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In triangle ABC, M is the midpoint of BC and N is the mid point of AM. BN when extended intersect AC at D. If area of triangle ABC is 20 sq. units then what is the area of \(\triangle AND\)?

A) 1.67 sq.units
B) 1.5 sq. units
C) 2 sq.units
D) 3 sq. units

Correct Answer:

A) 1.67 sq.units

Description for Correct answer:

\( \Large ar\triangle AMC=\frac{1}{2}ar\triangle ABC\)\(\because \) M is the the midpoint

Draw ME || BD

\( \Large In\triangle BCD\ and\ \triangle CEM \)

\( \because ME||BD \)

\( \Large \Rightarrow \triangle BCD \sim \triangle CEM \)

and

\(\because\) M is the mid-point.

\(\therefore DE = EC..........(i)\)

\( \Large In\triangle AME\ and\ \triangle AND \)

\(\because\) ME || ND

\( \Large \Rightarrow \triangle AME \sim \triangle AND \)

and \(\because\) N is the mid point.

\( \Large \Rightarrow AD=DE.......(ii) \)

and\( \Large \frac{ar\triangle AND}{ar\triangle AME}=\frac{AD^{2}}{AE^{2}} \)

\( \Large = \left(\frac{1}{2}\right)^{2}=\frac{1}{4}.........(iii) \)

From (i) and (ii)

\( \Large AD=DE=EC \)

\(\Rightarrow\) ME bisects Ac in the ratio 2 : 1

\(\Rightarrow \) ME also bisects the area of \(\triangle AMC\) in the ratio 2 : 1

\(\Rightarrow\) are of \( \Large \triangle AME=\frac{10}{3} \times 2 \)

\( \Large =\frac{20}{3} \)sq. units

From (iii)

\( \Large Area\triangle AND=\frac{20}{3} \times \frac{1}{4}=\frac{5}{3} \)

\( \Large =1.67 \)sq. units

Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry

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