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∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
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03 Oct 2018, 22:51
03 Oct 2018, 22:51
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43% (02:13) correct
56% (01:55) wrong based on 101 sessions
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∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Re: ∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
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04 Oct 2018, 04:47
04 Oct 2018, 04:47
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Carcass wrote:
Attachment:
triangles.jpg
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
In triangle ABC we have ∠ ABC = 45 ° , ∠ CAB = 90 °
Therefore ∠ ACB = 45 °
So the triangle is isosceles triangle
Similarly in triangle DEF, we have DF = DE and ∠ FDE = 90 °
Therefore ∠ DEF = ∠ DFE = 45 °
Now we can write as
\(\frac{1}{2} AC^2 = 4* \frac{1}{2} DF^2\)
or \(AC = 2*DF\)
In triangle DEF, since it is 90-45-45 the sides are distributed in
\(1 : 1 : \sqrt2\)
i.e. \(DF = DE = 1\) and \(EF = \sqrt2\)
Therefore AC = 2 * 1 = 2
Now \(QTY A =\frac{AC}{EF} = \frac{2}{{\sqrt2}}\) =\(2 *\frac{{\sqrt2}}{{\sqrt2 *\sqrt2}}\) = \(\sqrt2\)
Hence QTY A = QTY B
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∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
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30 Mar 2021, 07:47
30 Mar 2021, 07:47
1
Carcass wrote:
Attachment:
triangles.jpg
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Triangle ABC and DEF, both are Isosceles Right Triangles with ratio of their sides \(1 : 1 : \sqrt{2}\)
Also, both the triangles are Similar
i.e. \(\frac{AB}{ED} = \frac{AC}{DF} = \frac{BC}{EF} = \sqrt{\frac{Ar. (ABC)}{ Ar. (DEF)}
}\)
In Triangle ABC:
Let AB = AC = \(Y\)
So, BC = \(\sqrt{2}Y\)
Area = \(\frac{Y^2}{2}\)
In Triangle DEF:
Let ED = DF = \(X\)
So, EF = \(\sqrt{2}X\)
Area = \(\frac{X^2}{2}\)
Given, \(\frac{Y^2}{2} = 4(\frac{X^2}{2})\)
\(Y = 2X\)
Col. A: \(\frac{Y}{\sqrt{2}X}\)
Col. B: \(\sqrt{2}\)
Col. A: \(\frac{2X}{\sqrt{2}X}\)
Col. B: \(\sqrt{2}\)
Col. A: \(\sqrt{2}\)
Col. B: \(\sqrt{2}\)
Hence, option C
