Question

In ∆XYZ,XP is the bisector of angle X and XY=14cm, XZ=35cm.
A)what is XY:XZ?
B)What is the ratio of the length of YP and PZ?
C)What is the ratio of the areas ∆XPY and ∆XPZ?
D)If the area of ∆XYZ is 140cm^2 then what is the area of ∆XPZ?
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Answer 1

✪ANSWER✪

• $\boxed{\frac{XY}{XZ} = \frac{2}{5} }$

• $\boxed{\frac{YP}{PZ} = \frac{2}{5}}$

• $\boxed{\frac{ar(ΔXPY)}{ar(ΔXPZ)} = \frac{2}{5}}$

• $\boxed{ar(ΔXPZ) = 100 cm²}$

★EXPLAINATION★

☆GIVEN☆

• ∆XYZ with XP as the angle bisector of X and XY=14cm, XZ=35cm.

• ar(ΔXYZ) = 140 cm².

᯽THEOREM᯽

• Angle bisector of an angle of a triangle divides the opposite side in the ratio of the other two side.

✯CALCULATION✯

A)

☞︎︎︎$\frac{XY}{XZ} = \frac{14}{35}$

☞︎︎︎$\frac{XY}{XZ} = \frac{2}{5}$

B)

By the above theorem

☞︎︎︎$\frac{YP}{PZ} = \frac{XY}{XZ}$

☞︎︎︎$\frac{YP}{PZ} = \frac{2}{5}$

C)

☞︎︎︎$\frac{ar(ΔXPY)}{ar(ΔXPZ)} = \frac{\frac{1}{2}YP\times (altitude)}{\frac{1}{2}PZ\times (altitude)}$

☞︎︎︎$\frac{ar(ΔXPY)}{ar(ΔXPZ)} = \frac{YP}{PZ}$

☞︎︎︎$\frac{ar(ΔXPY)}{ar(ΔXPZ)} = \frac{2}{5}---(1)$

D)

Let 2x, 5x be the area of ΔXPY and ΔXPZ respectively using (1). Now

☞︎︎︎$ar(ΔXYZ) = 140 cm²$

☞︎︎︎$ar(ΔXPY) + ar(ΔXPZ) = 140$

☞︎︎︎$2x + 5x = 140$

☞︎︎︎$7x = 140$

☞︎︎︎$x = 20 cm²$

Now,

☞︎︎︎$ar(ΔXPZ) = 5x$

☞︎︎︎$ar(ΔXPZ) = 5\times 20$

☞︎︎︎$ar(ΔXPZ) = 100 cm²$