SOLUTION: A right angled triange PQR is drawn such that QR is hypotenuse, QP is Base and RP is Perpendicular. Equilateral triangles are drawn on the sides of right angled triangle. Is area Q

Algebra ->  Pythagorean-theorem -> SOLUTION: A right angled triange PQR is drawn such that QR is hypotenuse, QP is Base and RP is Perpendicular. Equilateral triangles are drawn on the sides of right angled triangle. Is area Q      Log On


   

Question 91719: A right angled triange PQR is drawn such that QR is hypotenuse, QP is Base and RP is Perpendicular. Equilateral triangles are drawn on the sides of right angled triangle. Is area Q = Area P + Area R?
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
The equilateral triangle with side PQ is referred Tri_R (opposite to vertex R).
The equilateral triangle with side QR is referred Tri_P (opposite to vertex P).
The equilateral triangle with side RP is referred Tri_Q (opposite to vertex Q).

From Pythagoras' theorem in right angled triangle RPQ,
QR%5E2+=+PQ%5E2+%2B+RP%5E2 ________ (1)

Now, area of an equilateral triangle with side equal to a is A+=+sqrt%283%29a%5E2%2F4.
So, area of Tri_P is A_P+=+sqrt%283%29QR%5E2%2F4 _____ (2)
So, area of Tri_Q is A_Q+=+sqrt%283%29RP%5E2%2F4 _____ (3)
So, area of Tri_R is A_R+=+sqrt%283%29PQ%5E2%2F4 _____ (4)

So, A_Q+%2B+A_R+=+sqrt%283%29RP%5E2%2F4+%2B+sqrt%283%29PQ%5E2%2F4 [from (3) and (4)]
or A_Q+%2B+A_R+=+sqrt%283%29%28RP%5E2+%2B+PQ%5E2%29%2F4
or A_Q+%2B+A_R+=+sqrt%283%29QR%5E2%2F4 [from (1)]
or A_Q+%2B+A_R+=+A_P [from (2)]