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.

Our approach will be to find the area of ∆ABC, then determine the fraction of that comprising ∆MNC. Multiplying the area of ∆ABC by that fraction will give the area of ∆MNC.

The area of a right triangle is half the product of the leg lengths. Then the area of ∆ABC is ...

... A = (1/2)(6 cm)(8 cm) = 24 cm²

The segment AB is divided into 3 parts having the ratios ...

... 1 : 2.5 : 1.5

where segment MN is the segment corresponding to 2.5 out of the total of 5 ratio units. That is, the length of MN is half the length of AB.

If we consider the "base" of ∆ABC to be segment AB, then the altitude of that triangle is the perpendicular distance from C to AB. We don't need to know what that altitude is. We just need to know that ∆MNC has the same altitude but half the base length of ∆ABC. Thus ∆MNC will have half the area of ∆ABC.

Area ∆MNC = (1/2) × (Area ∆ABC) = (1/2)(24 cm²)

Area ∆MNC = 12 cm²

i can't solve unless i know what the line is

step-by-step explanation:

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linear pair postulates