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P is a point on side AB of a rectangle ABCD, which divides AB in the ratio 5: 3. Q is a point on CD, such that CQ: QD = 3: 1. If the area of rectangle ABCD is 480 sq. units, then what is the area of the quadrilateral ADQP?

    A. 105 sq. units
    B. 150 sq. units
    C. 175 sq. units
    D. 210 sq. units
    E. 270 sq. units

I think answer is D.
Please see the attached image for the solution.its tough to explain via typing.
I do not know what is the toughness level of this question but it did take me good amount of time to think this way.
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Archit3110


Hello I tried doing the question but I am not able to get correct answer , my doubt is on the points P & Q are they both dividing the rectangle on same side i.e lengths/ breadth of the rectangle? without figure its becoming really difficult to understand the question... please provide solution

Hi Archit3110

Yes, P and Q are points on the opposite sides of rectangle, ABCD. Since, AB and CD will be opposite to each other when you form a rectangle by marking the points, A, B, C and D either in clockwise or in anti-clockwise direction.

Regards,

I was doing a calculation mistake... solution attached...

Posted from my mobile device
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Solution


Given:
    • P is a point on side AB of rectangle ABCD
      o P divides AB in the ratio 5: 3

    • Q is a point on side CD of rectangle ABCD
      o Q divides CD in the ratio 3: 1

    • The area of triangle ABCD = 480 sq. units

To find:
    • The area of quadrilateral ADQP

Approach and Working:



Let us assume that the length of AB = CD = x units, and length of BC = DA = h units
    • This implies,
      o Length of AP = \(\frac{5x}{(5 + 3)} = \frac{5x}{8}\)
      o Length of PB = \(\frac{3x}{(5 + 3)} = \frac{3x}{8}\)
      o Length of CQ = \(\frac{3x}{(1 + 3)} = \frac{3x}{4}\)
      o Length of QD = \(\frac{x}{(1 + 3)} = \frac{x}{4}\)

Area of quadrilateral, ADQP =\(\frac{1}{2} * h * (AP + QD) = \frac{1}{2} * h * (\frac{5x}{8} + \frac{x}{4}) = (\frac{7}{16}) * x * h\)
    o We are given that area of rectangle ABCD = 480 = AB * BC = x * h

Therefore, area of quadrilateral ADQP = \((\frac{7}{16}) * 480 = 210\) sq. units

Hence the correct answer is Option D.

Answer: D

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Solution


Given:

We are given that,
    • AB and CD are two parallel chords of a circle
    • O is the center of the circle
    • AB = 12 and CD = 6
    • Diameter of the circle = 6√5

To find:
    • The distance between the two chords, AB and CD

Approach and Working:

From the above figure, we can infer that we need to find the value of EF

In triangle, OAE, we know that,
    • \(AE = \frac{AB}{2} = \frac{12}{2} = 6\)
    • OA = radius of the circle =\(\frac{6√5}{2} = 3√5\)
    • \(∠OEA = 90^o\)

So, we can write, \(OA^2 = OE^2 + AE^2\)
    • Implies, \((3√5)^2 = OE^2 + 6^2\)
    • \(OE^2 = 45 – 36 = 9\)
    • Thus, OE = √9 = 3

Similarly, in triangle DOF, we ca say that
    • \(OD^2 = OF^2 + FD^2\)
      o \((3√5)^2 = OF^2 + (\frac{6}{2})^2\)
      o \(OF^2 = 45 – 9 = 36\)

    • Thus, OF = √36 = 6

Therefore, EF = OF = OE = 6 – 3 = 3

Hence the correct answer is Option B.

Answer: B

i think you have misplaced this solution here.please check
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vanam52923

i think you have misplaced this solution here.please check

Hi vanam52923,

Yeah. Thanks for letting us know.
We have noticed it and changed the solution.

Regards,
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There's another easier way. Since the area of the whole rectangle is 480, let AB = DC = 80 (length of rectangle) and BC = AD = 6 (height).

AP:PB = 5:3. hence AP is (5/8)*80 = 50. same for DQ:QC = 1:3. DQ = (1/4)*80 = 20.

Area of ADQP (trapezoid) is (b+B)*h/2=(50+20)*6/3=70*3 = 210.

Hence D.
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Fastest way to do this is to recognize that the sides whos ratios are given are of the same part of the rectangle
5:3
3:1

Set side AB and DC length equal to 8
5x+1x = 8
x=1
AP=5*1 = 5
PB=3*1 = 3

3x+1x = 8
x=2
DQ = 1*2=2
QC = 3*2 = 6

Solve for height (or missing side of rectangle)
480 = 8*y
60 = y

solve the trapezoid
[(AP+DQ)/2]*h
[(5+2)/2] * 60
7*30
210
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Solution


Given:
    • P is a point on side AB of rectangle ABCD
      o P divides AB in the ratio 5: 3

    • Q is a point on side CD of rectangle ABCD
      o Q divides CD in the ratio 3: 1

    • The area of triangle ABCD = 480 sq. units

To find:
    • The area of quadrilateral ADQP

Approach and Working:



Let us assume that the length of AB = CD = x units, and length of BC = DA = h units
    • This implies,
      o Length of AP = \(\frac{5x}{(5 + 3)} = \frac{5x}{8}\)
      o Length of PB = \(\frac{3x}{(5 + 3)} = \frac{3x}{8}\)
      o Length of CQ = \(\frac{3x}{(1 + 3)} = \frac{3x}{4}\)
      o Length of QD = \(\frac{x}{(1 + 3)} = \frac{x}{4}\)

Area of quadrilateral, ADQP =\(\frac{1}{2} * h * (AP + QD) = \frac{1}{2} * h * (\frac{5x}{8} + \frac{x}{4}) = (\frac{7}{16}) * x * h\)
    o We are given that area of rectangle ABCD = 480 = AB * BC = x * h

Therefore, area of quadrilateral ADQP = \((\frac{7}{16}) * 480 = 210\) sq. units

Hence the correct answer is Option D.

Answer: D




EgmatQuantExpert,

It appears that the image of the rectangle is misleading. CQ:QD is not appearing in the ration of 3:1.

Thank you.
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In the figure we can see that : 4y = 8x. Therefore, y = 2x.

We need area of the two red areas combine. This can be calculated as :

Area of rectangle + area of triangle = AD*2x + 0.5*3x*AD = 3.5 x * AD

Also, area of ABCD = 480 = 8x * AD. -------> x * AD = 60.


Putting this value of x*AD in above figure we get : 3.5 * 60 = 210.


Kudos if helps :)

Regards.
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