Question

In figure 12.35, the ratio of AD to DC is 3 to 2. If the area of ΔABC is 40 cm2, what is the area of ΔBDC?
A. 16 cm2
B. 24 cm2
C. 30 cm2
D. 36 cm2

Answer 1

Area of ΔBDC = 16 $cm^2$

Step-by-step explanation:

Given:

Area of ΔABC = 40 $cm^2$

AD : DC = 3 : 2

Let AD = 3x and DC = 2x        [1]

Now,

AC = AD + DC

AC = 3x + 2x            [from 1]

AC =5x

Let us draw BE ⊥ AC and also let BE = h

Area of ΔABC = $\frac{1}{2}\times base \times height$

$40 =\frac{1}{2}\times AC\times h$                                      [1]

We know that height of ΔABC and ΔABC are same.

So, Area of ΔABC = $\frac{1}{2}\times base \times height$

                              = $40 =\frac{1}{2}\times DC\times h$    [2]

Dividing Eq (2) from (1), we get

$\frac{ar(\triangle BCD)}{40}=\frac{\frac{1}{2}\times h \times DC }{\frac{1}{2}\times h \times AC }$

$\frac{ar(\triangle BCD)}{40}=\frac{DC}{AC}$

$\frac{ar(\triangle BCD)}{40}=\frac{2x}{5x}$

$ar(\triangle BCD)=\frac{2\times 40}{5}$

ar(ΔBDC) = $16 cm^2$

Answer 2

Answer:

let AD be 3cm and cd be 2cm....

are of triangle BAC =40cm^2=1/2*h*b

i.e....40=1/2*5*h

h=16cm

area of triangle BCD =1/2*2*h

=1*16=16cm^2