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ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25

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ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 10 May 2018, 03:58
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A
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  95% (hard)

Question Stats:

45% (02:39) correct 55% (02:49) wrong based on 44 sessions

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ABCD is square having side length 27 cm. E is point on the side CD such that DE= 20.25 cm. A circle is drawn such that it touches the sides AB,BC and line Segment AE.
Find the radius of circle

[1]20
[2]18
[3]15
[4]9
[5]6

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ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 10 May 2018, 05:01
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1
Please refer to the diagrams in the attachment.

Refer to diagram 1: DE measures 20.25. Therefore, E divides DC as 20.25 (DE) and 6.75 (EC) i.e., in the ratio 3: 1.

Refer to digram 2: Extend BC and AE to meet at F.
ABF is a right triangle and the circle drawn touching the 3 sides AB, BC, and AE is nothing but the circle inscribed in this right triangle.
For a right triangle the quickest formula to find the radius of the inscribed circle, r = s - h, where s is the semi perimeter and h is the hypotenuse (I will provide the derivation for this as a reply to this post)
We know one of the sides AB. We have to compute the remaining two sides of the right triangle i.e., hypotenuse AF and side FB.

Triangles ABF and ECF are similar triangles (AA or AAA rule).
Therefore, the ratio of corresponding sides will be same for all 3 corresponding sides.
Of the values available, AB and EC are corresponding sides. Ratio of AB : EC = 27 : 6.75 = 4 : 1
i.e., if the sides of ABF measure 4k units, that of ECF will measure k units.
Therefore, ratio of BF : CF will also be 4 : 1
If CF = x, then (27 + x) : x = 4: 1
Therefore, x = 9.

Hence, BF = 4 * 9 = 36.
AB = 27 (which is nothing but 3*9), BF = 36 (which is nothing but 4*9). Therefore, the hypotenuse will be 5*9 = 45 units.

Semi perimeter of the triangle = (27 + 36 + 45)/2 = 108/2 = 54 units.
Radius of inscribed circle = s - h = 54 - 45 = 9 units.
Attachments

Square-right-triangle.png
Square-right-triangle.png [ 33.28 KiB | Viewed 667 times ]


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Re: ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 10 May 2018, 05:47
WizakoBaskar wrote:
For a right triangle the quickest formula to find the radius of the inscribed circle, r = s - h, where s is the semi perimeter and h is the hypotenuse (I will provide the derivation for this as a reply to this post)


Please refer to the diagram 1 in the attachment

ABC is a right triangle right angled at B. AC is the hypotenuse.
Objective to find radius of inscribed circle.
PQR are points where the inscribed circle touches the sides of the triangle.

AR and AQ are equal tangents drawn to the circle from A.
Similarly, BP and BR are equal tangents drawn to the circle from B and CP and CQ are equal tangents drawn to the circle from C.
Hypotenuse AC, h = AQ + QC
We can replace AQ with AR and QC with CP. So, h = AR + CP ... (1)

Refer to diagram 2.
OR is perpendicular to AB and OP is perpendicular to BC (tangent and radius will be at right angles at the point of tangency)
The triangle is right angled at B. So, angle B is 90 degrees. So, BROP is a square whose side equals radius of the inscribed circle.
So, BR = BP = r (the radius of the inscribed circle).

If side AB measures c units, AR = AB - BR = c - r ... (2)
If side BC measures a units, CP = BC - BP = a - r ... (3)

Substitute (2) and (3) in (1)
h = (c - r) + (a - r)
Or 2r = a + c - h
Add and subtract h on the right side
2r = a + c + h - h - h
So, r = (a + c + h)/2 - (2h)/2
(a + c + h)/2 is the semi perimeter of the triangle.

Or r = s - h
Attachments

right-triangle-inradius-proof.png
right-triangle-inradius-proof.png [ 29.54 KiB | Viewed 654 times ]


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Re: ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 10 May 2018, 10:45
1
Refer to diagram 2 above of first post shared by wizakobaskar

We need to get sides of triangle ABF
let's calculate CF FIRST

∆EDA~∆ECF (by AA)
So AD/CF=ED/EC
27/CF=20.25/6.75
CF=9

Let's calculate AF=AB^2+ BF^2
=36^2+27^2
=45^2
AF=45

For right angle ∆, in which circle is inscribed radius r=
(Sum of other two sides -hypotenuse)/2

= 27+36-45/2
=18/2
=9

Give kudos if it helps!!

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Re: ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 11 May 2018, 04:50
OA: D

Point E is on side CD, Radius of circle is dependent on the position of E on CD.
It can be seen from below sketch( not drawn to scale).
Attachment:
Capture AC.PNG
Capture AC.PNG [ 23.63 KiB | Viewed 479 times ]

When E is coincident with C , circle will have minimum radius.
As E starts moving towards D, radius of circle will keep on increasing until E coincides with D.

Case1 : Minimum radius,\(r_{min}\)
Using 1st figure of above sketch
\(AB=BC=a\)

\(AC= \sqrt{2}a\) , (using Pythagoras theorem in \(\triangle\)ABC)

\(BH=BI=r_{min}\)
\(AH=IC=a-r_{min}\)
\(AG=AH=a-r_{min}; GC=IC=a-r_{min}\) (Tangents from a point outside the circle are equal in length.)
\(AC = AG+GC\)
\(\sqrt{2}a= 2(a-r_{min})\)
\(r_{min}=\frac{{(2-\sqrt[]{2})a}}{{2}}\)

putting \(a = 27 cm\), we get \(r_{min}\approx{7.9}\)

Case 2: considering last figure of sketch
\(r_{max}= \frac{a}{2}=\frac{27}{2}=13.5\)

Only Option D lies between this range, hence it is the answer.
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ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 11 May 2018, 14:30
Since the range of options is large we can use "easy peasy" approach :-)
I draw the figure and saw that are of the circle is about twice smaller than that of the trapezoid. See my attachment.

Answer (D)
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1.png
1.png [ 9.54 KiB | Viewed 433 times ]

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Re: ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25  [#permalink]

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New post 11 May 2018, 22:24
OA : D

First we should simplify the dimensions specified in question stem.
Let take side dimension 27 cms = a, then DE(20.25cm) becomes =\(\frac{3}{4}a\)
Attachment:
Radius.PNG
Radius.PNG [ 14.76 KiB | Viewed 373 times ]


\(AF=GC=a-r\)
\(AI=AF=a-r\) (Tangents from a point outside the circle are equal in length)

In right angle \(\triangle\)ADE
\(AE^2=AD^2+DE^2\)
\(AE^2 = a^2+\frac{{9a^2}}{16}\)= \(\frac{{25a^2}}{16}\)
\(AE = \frac{5a}{4}\)
\(IE = AE- AI = \frac{5a}{4}-a+r =\frac{a}{4}+r\)

As \(\triangle\)ADE and \(\triangle\)ECH are similar.
\(\frac{CH}{AD}=\frac{HE}{AE}=\frac{EC}{DE}\)

\(\frac{CH}{a}=\frac{HE}{\frac{5a}{4}}=\frac{\frac{a}{4}}{\frac{3a}{4}}\)
\(CH=\frac{a}{3} ; HE = \frac{5a}{12}\)

\(HI = GH\) (Tangents from a point outside the circle are equal in length)
\(IE+EH=GC+CH\)
\(\frac{a}{4}+r+\frac{5a}{12}=a-r+\frac{a}{3}\)
\(2r=a+\frac{a}{3}-\frac{a}{4}-\frac{5a}{12}\)
\(2r=\frac{{12a+4a-3a-5a}}{12}\)
\(2r=\frac{{8a}}{12}\)

\(r=\frac{{a}}{3}\)

putting \(a =27 cm\), we get \(r =9 cm\)
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Re: ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25 &nbs [#permalink] 11 May 2018, 22:24
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