Sir/Madam can you please solve question no. 20
Dear student,
(i)
Here Triangle BEC and parallelogram ABCD are lying on same base BC and in between same parallel pair AD and BC, so,
ar(Triangle BEC)=(1/2)ar(ABCD)
=>ar(Triangle BEC)={(1/2)x36} sq.cm=18 sq.cm.
(ii) Here according to the figure, since M and N are mid points of DC and AB respectively, so,
ar(parallelogram MCBN)=(1/2)ar(parallelogram ABCD)
=>ar(MCBN)=ar(Triangle BEC) [By (i)]
(i)
Here Triangle BEC and parallelogram ABCD are lying on same base BC and in between same parallel pair AD and BC, so,
ar(Triangle BEC)=(1/2)ar(ABCD)
=>ar(Triangle BEC)={(1/2)x36} sq.cm=18 sq.cm.
(ii) Here according to the figure, since M and N are mid points of DC and AB respectively, so,
ar(parallelogram MCBN)=(1/2)ar(parallelogram ABCD)
=>ar(MCBN)=ar(Triangle BEC) [By (i)]