Question

In the right triangle ?ABC, leg AC=6 cm and leg BC=8 cm. Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5. Find area of ?MNC.

Answer 1

Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.

To find : Area (ΔMNC)

Solution: In Δ ABC, right angled at C,

AC= 6 cm, BC= 8 cm

Using pythagoras theorem

AB² =AC²+ BC²

      =6²+8²

     = 36 + 64

→AB²  =100

→AB²  =10²

 →AB  =10

Also, AM:MN:NB=1:2.5:1.5

Then AM, MN, NB are k, 2.5 k, 1.5 k.

→2.5 k + k+1.5 k= 10

→ 5 k =10

Dividing both sides by 2, we get

→ k =2

MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm

As Δ ACB and ΔMNC are similar by SAS.

So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.

$\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}$

$\frac{Ar(ACB)}{Ar(MNC)}=4$

But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]

$\frac{24}{Ar(MNC)}=4$

→ Area(ΔMNC)=24÷4

→Area(ΔMNC)=6 cm²