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11 Dec 2017, 23:06
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:43) correct
26%
(02:09)
wrong
based on 54
sessions
History
Date
Time
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Not Attempted Yet
In the figure above, what is the perimeter of triangle OPQ?
(A) 4 + 2√2
(B) 8 + 4√2
(C) 6 + 2√5
(D) 6 + 6√2
(E) 6√2 + 2√10
Attachment:
2017-12-12_1001_001.png [ 8.53 KiB | Viewed 5276 times ]
12 Dec 2017, 00:24
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Since the triangle passes through the origin, the triangle has the co-ordinates (0,0) , (2,2) and (-4,4)
The distance between the points can be calculated using the formula
\(\sqrt{(x2-x1)^2 + (y2-y1)^2}\) where the two points are (x1,y1) and (x2,y2)
Here, the perimeter is the sum of the distance between the points.
Now, the perimeter of the triangle is calculated as follows:
Perimeter = \(\sqrt{(-4-0)^2 + (4-0)^2} + \sqrt{(-4-2)^2 + (4-2)^2} + \sqrt{(2-0)^2 + (2-0)^2}\) = \(\sqrt{32} + \sqrt{40} + \sqrt{8}\)
= \(4\sqrt{2} + 2\sqrt{10} + 2\sqrt{2}\) = \(6\sqrt{2} + 2\sqrt{10}\) (Option E)
13 Dec 2017, 18:26
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Bunuel
Attachment:
box.method.png [ 22.23 KiB | Viewed 4656 times ]
(This problem took one minute.)
Draw a box / rectangle around the triangle.
There are three new right triangles around the original triangle.
If you can calculate the lengths of the legs of the new triangles in your head, no need for coordinates.
If not, see footnote.*
The hypotenuse of each new right triangle is one side of the original triangle.
Find the length of those original sides with the Pythagorean theorem or special triangle properties.
1) Length of side OP
Side lengths of right ∆ KPO
KO = 4
KP = 4
OP = ?
A right isosceles triangle has side lengths in ratio \(x : x: x\sqrt{2}\)
\(4\) corresponds with \(x\)
OP corresponds with \(x\sqrt{2}\)
Length of OP is \(4\sqrt{2}\)
OR:
\(4^2 + 4^2 = OP^2\)
\(32 = OP^2\)
\(\sqrt{16 * 2} = \sqrt{OP^2}\)
\(OP = 4\sqrt{2}\)
2) Length of side OQ
Right ∆ MOQ is isosceles, too
Side lengths in ratio \(x : x: x\sqrt{2}\)
Its sides MO and MQ = \(2\)
Side \(OQ = 2\sqrt{2}\)
OR
\(2^2 + 2^2 = OQ^2\)
\(8 = OQ^2\)
\(\sqrt{4 * 2} = \sqrt{OQ^2}\)
\(OQ = 2\sqrt{2}\)
3) Length of side PQ
One leg, LQ, of right ∆ LPQ = \(2\)
Other leg, LP = \(6\)
PQ = ?
\(2^2 + 6^2 = PQ^2\)
\(4 + 36 = PQ^2\)
\(40 = PQ^2\)
\(\sqrt{4 * 10} = \sqrt{PQ^2}\)
\(PQ = 2\sqrt{10}\)
Perimeter of original ∆ OPQ = sum of lengths of three sides calculated above
\(4\sqrt{2} + 2\sqrt{2} + 2\sqrt{10}=\)
\(6\sqrt{2} +2\sqrt{10}\)
ANSWER E
*
Coordinates of vertices of drawn rectangle
K has the same x-coordinate as P, and the same y-coordinate as O
--K is on the same vertical line as P
--K is on the same horizontal line as O
L has the same x-coordinate as Q, and the same y-coordinate as P
--L is on the same vertical line as Q
--L is on the same horizontal line as P
M has the same x-coordinate as Q, and the same y-coordinate as O
-- M is on the same vertical line as Q
-- M is on the same horizontal line as O
To calculate side lengths, generally:
Each end point is a vertex, and each vertex has one coordinate that is equal to the corresponding coordinate for the other vertex.
Ignore the identical number (it means they're on the same horizontal or vertical line).
Subtract the coordinate numbers that are different - the absolute value of the difference is the length of the line
Length of KO: Its endpoints lie on the x-axis, y = 0.
Both have y-coordinates of 0. Ignore them.
Subtract x-coordinate of K from x-coordinate of O
0 - (-4) = 4 = KO
Length of KP
Both have x-coordinates of -4. Ignore them.
Subtract: (4 - 0) = 4
Length of KP = 4
And so on . . .
