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Re: What is the area of rectangular region R?
[#permalink]
21 Apr 2020, 06:06
21 Apr 2020, 06:06
Bunuel wrote:
unceldolan wrote:
Bunuel,
isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...
isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...
This is a common trap.
Knowing that the hypotenuse equals to 5 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: if \(x^2+y^2=5^2\) DOES NOT mean that \(x=3\) and \(y=4\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=5^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=3\) and \(y=5\).
For example: \(x=1\) and \(y=\sqrt{24}\) or \(x=2\) and \(y=\sqrt{21}\)...
Hope this helps.
Hi Bunuel - i thought the answer might be A because
-- the diagonal breaks up the 90 degrees into 2 equal halves (45 degrees each)
hence i thought the diagonal breaks up the rectangle into 2 triangles that are 45-45-90 triangles
hence, i thought this was a special triangle with the ratio of the sides : 1:1 : square root 2
hence you can find the length and breadth
given area is length * breadth = you can then find the area of the rectangle
hence i marked (A) as sufficient
Could you let me know where is the break in my logic
Re: What is the area of rectangular region R?
[#permalink]
21 Apr 2020, 07:06
21 Apr 2020, 07:06
Expert Reply
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
unceldolan wrote:
Bunuel,
isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...
isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...
This is a common trap.
Knowing that the hypotenuse equals to 5 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: if \(x^2+y^2=5^2\) DOES NOT mean that \(x=3\) and \(y=4\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=5^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=3\) and \(y=5\).
For example: \(x=1\) and \(y=\sqrt{24}\) or \(x=2\) and \(y=\sqrt{21}\)...
Hope this helps.
Hi Bunuel - i thought the answer might be A because
-- the diagonal breaks up the 90 degrees into 2 equal halves (45 degrees each)
hence i thought the diagonal breaks up the rectangle into 2 triangles that are 45-45-90 triangles
hence, i thought this was a special triangle with the ratio of the sides : 1:1 : square root 2
hence you can find the length and breadth
given area is length * breadth = you can then find the area of the rectangle
hence i marked (A) as sufficient
Could you let me know where is the break in my logic
The diagonal of e rectangle bisects the angle ONLY IF the rectangle is a square. If a rectangle is NOT a square, then the diagonal can create an angle with an adjacent side, from 0 to 90, not inclusive.
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Re: What is the area of rectangular region R?
[#permalink]
03 Jun 2020, 04:40
03 Jun 2020, 04:40
Bunuel wrote:
Rajeet123 wrote:
Bunuel wrote:
Rajeet123 wrote:
I had read a formula somewhere which said if any quadrilateral has diagonals intersecting at 90 degrees, then their area can be found out by 1/2 * d1 * d2.
If we use this formula, then A alone is sufficient to answer the question.
Where am I going wrong Bunuel ? Is that formula right?
If we use this formula, then A alone is sufficient to answer the question.
Where am I going wrong Bunuel ? Is that formula right?
Do we know that the diagonals intersect at 90 degrees from (1)?
You're right, "rectangular" doesn't necessarily mean a rectangle. Missed that.
No, that's not the point. Rectangular region does mean that the figure is a rectangle. But the diagonals of a rectangle do not always intersect at 90 degrees. This happnes only when a rectangle is a square.
If the two diagonal are the same does not mean the rectangular is a square?
