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03 Apr 2020, 04:34
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D
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Difficulty:
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Question Stats:
53% (02:22) correct
47%
(03:13)
wrong
based on 62
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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
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
A. 10
B. 12
C. 14
D. 16
E. 18
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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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
