Plz 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.

56 m²

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

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

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.