Find the area of triangle BCD in the following figure

Question

I have tried the following:

Since CD is median, area of Triangle BCD is half the area of triangle BCQ.Area of BCQ is 1/2*BQ*CD' where CD' is the altitude. Now we have to find CD'. This is where I am stuck.

How to find out the height of Traingle BCQ in this? Do we apply the concept of triangles on same base? Posted here only after trying my best.

Answer 1

As in the comments, the answer must be (a).

However, to prove this if the answers had not been given, drop a perpendicular, $CE$, of height $h$ from $C$ onto $BQ$ where $BE=x$.

By similar triangles, $\frac{10}{h}=\frac{100}{100-x}$ and $\frac{40}{h}=\frac{100}{x}$.

Then $x=20,h=8$ and so the required area is $\frac{1}{2}\times50\times h=200$.

N.B. A neat general result for this set up is that the reciprocal of $h$ is the sum of the reciprocals of the two given heights i.e.

$$\frac{1}{h}=\frac{1}{10}+\frac{1}{40}.$$

Proof

Let the given heights be $L,M$ and the base $D$. By similar triangles: $$\frac{L}{h}=\frac{D}{D-x}\text { and }\frac{M}{h}=\frac{D}{x}.$$ Then $$\frac{D-x}{hD}=\frac{1}{L}\text { and }\frac{x}{hD}=\frac{1}{M}.$$ Now add the two LH terms: $$\frac{D-x}{hD}\text +\frac{x}{hD}=\frac{D}{hD}=\frac{1}{h}$$ and so $$\frac{1}{h}=\frac{1}{L}+\frac{1}{M}.$$

Answer 2

Let [.] denote areas. Then,

$$[ABC]+[CBQ]=[ABQ] = \frac12\cdot AB \cdot BQ = 500\tag 1$$

$$\frac{[ABC]}{[CBQ]}= \frac{AC}{CQ} = \frac{AB}{PQ} = \frac14\tag 2$$

Substitute (1) into (2) to obtain $[CBQ] = 400$. Thus, $[BDC] = \frac12 [CBQ] = \frac{400}2= 200$.