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06 Sep 2022, 22:36
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (02:36) correct
44%
(02:45)
wrong
based on 16
sessions
History
Date
Time
Result
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A semicircle with centre O is drawn inside rectangle ABCD such that the longer side BC is the diameter of the circle. Point E divides AB in the ratio 5:4. Line DE is tangent to the semicircle. If the area of the triangle DEO is 65 square units, then the area (in square units) of the rectangle will be
A. 360
B. 300
C. 240
D. 180
E. 120
Source : Varsity Tutors
A. 360
B. 300
C. 240
D. 180
E. 120
Source : Varsity Tutors
Archived Topic
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11 Sep 2022, 10:26
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pintukr
Please draw to understand better:
AB = 5x + 4x
BC = 2r where r is the radius of the circle
CD = 9x (same as AB)
M is say the point at which DE is tangent to the semi-circle. So,
1. DM = DC property of a circle = 9x
2. EM = EB property of a circle = 4x
I know: 1[fraction][/2fraction]*OM*ED=65
So, DE.r=130
DE = DM + EM = 9x + 4x = 13x
So, DE.r=13x.r=130
xr=10
We need: area of the rectangle: 2r*9x = 18*rx = 18*10 = 180
11 Sep 2022, 12:10
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Refer to the attached figure -
Given:\(\frac{AE}{EB}\) = \(\frac{5}{4}\)
Let the length of the rectangle be 9x and the breadth be 2y.
Question : Find the area of the rectangle
Area of rectangle = 18xy
Solution :
As O is the center of the circle, BO = OC = y
OM is the radius, therefore OM = y and \(\angle OME = 90\) (angle between radius and point of tangency = 90)
\( \triangle BOE\)
Using pythagoras theorem -
\(EO^2 = EB^2 + BO^2\)
\(EO^2 = 16x^2 + y^2\)
\( \triangle DCO\)
Using pythagoras theorem -
\(DO^2 = DC^2 + CO^2\)
\(DO^2 = 81x^2 + y^2\)
\( \triangle MOD\)
Using pythagoras theorem -
\(DO^2 = MO^2 + MD^2\)
\(MD^2 = 81x^2 + y^2 - y^2\)
\(MD = 9x\)
\( \triangle EMO\)
Using pythagoras theorem -
\(EO^2 = MO^2 + EM^2\)
\(EM^2 = 16x^2 + y^2 - y^2\)
\(EM = 4x\)
\( \triangle EOD\)
\(Area = \frac{1}{2} * ED * MO\)
\(Area = \frac{1}{2} * 13x* y\)
Given :\(\frac{1}{2} * 13x* y\) = 65
Solving x * y = 10
Area of rectangle = 18xy = 180
Option D
Given:\(\frac{AE}{EB}\) = \(\frac{5}{4}\)
Let the length of the rectangle be 9x and the breadth be 2y.
Question : Find the area of the rectangle
Area of rectangle = 18xy
Solution :
As O is the center of the circle, BO = OC = y
OM is the radius, therefore OM = y and \(\angle OME = 90\) (angle between radius and point of tangency = 90)
\( \triangle BOE\)
Using pythagoras theorem -
\(EO^2 = EB^2 + BO^2\)
\(EO^2 = 16x^2 + y^2\)
\( \triangle DCO\)
Using pythagoras theorem -
\(DO^2 = DC^2 + CO^2\)
\(DO^2 = 81x^2 + y^2\)
\( \triangle MOD\)
Using pythagoras theorem -
\(DO^2 = MO^2 + MD^2\)
\(MD^2 = 81x^2 + y^2 - y^2\)
\(MD = 9x\)
\( \triangle EMO\)
Using pythagoras theorem -
\(EO^2 = MO^2 + EM^2\)
\(EM^2 = 16x^2 + y^2 - y^2\)
\(EM = 4x\)
\( \triangle EOD\)
\(Area = \frac{1}{2} * ED * MO\)
\(Area = \frac{1}{2} * 13x* y\)
Given :\(\frac{1}{2} * 13x* y\) = 65
Solving x * y = 10
Area of rectangle = 18xy = 180
Option D
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
