Find the area of parallelogram if side AB = 10 cm,BC=15 cm,Diagonal =20 cm.
4 answers
15/2 √279
Steve, what was that formula?
I don't know how Steve got those numbers, it comes out to appr 125.3
I made a sketch.
I extended BC to E so the DE is an altitude and triangle DCE is right-angled.
I let CE = x and DE = h
x^2 + h^2 = 100, and
(x+15)^2 + h^2 = 400
subtract them
(x+5)^2 - x^2 = 300
30x + 225 = 300
x = 75/30 = 5/2
then h^2 = 100 - 25/4 = 375/4
h = 5√15/2
area = base x height
= 15(5√15)/2 = 75√15/2 cm^2 or appr 145.2 cm^2
or
look at triangle BCD, it is half the area of the parallogram
by the cosine law:
400 = 225 + 100 - 2(10(15)cosC
cosC = (225+100-400)/300 = -1/4
then sinC = √ (1 - 1/16) = √15/4
area of triangle BCD
= (1/2)(10)(15)sinC
= 75√15/4
area of ||gram = 2(75√15/4) = 75√15/2
same as above
I made a sketch.
I extended BC to E so the DE is an altitude and triangle DCE is right-angled.
I let CE = x and DE = h
x^2 + h^2 = 100, and
(x+15)^2 + h^2 = 400
subtract them
(x+5)^2 - x^2 = 300
30x + 225 = 300
x = 75/30 = 5/2
then h^2 = 100 - 25/4 = 375/4
h = 5√15/2
area = base x height
= 15(5√15)/2 = 75√15/2 cm^2 or appr 145.2 cm^2
or
look at triangle BCD, it is half the area of the parallogram
by the cosine law:
400 = 225 + 100 - 2(10(15)cosC
cosC = (225+100-400)/300 = -1/4
then sinC = √ (1 - 1/16) = √15/4
area of triangle BCD
= (1/2)(10)(15)sinC
= 75√15/4
area of ||gram = 2(75√15/4) = 75√15/2
same as above
Hmmm. Better go with Reiny. I thought my answer was a bit odd.