MATHS dilemma !?

Question

1. A Point O lies inside the parallelogram ABCD. Show that the sum of the ares of the triangle AOB and COD is equal to the sum of the ares of the triangles DOA and COB.

2. G is the point of intersection of the medians of the triangle ABC. Show that the triangles CGB, ABG and AGC are all equal in area.

Please ... 10 points !!!!!!!!! Please help

Answer 1

I do not know how much you know. Do you know what are similar triangles?
question 2: Let T be midpoint of BC,
let GM be perpendicular to BC and AS perpendicular to BC.
Triangle GTM is similar to triangle ATS(why?)
hence GT/AT= GM/AS = 1/3 (a property of centroid). ie GM = 1/3AS
This means that area Of GBC = 1/2 BC*GM
1/2(BC)(1/3AS)= 1/3(1/2 BC*AS)=1/3 area of ABC. Similarly Area of AGC = 1/3 area of ABC and area of AGB = 1/3 area of ABC.
hence they are equal to each other.

Question 1:
Do you know that the area of a parallelogram = b* corresponding height?
LET M, N, P and Q be foot of perpendicular from O to AD, DC, B C and AB respectively.
we know AB=DC and AD=BC
Area AOB + BOC = 1/2 AD * OM + 1/2 BC* OP = 1/2 AD * OM + 1/2 AD * OP =
1/2AD(OM + OP) = 1/2 area of parallelogram ABCD. similarly Area AOB + area DOC = 1/2 area of parallelogram. Hence they two sums are equal.

Answer 2

1.)
Area (AOB) = ½ (AB) h'
Area (DOC) = ½ (DC) h''

AB = DC since the figure is a parallelogram and opposite sides of a parallelogram are equal in length.

h' + h'' = h, the constant distance between AB and DC and hence the height of the parallelogram with respect to those two sides.

Area (AOB) + Area (DOC) = ½ (AB) h' + ½ (DC) h''

Since AB = DC:

Area (AOB) + Area (DOC) = ½ (AB) h' + ½ (AB) h''
½ (AB) h' + ½ (AB) h'' = ½ (AB)(h' + h'')
½ (AB)(h' + h'') = ½ (AB)(h)

Notice that the right side of the last equation happens to be the area formula for the entire parallelogram.

Now, let's look at the other two triangles, whose bases are the other equal pair of sides of the figure. We'll let k' and k'' represent the heights of the two triangles, and k be the sum of their heights, and hence the height of the parallelogram with respect to those two sides.

Area (DOA) = ½ (DA) k'
Area (COB) = ½ (CB) k''

DA = CB for the same reason the other pair of sides were equal in length.

k' + k'' = k for the same reason h' + h'' = h.

Area (DOA) + Area (COB) = ½ (DA) k' + ½ (CB) k''

Since DA = CB:

Area (DOA) + Area (COB) = ½ (DA) k' + ½ (DA) k''
½ (DA) k' + ½ (DA) k'' = ½ (DA)(k' + k'')
½ (DA)(k' + k'') = ½ (DA)(k)

The right side of the last equation gives the area of the parallelogram with respect to the other pair of sides. But, since the parallelogram has only one area, the following must be the case:

Area (AOB) + Area (DOC) = Area (DOA) + Area (COB).

So the sums of the two pairs of triangles are equal.

2.)
This problem has a very elegant solution.

The point of intersection of the medians of a triangle is called the centroid. The centroid divides the medians in the ratio of 2/1. Now, look at each side of triangle ABC individually. If we extend a median from A, B, and C to the midpoints of the opposite sides, and label those points of intersection with the sides A', B' and C', we can create two similar triangles on each side by dropping a perpendicular from A, B and C to the opposite side and by dropping a perpendicular from G to each side. The perpendiculars from A, B and C intersect the opposite sides at A", B" and C". The perpendiculars dropped from G intersect the sides at A'", B"' and C"'. Looking specifically at side BC, with median dropped from A, the two similar triangles formed are GA'A"' and AA'A". Now GA' = ½ AG or 2 GA' = AG, and since AA' = AG + GA', then AA' = 2 GA' + GA' = 3GA'. That means that AA' / GA' = 3GA' / GA' = 3 / 1. Since triangles GA'A"' and AA'B" are similar, all their linear pieces are in the same ratio as their sides, so their altitudes are in the same ratio, 3 / 1. That means GA"', which is also the altitude of triangle CGB, is 1/3 the height of the altitude of triangle ABC, which happens to be AA". Now, since triangles ABC and CGB share the same base, then the area of triangle CGB must be 1/3 that of triangle ABC. Here's proof:

Area (ABC) = ½ b h = ½ (BC)(AA")
Area (CGB) = ½ b h = ½ (BC)(GA"') = ½ (BC)[(1/3)(AA")]
½ (BC)[(1/3)(AA")] = (1/3) [½ (BC)(AA")] = 1/3 [Area (ABC)]

This same argument can be used for the other two sides. And, since the area of the larger triangle remains constant regardless of which side is chosen to calculate that area, the areas of each of the smaller triangles, regardless of which is chosen, will be 1/3 the area of the larger triangle. Therefore, triangles CGB, ABG and AGC are all equal in area.

Answer 3

http://education.yahoo.com/homework_help/math_help/geometry;_ylt=AiOPoX_2yF9yHY74rOsgn2TWZcgF


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