In △abc, cm is the median to ab and side bc is 12 cm long. there is a point p∈cm and a line ap intersecting bc at point q. find the lengths of segments cq and bq, if p divides cm into cp: pm=1: 2.

Step-by-step explanation:

This development is ugly, but it works.

Let area ΔABC = 6a. Then area ΔAMC = 3a, and area ΔAPC = a. Similarly, area ΔBMC = 3a, and area ΔBPC = a.

Let area ΔCPQ = x. Then ...

... area ΔACQ = area ΔAPC + area ΔCPQ

... area ΔACQ = a + x

Define k such that BQ : CQ = k : 1. Then ...

... area ΔBPQ + area ΔCPQ = area ΔBPC

... kx + x = a = (k +1)x

The division of BC into parts of ratio k : 1 means ...

... area ΔABQ = k × area ΔACQ

... area ΔABQ + area ΔACQ = area ΔABC

... k × area ΔACQ + area ΔACQ = area ΔABC

... (k +1)(x +a) = 6a . . . . . . . area ΔACQ = x+a

... (k +1)x +(k +1)a = 6a . . . . . distributive property

... a + (k +1)a = 6a

... k +2 = 6 . . . . . . . . . . . . . . . divide by a, simplify

... k = 4

Now, we know BQ : CQ = 4 : 1, so ...

... BQ = 4/5 × 12 cm = 9.6 cm

... CQ = 1/5 × 12 cm = 2.4 cm

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm