Question

PLZ HELP 80 POINTS

In △ABC, point P∈
AB
is so that AP:BP=1:3 and point M is the midpoint of segment
CP
. Find the area of △ABC if the area of △BMP is equal to 21m2.

Answer 1

56 m²

Step-by-step explanation:

Since M is the midpoint of PC, area of PBC = 2(21) = 42 m²

(Same height, double base)

PC is the height of the triangle ABC

Base AB:

AB : PB

4 : 3

Areas will be in the same ratio

Area of ABC:

(4/3) × 42

56 m²

Answer 2

56 m²

Step-by-step explanation:

The altitude from point B to segment CP is the same for ∆BMP as for ∆BMC. Since both have the same base length (MP = MC), both have the same area, 21 m². Hence the area of ∆CPB is (21+21) m² = 42 m².

The altitude from point C to segment AB is the same for ∆CPA as for ∆CPB, so the areas of those triangles will have the same proportion as the base segments AP and BP. That is, ...

... AACPA : ACPB = PA : PB = 1 : 3

The ratio of ACPB to the total is then ...

... ACPB : (ACPA +ACPB) = 3 : (1+3) = 3 : 4

The area of ∆ABC is the total of the areas of the smaller triangles CPA and CPB, so we have

... ACPB : AABC = 3 : 4

... AABC/42 m² = 4/3 . . . . . rearranging slightly and substituting for ACPB

... AABC = (42 m²)×(4/3) . . . . multiply by the denominator

... AABC = 56 m²