Question

AB = 7cm, AC = 8cm, BC = 6cm. CD is the bisector of C. ∠

a) What is AD : DB?

b) What is the ratio of areas of ΔADC and ΔBDC?

c) If area of ΔADC is 20cm², what is the area of ΔBDC?​

Answer 1

difficult I think so anyone could answer

Answer 2

Given:

AB = 7cm, AC = 8cm, BC = 6cm

CD is the bisector of ∠C

To find:

Ratio of sides AD and DB

Ratio of areas ΔADC and ΔBDC

Area of ΔBDC if area of ΔADC = 20cm²

Solution:

a) Since CD is the angle bisector of ∠C, it divides the angle in two equal measures. CD also divides the opposite side AB into two parts that are proportional to the other two sides BC and AC of ΔABC.

i.e., $\frac{CA}{CB}=\frac{AD}{DB}$

Length of sides AC and BC are given. Substituting the values, we get the ratio of sides AD and DB.

$\frac{8}{6}=\frac{AD}{DB}$

which can be further simplified as $\frac{4}{3}$

∴ AD:DB = 4:3

B) Area of a triangle = $\frac{1}{2}$ × base × height

Area of ΔADC = $\frac{1}{2}$ × AD × CD

Area of ΔBDC = $\frac{1}{2}$ × BD × CD

$\frac{a(ADC)}{a(BDC)}=\frac{\frac{1}{2}*AD*CD }{\frac{1}{2}*BD*CD }$

$\frac{a(ADC)}{a(BDC)}=\frac{AD}{BD}=\frac{4}{3}$

Since ratio of sides AD and BD is 4:3, then ratio of the areas of triangles is given by 4:3

c) Area of ΔADC = 20cm²

$\frac{a(ADC)}{a(BDC)}=\frac{AD}{BD}=\frac{4}{3}$   (proved)

$\frac{a(ADC)}{a(BDC)}=\frac{20}{a(BDC)}=\frac{4}{3}$

$\frac{20}{a(BDC)}=\frac{4}{3}$

$4*a(BDC) = 3*20$

$a(BDC) = \frac{60}{4}$

$a(BDC) = 15cm^{2}$

Ratio of sides AD and DB = AD:DB = 4:3

Ratio of areas ΔADC and ΔBDC = $\frac{a(ADC)}{a(BDC)}=\frac{4}{3}$

Area of ΔBDC = 15cm²