The perimeter of triangle ABC is 120. LIne segment BD bisects <B. Line segment AD= 15, and line segment DC= 21. FInd AB and BC.

Question

Thanks.

Answer 1

This is a very tough question for grade 8, I am totally impressed.

make your sketch, let
AB = a
BC = b
BD = x
Angle B = 2Ø , so that angle ABD = Ø and angle BCD = Ø

I am going to use the fact that if we have triangle with sides a and b with Ø the angle between them
the area = (1/2)ab sinØ

area of triangle ABD= (1/2)ax sinØ
area of triangle BCD=(1/2)bx sinØ

But if we consider the bases along AC , the both have the same measured from B
so they are in the ratio of 15:21 or 5:7

then:
(1/2)axsinØ/( (1/2)bxsinØ ) = 5/7
canceling reduces this to
a/b = 5/7
5b = 7a
b = 7a/5 = 1.4a ---- ******

now let's use the perimeter:
a + b + 15+21 = 120
a+b =84 -----******

sub in b = 1.4a into the above equation
a+b = 84
a + 1.4a = 84
2.4a = 84
a = 84/2.4 = 35
then b = 1.4(35) = 49

great question !!!!

Answer 2

or, using the angle bisector theorem,

AD/DC = AB/BC
so, as Reiny calculated,

AB/BC = 5/7
AD+BC = 84
5x+12x=84
x=7

so, AB = 35
BC = 49

Answer 3

oops
5x+7x=84
x=7

Answer 4

Thanks.