Question

In ΔABC, B-D-C and BD=7,BC=20 then find following ratios.
(1) A(ΔABD)/A(ΔADC)
(2) A(ΔABD)/A(ΔABC)
(3) A(ΔADC)/A(ΔABC)

Answer 1

a)1:1
b)1:2
c)1:2
will be the answer

Answer 2

Answer:

(1). $\frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}$

(2). $\frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}$

(3). $\frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}$

Step-by-step explanation:

From the given figure,

BDC is a straight line.

BD = 7 and BC = 20

So,

DC = BC - BD = 20 - 7 = 13

DC = 13

So,

Area of triangle ABD is,

$Area(ABD)=\frac{1}{2}\times BD \times h$

and,

Area of triangle ADC is,

$Area(ADC)=\frac{1}{2}\times DC \times h$

where, h is the height of the triangle which is same for all the triangles.

So,

$\frac{ar(ABD)}{ar(ADC)}=\frac{\frac{1}{2}\times BD\times h}{\frac{1}{2}\times DC\times h}=\frac{7\times h}{13 \times h}\\\frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}$

Therefore, the ratio of area of triangles ABD and ADC is given by,

$\frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}$

(2).

Area of triangle ABD is,

$Area(ABD)=\frac{1}{2}\times BD \times h$

and,

Area of triangle ABC is,

$Area(ABC)=\frac{1}{2}\times BC \times h$

So,

$\frac{ar(ABD)}{ar(ABC)}=\frac{\frac{1}{2}\times BD\times h}{\frac{1}{2}\times BC\times h}=\frac{7\times h}{20 \times h}\\\frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}$

Therefore, the ratio of area of triangles ABD and ABC is given by,

$\frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}$

(3).

Area of triangle ADC is,

$Area(ADC)=\frac{1}{2}\times DC \times h$

and,

Area of triangle ABC is,

$Area(ABC)=\frac{1}{2}\times BC \times h$

So,

$\frac{ar(ADC)}{ar(ABC)}=\frac{\frac{1}{2}\times DC\times h}{\frac{1}{2}\times BC\times h}=\frac{13\times h}{20 \times h}\\\frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}$

Therefore, the ratio of area of triangles ADC and ABC is given by,

$\frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}$