Question

N is the midpoint of AB, NM||BC and ar (△ABC) = 20cm2, then ar(△ANM) is equal to​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• In △ ABC,

• N is the midpoint of AB

• NM || BC

• ar ( △ ABC ) = 20 sq. cm

Since, In △ ABC,

⟼ N is the midpoint of AB and NM || BC

⟼ AB = 2AN ------------- [ 1 ]

Also, In △ ABC and △ ANM

⟼ ∠ ANM = ∠ABC [ Corresponding angles ]

⟼ ∠ AMN = ∠ ACB [ Corresponding angles ]

⟼ So, △ABC ~ △ANM

⟼ So, By Area Ratio Theorem,

Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

So,

$\rm\implies \:\dfrac{ar( \triangle \: ABC)}{ar( \triangle \:ANM)} = \dfrac{ {AB}^{2} }{ {AN}^{2} }$

$\rm\implies \:\dfrac{20}{ar( \triangle \:ANM)} = \dfrac{ {(2AN)}^{2} }{ {AN}^{2} }$

$\rm\implies \:\dfrac{20}{ar( \triangle \:ANM)} = \dfrac{ {4AN}^{2} }{ {AN}^{2} }$

$\rm\implies \:\dfrac{20}{ar( \triangle \:ANM)} = 4$

$\bf\implies \:ar( \triangle \:ANM) = 5 \: {cm}^{2}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Basic Proportionality Theorem :-

This theorem states that :- If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

Given,

Area of △ABC = 20$cm^{2}$,

N is the midpoint of AB and,

NM || BC

To Find,

Area of △ANM

Solution,

Since it is given that N is the midpoint of AB,

We can say that,  

     AB = 2AN        ⇒ (1)

Also, In △ABC and △ANM

     ∠ ANM = ∠ ABC          [Corresponding Angles]

     ∠ AMN = ∠ ACB          [Corresponding Angles]

Therefore, By a Similar Triangle Theorem,

     △ABC  ∼  △ANM

So, by Area Ratio Theorem, which states,

'The ratio of the area of two similar triangles is equal to the squares of corresponding sides.'

Hence,

     $\frac{ar(ABC) }{ar(ANM) } = \frac{AB^2}{AN^2}$

From equation (1), we can write,

     $\frac{20}{ar(ANM)} = \frac{(2AN)^2}{AN^2}$  

     $\frac{20}{ar(ANM)} = \frac{(4AN)}{AN^2}$

     $\frac{20}{ar(ANM)} = 4$

     $ar(ANM) = 5 cm^2$

Therefore, ar(△ANM) is equal to​ 5 cm2.