Question

There are ‘p’ number of isosceles triangle with integer sides, whose sum of the equal sides is
14. There is another triangle ABC with medians AD and BE which are perpendicular to each
other having lengths 15 cm and 18cm, respectively. The area of this triangle ABC (in sq. cm)
is 'q'. The value of
q/p+3
is :​

Answer 1

AD is the median

In triangle AXB & AXC, by pythagoras theorem,

ABP = AX2 + BX2

and AC2 = AX+ CX2

Adding (1) & (2)

AB2 + AC2 = 2AX2 + BX? + CX?

but BX = BD - DX and CX= DX +

DC = BD + DX

. BX? + CX? = (BD DX)? +

(DX + BD)

BX2 + CX2 (BD2 + DX? - 2BD x

DX) + (DX² + BD2 – 2BD x DX)

BX? + CX? = 2BD? + 2 DX?

Now, AB? + AC2 CX? 2AX? + BX? +

AB? + AC2 = 2AX2 + 2BD2 + 2DX2

AB? + AC? = 2(AX? + DX?) + 2BD? :: AX? + DX? = AD? =

AB? + AC? = 2AD? + 2BD² =

AB? + AC? = 2AD? = BC2 2 + (3)

similarly, BC2 + ABP = 2BE2 +

(4)

AC? + BC2 2CF2 + AB2 2

Adding (3),(4) and (5)

2(AB? + BC? + AC?) = 2(AD? +

BE2 1 +CF2)+(ABP + BC2 + AC?) 2

multiplying throughout by 2,

4(AB? + BC? + AC?) = 4(AD? + = BE? + CF2) + (AB? + BC? + AC?)

3(AB? + BC? + AC?) = 4(AD? + BE+ CF2)

Hence Proved.