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.

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.

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

→ Area(ΔMNC)=24÷4

→Area(ΔMNC)=6 cm²

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)