Question

In the given figure, ABCD is a parallelogram whose diagonals intersect at O. The areas of $\Delta AOD,\Delta DOC,\Delta COB$ and $\Delta BOA$ are $pc{m}^2,qc{m}^2,rc{m}^2$ and $sc{m}^2$ respectively. Then is the statement
$p=q\ne r=s$ true?

• A. True
• B. False

Answer 1

The correct option is B

False

(i) We know that diagonals of parallelogram bisect each other.

(ii) Therefore AO = OC

(iii) area $\Delta ADC$ = area $\Delta ABC=\frac{1}{2}$ area parallelogram ABCD ..... ( Also diagonal divides a parallelogram into two triangles of equal area )

(iv) area $\Delta AOD$ = area $\Delta COD=\frac{1}{2}$ area $\Delta ADC$ ...... (Median divides a triangle into two triangles of equal area)
$=\frac{1}{4}$ area parallelogram ABCD

(v) area $\Delta AOB$ = area $\Delta COB=\frac{1}{2}$ area $\Delta ABC$… (Median divides a triangle into two triangles of equal area)
$=\frac{1}{4}$ area parallelogram ABCD

(vi) area $\Delta AOD$ = area $\Delta COD$ = area $\Delta COB$= area $\Delta AOB$…… (from (iv) and (v))
p = q = r = s

Hence the statement is false