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03 Apr 2020, 06:48
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:33) correct
15%
(01:44)
wrong
based on 47
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Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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03 Apr 2020, 07:35
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Since the radius of the circle is 13 cm. The length of diagonal is 26cm.
Using Pythagoras theroam.
(Diagonal)^2=(length)^2+(breadth)^2
(26)^2=(L)^2+(B)^2
676=l^2+b^2---------(1)
Now the possible length of length and breadth of rectangle is
1. 25^2+12^2=625+144=769 (not equal to 676)
2. 25^2+10^2=625+100=725 (not equal to 676)
3. 25^2+9^81=625+81=706 (not equal to 676)
4. 24^2+12^2=576+144=720 (not equal to 676)
5. 24^2+10^2=576+100=676 (equal to 676(26)^2)
So OA is E
Posted from my mobile device
Using Pythagoras theroam.
(Diagonal)^2=(length)^2+(breadth)^2
(26)^2=(L)^2+(B)^2
676=l^2+b^2---------(1)
Now the possible length of length and breadth of rectangle is
1. 25^2+12^2=625+144=769 (not equal to 676)
2. 25^2+10^2=625+100=725 (not equal to 676)
3. 25^2+9^81=625+81=706 (not equal to 676)
4. 24^2+12^2=576+144=720 (not equal to 676)
5. 24^2+10^2=576+100=676 (equal to 676(26)^2)
So OA is E
Posted from my mobile device
03 Apr 2020, 08:22
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Bunuel
The rectangle is inside the circle, the diameter of the circle is equal to the diagonal of the rectangle.
Then the rectangle must have a diagonal of exactly 13*2 = 26 cm.
Therefore we want to find the combination of length/width that gives a diagonal of 26 cm.
If you know your Pythagorean integer solutions well, then \(13^2 = 12^2 + 5^2\) should come to mind. We can double it to get \(26^2 = 24^2 + 10^2\), hence the dimensions of the rectangle must be 24 * 10.
Ans: E
