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

ABC = leg x leg / 2

ABC = 6 x 8 / 2

ABC = 24

MNC = (MN / (AM+MN+NB)) x ABC

MNC = (2.5 / (1 + 2.5 + 1.5) )x 24

MNC = (1/2) x 24

MNC = 12

really hope this helps :)

Answer 2

Answer: Area of
`\triangle MNC= 12` centimeter square .

Explanation:

According to the question,
`\triangle ABC` is a right angled triangle with edges AB=8 cm and AC=6 cm

And, M and N are two points in side AB Such that, AM:MN:NB=1:2.5:1.5

So, let AM=1x MN=2.5x and NB=1.5x

Since, AM+MN+NB=AB ⇒1x+2.5x+1.5x=5x

But AB=8 cm ⇒5x=8⇒x=1.6

Therefore, MN=4 cm

Now, In
`\triangle MNC` height= 6 cm ( shown in diagram)

while, base=4 cm

Thus, area of
`\triangle MNC` = 1/2×6×4=12
`cm^2` ( because area of a triangle = 1/2×base×height)