10 years ago
well all the angles of a quadrilateral add up to 360. So 8x-10+x^2+10+2x+30+x^2+2x+10=360
2x^2+12x+40=360
2x^2+12x-320=0
2(x^2+6x-160)=0
2(x-10)(x+16)=0
so x=10 or x=-16 but can't be negative so x=10
plug that x in.
E=8*10-10 or 70
etc....
Source(s): im a math major - 10 years ago
Well then the sum of the angles would be equal to 360;
8x-10+x^2+10+2x+30+x^2+2x+10=360
2x^2+14x+40=360
2x^2+14x-320=0
2(x^2+7x-160)=0
-7+/- sqrt(49-4(1)(-160)=x
-7+/- sqrt(689)
-7+ 26.248 -7-26.248
x=19.24 x=-33.248
only 19.24 makes sense because -33 cannot be a measurement is a positive plane.
DISREGARD THE ANSWER BEFORE MINE ONE OF THE 2x's WAS NOT ADDED
