looks like the same problem here:
http://www.jiskha.com/display.cgi?id=1493855490
Given: KLMN is a trapezoid, m∠K = 90°, LM = 9, KN = 18, MN = 15 Find: Area of KLMN
LM and KN are parallel
4 answers
No, but the answer is not twelve for this, so can you please solve it
The answer is 162 square units. I got it right and I am in RSM. You can do this by:
1) You can find that KH is 9 because there is a rectangle.
2)You know that HN is 9 because KN = 18, and if KH is 9, then HN is 9. 18-9 = 9
3)Use Pythagorean theorem to find the hight of the triangle.
4)9^2 = 81, 15^2= 225, so 225-81=144, and the root of 144 is 12.
5) Now you know the height, so you can find the area of the rectangle.
6) 9*12 = 108 square units for the rectangle.
7) Find the area of the triangle. (9*12)÷2. 108÷2 = 54.
8) 54+108 = 162 square units.
1) You can find that KH is 9 because there is a rectangle.
2)You know that HN is 9 because KN = 18, and if KH is 9, then HN is 9. 18-9 = 9
3)Use Pythagorean theorem to find the hight of the triangle.
4)9^2 = 81, 15^2= 225, so 225-81=144, and the root of 144 is 12.
5) Now you know the height, so you can find the area of the rectangle.
6) 9*12 = 108 square units for the rectangle.
7) Find the area of the triangle. (9*12)÷2. 108÷2 = 54.
8) 54+108 = 162 square units.
KN =KS +SNKS =10Δ MSN is isosceles, because m∠MSN=90⁰, m∠N=45⁰, so m∠NMS=45⁰,
and MS=SN=10.KN =KS +SN =10+10=20
KN=20Area of a trapezoid = (1/2)*(b1+b2)*h
b1=10, b2=20, h=10Area of a trapezoid = (1/2)*(10+20)*10=150Area of trapezoid KLMN =150KN=20
and MS=SN=10.KN =KS +SN =10+10=20
KN=20Area of a trapezoid = (1/2)*(b1+b2)*h
b1=10, b2=20, h=10Area of a trapezoid = (1/2)*(10+20)*10=150Area of trapezoid KLMN =150KN=20