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03 Apr 2020, 04:34
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D
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Difficulty:
65%
(hard)
Question Stats:
53% (02:22) correct
47%
(03:13)
wrong
based on 62
sessions
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Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance from P to each of AB, BC, and CA is 4(√2 - 1) cm, then the area, in sq. cm, of the triangle ABC is
A. 10
B. 12
C. 14
D. 16
E. 18
Are You Up For the Challenge: 700 Level Questions
A. 10
B. 12
C. 14
D. 16
E. 18
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 05:27
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Bunuel
Rule: Length of the tangents drawn from a point outside the circle are equal
Note that, Point 'P' which is equidistant from all the perpendiculars to sides is Incentre of the triangle ABC
Since AC = BC, It follows that AF = AD = BD = BE
Let AF = AD = BD = BE = a(assume)
For right angled triangle ABC, \(AB^2 = AC^2 + BC^2\)
--> \((2a)^2 = (a + r)^2 + (a + r)^2\)
--> \(4a^2 = 2(a + r)^2\)
--> \((a + r)^2 = 2a^2\)
--> \(a + r = \sqrt{2}a\)
--> \(r = a(\sqrt{2} - 1)\)
Given, \(r = 4(\sqrt{2} - 1) = a(\sqrt{2} - 1)\)
--> \(a = 4\)
Side AC = BC = \(a + r = 4 + 4\sqrt{2} - 4 = 4\sqrt{2}\)
--> Area of triangle ABC = \(\frac{1}{2}*AC*BC = \frac{1}{2}*4\sqrt{2}*4\sqrt{2} = 16\)
Option D
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General Discussion
sting8
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Joined: 03 Dec 2020
Last visit: 12 Aug 2023
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Location: India
Debrief: https://gmatclub.com/forum/740-v40-q50-second-attempt-self-preparation-375379.html
Concentration: Technology, Strategy
Schools: ISB '23 (A) Tepper '24 (WL) Ross '24 (D)
GMAT 1: 720 Q49 V39
GMAT 2: 740 Q50 V40
WE:Engineering (Computer Hardware)
30 Jun 2021, 10:47
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We can use the following property of triangles
Area = Inradius * Semiperemeter
Solve the equation by taking sides as x,x, \(\sqrt{{2}} x\)
I think this process can save some time in exam
Area = Inradius * Semiperemeter
Solve the equation by taking sides as x,x, \(\sqrt{{2}} x\)
I think this process can save some time in exam
