Triangle. Its longest side is 10 and another side had a length of 7.Area is 20. Whats the 3rd side? ?

Question

Can you tell me the formula?

And show your logic if you can..please...it would help i am confused

Answer 1

c=10, b=7, A=20

A=0.5*b*c*sinα

sinα=2A/(b*c)=2*20/(7*10)=4/7

a<c ==> α<90⁰ ==> cosα>0

cosα=√(1-sin²α)=√[1-(4/7)²]

cosα=√(33/49)=(1/7)√33

a²=b²+c²-2bc*cosα

a²=7²+10²-2*7*10*(/7)√33

a=√(149-20√33)~5.84 units of length

Answer 2

area of triangle = 1/2 length X height

20 = 1/2 * 10 * height

therefore height = 4

the height is perpendicular to the hypotenuse (the side with 10)

this created two right angle triangles with the height as one side in each.  we know one side of the triangle is 7 - this will be the hypotenuse to one of the new right angle triangles.

hyp ^2 = sideA^2 + side B^2

7^2 = 4^2 + side b^2

49 = 16 + side b^2

side b = square root of 33 = 5.75

Therefore the side which is 10 has 5.75 on one side of the height and therefore has 4.25 on the other side. in the second triangle, the hypotenuse will be the original 3rd side whose length we need to determine.  

3rd side ^2 = 4.25^2 + 4^2 = 18+16 = 34

therefore the third side = square root of 34 = 5.83

AM