Question

In fig. Xp/py=xq/qz=3,if the area of triangle xyz is 32cm square ,then find the area of the quadrilateral pyzq.

Answer 1

Xp/py=xq/qz=3,------------> A

pq||yz

Taking reciprocals of eq A

py/xp=qz/xq=1/3

py/xp+1=qz/xq+1=1/3+1

py+xp/xp=qz+xq/xq=4/3

xy/xp=xq/xz=4/3 --------B

again taking reciprocals of above term,

xp/xy=xz/xq=3/4

triangle(xpq)/triangle(xyz)=(xp/xy)^2=9/16

trianlge xpq=9/16*triangle (xyz)

triangle (xyz)=32cm

triangle(xpq)=18 square cm

Quad(pyzq)=triangle(xyz)-triangle(xpq)

=32-18=14 square cm. Answer

Answer 2

Answer:

Let x/a = y/b = z/c = k, [By k method]

x = ak, y= bk and z=ck

L.H.S. = a3k3/a2 + b3k3/b2 + c3k3/c2 > k3[a + b + c]

R.H.S. = [ak + bk + ck]3/[a + b + c)2 → k3[a + b + c]3/[a + b + c)2

= k3(a + b + c)

L.H.S. = R.H.S. =

Hence proved.