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Slougi

New member
I know this one is easy, it is just somehow escaping me atm (I blame summer holidays!). If someone could help me I'd be grateful.

In the parallelogram OABC, OA = a, OC = c, and M is a point on the line segment AB such that AM : MP = 2 : 3. OM meets AC at the point P.

(a) Find an expression for OP in terms of a and c.
(b) Find the value of the ratio OP : PM
(c) Show that AP : PC = 2 : 5
(d) Find the ratio of the areas of the triangle AMP and the quadrangle PMBC.

I just want help with (a), since (b) and (c) can be easily obtained using the result from that and (d) is easy.
 
R

ra5555

N64 Newbie
Slougi said:
I know this one is easy, it is just somehow escaping me atm (I blame summer holidays!). If someone could help me I'd be grateful.

In the parallelogram OABC, OA = a, OC = c, and M is a point on the line segment AB such that AM : MP = 2 : 3. OM meets AC at the point P.

(a) Find an expression for OP in terms of a and c.
(b) Find the value of the ratio OP : PM
(c) Show that AP : PC = 2 : 5
(d) Find the ratio of the areas of the triangle AMP and the quadrangle PMBC.

I just want help with (a), since (b) and (c) can be easily obtained using the result from that and (d) is easy.
Click to expand...

I believe you mean AM:MB= 2:3
anyway can't think about how to do it right now, its been a year since I have taken geometry and trig in high school. GL though
 
The Khan Artist

The Khan Artist

Warrior for God
Slougi said:
In the parallelogram OABC, OA = a, OC = c, and M is a point on the line segment AB such that AM : MP = 2 : 3. OM meets AC at the point P.
Click to expand...

Whoa there... what is P? A diagram of this parallelogram would be most helpful.
 
OP
S

Slougi

New member
This one is actually very easy...


Triangle[MPA] ~ Triangle[CPO] (2 equal angles)
MA/MP = CO/OP = 2/3 => OP = 3c/2

Similarly you can solve all other questions.
 
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