Question

In figure,the ratio of AD to DC is 3 to 2. if the area of triangle ABC is 40 cm square .what is the area of triangle BDC

Answer 1

area of triangle ABC= 1/2 base x height ==  40

base= AD+DC AD/DC= 3/2        AD= 3DC/2

80= (3DC/2+DC)Height

DC X Helight= 32

area of triangle BDC= 1/2 Dc xheight= 32/2=16cmsquare

Answer 2

Answer:

16 cm²        

Step-by-step explanation:

Since the ratio of AD to DC is 3:2

Also, the Area of ΔABC = 40 cm²

We need to find the area of ΔBDC.

As we can see that

DC =$\dfrac{2}{5}AC$  

So, it implies that

Area of ΔBDC $\dfrac{2}{5}\times \text{Area of ABC}$

So, Area of ΔBDC = $\dfrac{2}{5}\times 40=2\times 8=16\ cm^2$

Hence, Area of ΔBDC = 16 cm².