In parallelogram ABCD, P is mid-point of AB.CP and BD intersect each other at point O. If area of triangle POB = 40 cm^2,
find : (i) OP :OC
(ii) areas of triangle BOC and PBC
(iii) areas of /\ ABC and parallelogram ABCD ? ?
Did you make your sketch?
Notice that ∆PBO is similar to ∆CDO
let PB = 1, then CD = 2 (remember AB was bisected)
then OP : OC = 1 : 2
let's fill in areas:
∆POB = 40 (given)
∆PBO = 40 , (same base, same height)
∆COD = 160 ( areas proportional to square of sides)
let ∆BCO = x
let ∆AOD = y
x + 160 = y + 80
y = x+80
now look at ∆PBC and ABD , they have the same height, but bases of 1 and 2
so x+40 = (1/2)(y+80)
2x + 80 = y+80
2x = y , ahhhh!
then 2x = x+80
x = 80 and y = 160
Now we know it all, take it from here.
To solve this problem, we can use the properties of parallelograms and triangles. Here's how we can find the answers to each part of the question:
(i) To find the ratio OP : OC, we need to find the lengths of OP and OC.
Since P is the midpoint of AB, we can divide AB into two equal segments, AP and PB. Since P is the midpoint, AP = PB.
Now, let's consider triangle OPB. We are given that the area of triangle POB is 40 cm^2. We know that the area of a triangle is given by the formula: Area = (base × height) / 2.
In this case, the base of triangle POB is PB, and the height is the length of the perpendicular from point O to the line PB.
Let's assume the length of OP = a and the length of OC = b. Then, the length of PB is 2a (since P is the midpoint), and the height is b.
So, the area of triangle POB can be expressed as: 40 = (2a × b) / 2. Simplifying this equation, we get: 40 = ab.
Now, we need to find the ratio OP : OC. Dividing both sides of the equation ab = 40 by b, we get: a = 40/b.
Therefore, the ratio OP : OC is a/b = 40/b : b = 40 : b^2.
(ii) To find the areas of triangles BOC and PBC, we can use the fact that triangles with the same base and height have equal areas.
Triangle BOC has the same base as triangle POB, which is PB = 2a, and the same height b. So, the area of triangle BOC is also 40 cm^2.
Triangle PBC has the same base as triangle POB, which is PB = 2a, and the same height b. So, the area of triangle PBC is also 40 cm^2.
(iii) To find the areas of triangle ABC and parallelogram ABCD, we can use the fact that the area of a triangle is half the area of a parallelogram with the same base and height.
Since triangle POB has an area of 40 cm^2, the parallelogram ABCD must have an area of 2 × 40 = 80 cm^2.
Since triangle ABC has the same base as triangle POB (which is PB = 2a), and the same height b, the area of triangle ABC is (2a × b) / 2 = a × b. Therefore, the area of triangle ABC is ab.
So, the areas of triangle ABC and parallelogram ABCD are ab and 80 cm^2, respectively.