Question

In fig 1,xp/py=xq/qz=3,if the area of triangle xyz is 32 cm^2,then find the area of the quadrilateral pyzq

Answer 1

Answer:

14 $cm^{2}$

Step-by-step explanation:

We have, $\frac{XP}{PY}$ = $\frac{XQ}{QZ}$

then PQ || YZ.

By using AA similar condition, we have ΔXPQ ≅ ΔXYZ, therefore ΔXPQ and ΔXYZ have correspondent angles.

Therefore:

$\frac{XP}{PY}$ = $\frac{XQ}{QZ}$ = $\frac{3}{1}$

$\frac{PY}{XP}$ = $\frac{QZ}{XQ}$ = $\frac{1}{3}$

$\frac{PY}{XP}$ + 1 = $\frac{QZ}{XQ}$ + 1 = $\frac{1}{3}$ + 1

$\frac{PY + XP}{XP}$ = $\frac{QZ + XQ}{XQ}$ = $\frac{4}{3}$

$\frac{XY}{XP}$ = $\frac{XZ}{XQ}$ = $\frac{4}{3}$

$\frac{XP}{XY}$ = $\frac{XQ}{XZ}$ = $\frac{3}{4}$

16 x 2 = 32

Area of ΔXYZ = 32 $cm^{2}$

Then:

Area of the qaudrilateral PYZQ is equal to area of ΔXYZ - area of ΔXPQ.

Then:

32 - 18 = 14 $cm^{2}$

Hence, the area is equal to 14 $cm^{2}$.

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.