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In rectangle ABCD, side DC is divided into five equal segmen
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20 Dec 2017, 15:39
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In rectangle ABCD, side DC is divided into five equal segments by points P,Q,R and S
Quantity A
Quantity B
The area of ∆MCQ
The area of ∆NSP
A. Quantity A is greater. B. Quantity B is greater. C. The two quantities are equal D. The relationship cannot be determined from the information given.
Re: In rectangle ABCD, side DC is divided into five equal segmen
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20 Dec 2017, 17:07
1
So let's say that the lenghth of Dc is 10 and then it follows that each point is 2. The base of triangle NPS is comprised of 4 points inclusive so the length must be 6. The triangle QMC is also comprised of 4 points which gives us the same base: 6. Now we need to find the height. We can draw a height for triangle NPS which would be equal to the wide of the rectangle. For triangle QMC we can draw a height from the line DC to the point M for example and it would have the same wide as the rectangle.
Therefore C of I am right
If nothing is indicated regarding point M and N lying on segment AB, can we assume according to the graphic that they are effectively on the segment ???
Re: In rectangle ABCD, side DC is divided into five equal segmen
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20 Dec 2017, 19:10
1
We know if two triangle stands on base of same length and enclosed within two parallel lines than triangles are equal.hnsere both triangles have same base length. So ans is C
Re: In rectangle ABCD, side DC is divided into five equal segmen
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30 Dec 2017, 11:07
1
A very simple question if one has a critical thinking and knows that formula of triangle's area. / 1/2 a*h /. Both triangle's heights are the same, cause they both derive from the same origin and are perpendicular to the length of the rectangle. Moreover, the interval of their segments is also the same -- 3. Therefore, one can logically conclude that both triangles have the same area.
gmatclubot
Re: In rectangle ABCD, side DC is divided into five equal segmen [#permalink]