Find the intersection point between the 2 restraint equations:
Substitute back in to find y-value:
P is maximized at point (12,6)
Substitute back in to find y-value:
P is maximized at point (12,6)
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.
1 answer:
Answer:
- CQ = 2.4 cm
- BQ = 9.6 cm
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
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