In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).


Answer :

Let us draw AF, perpendicular to BC such that AF is the height of \(\triangle \)ABD, \(\triangle \)ADE and \(\triangle \)AEC



\(\therefore \) area of triangle =( \(\frac{1}{2} \)) × base ×Height
\(\therefore \) Area of \(\triangle \) ABD= (\(\frac{1}{2} \) ) × BD × AF
\(\therefore \) Area of \(\triangle \) ADE = (\(\frac{1}{2} \) ) × DE × AF
\(\therefore \) Area of \(\triangle \) AEC = (\(\frac{1}{2} \) ) × EC × AF
Since , BD= DE= EC (Given )
\(\therefore \) Area of \(\triangle \) ABD= Area of \(\triangle \) ADE= Area of \(\triangle \) AEC

NCERT solutions of related questions for Areas of parallelograms and triangles

NCERT solutions of related chapters class 9 maths

NCERT solutions of related chapters class 9 science