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.

Area of  ΔABC = 56m².

Explanation:

Since we have given that

point P∈ AB, such that AP:BP=1:3

Point M is the midpoint of segment CP.

Now, we have given that

Area of ΔBMP=21 m²

Since M is the mid-point of CP, so,

Area of ΔCMP= 21 m².

So, Area of ΔBCP = 21+21=42 m²

According to question,

So, area of ΔABC is given by

Hence, area of  ΔABC =56m².

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!