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10 Nov 2020, 03:30
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Difficulty:
95%
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Question Stats:
32% (02:32) correct
68%
(02:47)
wrong
based on 53
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In the diagram, four circles of radius 1 with centers P, Q, R and S are tangent to one another and sides of \(\triangle{ABC}\), as shown. Find the perimeter of \(\triangle{ABC}\).
(A) \(12+4\sqrt{2}\)
(B) \(12+4\sqrt{3}\)
(C) \(12+6\sqrt{2}\)
(D) \(12+6\sqrt{3}\)
(E) \(24\)
Attachment:
4circle-triangle 1.png [ 4.1 KiB | Viewed 10293 times ]
Attachment:
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Are You Up For the Challenge: 700 Level Questions
10 Nov 2020, 23:31
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Bunuel
Such questions almost always require some additional lines to be made.
STEP 1
Join P, R and S to get an equilateral triangle PRS.
So \(\angle PSR = 60\)
STEP 2
Join P, Q and S to get a \(\triangle PQS\).
\(\angle PSQ = 60\), and two sides are 2 and 4....2 and 4 come from the addition of radius 1.
Therefore \(\triangle PQS\) is a 30-60-90 triangle with sides \(2:2\sqrt{3}:4\)
Perimeter = \(2+2\sqrt{3}+4=6+2\sqrt{3}\)
STEP 3
Take \(\triangle ABC\) and \(\triangle PQS\)
All sides are parallel, so both are similar.
So the ratio of sides is same. Therefore, let us find the ratio between AC and PS.
\(\angle PSQ = \angle ACB=60\)
STEP 4
Join C, D and S to get aright angled triangle CDS.
Now, CS bisects the angle ACB, so \(\angle DCS=30\), and \(\triangle DCS\) is a 30-60-90 triangle with sides SD:DC:CS as \(1:\sqrt{3}:2\)
So AC=AD+DC=\(3+\sqrt{3}\)
STEP 5
The ratio of parallel sides.
\(PS:AC=2:3+\sqrt{3}\)
So ratio of perimeters of triangle \(PQS:ABC=2:3+\sqrt{3}=6+2\sqrt{3}:ABC\)
Perimeter of ABC = \(\frac{(3+\sqrt{3})(6+2\sqrt{3})}{2}=(3+\sqrt{3})(3+\sqrt{3})=(3+\sqrt{3})^2\)
\(9+3+6\sqrt{3}=12+6\sqrt{3}\)
D
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4circle-triangle 1.png [ 12.82 KiB | Viewed 9212 times ]
