Am is a median in △abc (m∈ bc ). a line drawn through point m intersects ab at its midpoint p. find areas of △apc and △pmc, if area of apm=35m2.

Given: AM is a median in ΔABC where M∈BC.A line drawn through point M intersects AB at its midpoint p.Also, area of ΔAPM = 35 m².

To Find: Area of △APC and △PMC

Solution: In ΔAMB

PM is median.

As you must keep in mind, Median of triangle divides it into triangles of equal area.

→Area(ΔAPM) = Area(BPM)= 35 m²

So, Area ( ΔABM)= Area(ΔAPM) + Area(BPM)= 35+35=70 m²

Now, Area (ΔABM) = Area(ΔAMC)=70 m²→[Median of triangle divides it into triangles of equal area.]

→Area (ΔABC)= Area (ΔABM) + Area(ΔAMC)=70+70=140 m²

Area(Trpzm BPMC) = Area(ΔABC) - Area(ΔBPM)

                                             = 140 - 35= 105 m²

PM ║AC→Line joining mid point of two sides of triangle is parallel to third side.

Area(ΔAMC) =Area(ΔPCA)= 70 m²→[Triangle on the same base and between the same parallels are equal in area.]

Area(ΔPMC)= Area(trpzmPMAC) - Area(ΔPAC)=135 - 70=35 m²

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

answer: 177/8

step-by-step explanation: