Question

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

Answer 1

Step-by-step explanation:

Consider the triangle ABC, M and N are the midpoints of the sides AB and AC respectively.

Let AP be the altitude from the vertex A to the side BC.

The line joining the midpoints of AB and AC meet the altitude AP at O.

Required to prove that AO = OP.

So, we have to prove that AO = ½AP

In triangle ABC.

M and N are the midpoints of the sides AB and AC.

So, we have

AM = ½AB; AN = ½ AC

According to the Midpoint Theorem,

MN = ½ BC

So, ΔABC ~ ΔAMN

⇒ Area(AMN) / Area(ABC) = MN2 / BC2

= ( BC/2 )2 / BC2

= 1/4

= AO2 / AP2

So,

AO/AP = 1/2

AO = ½ AP

Hence, AO = OP.

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Answer 2

Answer:

Area of the △ANM  = 5cm²

Step-by-step explanation:

Given,

N is the midpoint of AB

NM||BC

Area of △ABC = 20cm²

To find,

The Area of the △ANM

Solution:

The converse of mid point theorem

If a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, bisects the third side”.

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

a, b, and c are the sides of the triangle, and s = $\frac{a+b+c}{2}$

Solution:

Let the lengths of the sides of the △ABC, be AB = c, BC= a and AC = b respectively.

Since the area of the △ABC = 20cm², we have

$\sqrt{s(s-a)(s-b)(s-c)}$ = 20cm² -------------(1)

Since N is the midpoint of AB we have,

AN = $\frac{c}{2}$

Since  N is the midpoint of AB, NM||BC, by the converse of the mid-point theorem we have

M is the midpoint of AC

AM = $\frac{b}{2}$

Also, we have MN= $\frac{a}{2}$

Hence the sides of the triangle ANM are $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$

Area of the △ANM = $\sqrt{s_1(s_1-\frac{a}{2} )(s_1-\frac{b}{2})(s_1-\frac{c}{2})}$, where

s₁=  $\frac{\frac{a}{2} + \frac{b}{2} + \frac{c}{2}}{2}$ = $\frac{a+b+c}{4}$ = $\frac{s}{2}$

∴s₁ = $\frac{s}{2}$

Area of the △ANM = $\sqrt{\frac{s}{2} (\frac{s}{2}-\frac{a}{2} )(\frac{s}{2}- \frac{b}{2})(\frac{s}{2}-\frac{c}{2})}$

=  $\sqrt{\frac{s}{16} (s-a)(s-b)(s-c)}$

= $\frac{1}{4} \sqrt{s(s-a)(s-b)(s-c)}$

From equation (1) we have

= $\frac{1}{4}$Area of △ABC

=  $\frac{1}{4}$×20

= 5cm²

Area of the △ANM  = 5cm²

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