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.
[tex]\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}[/tex]
[tex]\frac{Ar(ACB)}{Ar(MNC)}=4[/tex]
But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]
[tex]\frac{24}{Ar(MNC)}=4[/tex]
→ Area(ΔMNC)=24÷4
→Area(ΔMNC)=6 cm²
