sonnco wrote:
Tadashi wrote:
What is the area of parallelogram \(ABCD\)?
1. \(AB = BC =CD = DA = 1\)
2. \(AC = BD = \sqrt{2}\)
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S_2 is not sufficient. By reducing side AB we can make the area of the parallelogram arbitrarily small.
From S_1 + S_2 it follows that ABCD is a well-defined square. Its area equals .
The correct answer is C.
I thought that the answer is B, since if AC and BD are equal, and since the shape is a parallelogram then it would be square making the area findable?
Can anyone explain?
I really don't understand how statement 2 is not sufficient.
If we know the parallelogram diagonals are equal the figure can be either a square or a rectangle. We are given the diagonal length of \sqrt{2}. If the diagonals are equal doesn't it make the the 4 angles inside the figure 90 degrees each?
Isosceles right triangle 1,1,\sqrt{2}
Area = 1?
If my reasoning is off please correct me or PM please. I am stumped.
Statement (2) says that the diagonals of parallelogram ABCD are equal, which implies that ABCD is a rectangle (square is just a special case of a rectangle). But knowing the length of the diagonals of a rectangle is not enough to calculate its area. So this statement is not sufficient.
Complete solution:
What is the area of parallelogram \(ABCD\)?(1) \(AB = BC =CD = DA = 1\) --> ABCD is a rhombus. Area of rhombus d1*d2/2 (where d1 and d2 are the lengths of the diagonals) or b*h (b is the length of the base, h is the altitude). Not sufficient.
(2) \(AC = BD = \sqrt{2}\) --> ABCD is a rectangle. Area of a rectangle L*W (length*width). Not sufficient.
(1)+(2) ABCD is rectangle and rhombus --> ABCD is square --> Area=1^2=1. Sufficient.
Answer: C.
Hope it's clear.