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09 Dec 2021, 09:45
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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:
75%
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Question Stats:
42% (02:10) correct
58%
(02:51)
wrong
based on 26
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Find the length of the common chord of two intersecting circles of radii 15 cm and 20 cm, given that the distance between the centers of the circles is 25cm.
A. 9
B. 12
C. 15
D. 18
E. 24
Are You Up For the Challenge: 700 Level Questions
A. 9
B. 12
C. 15
D. 18
E. 24
Are You Up For the Challenge: 700 Level Questions
25 Dec 2021, 12:40
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As we can see from the lengths that triangle formed by radius of both circle and the line between centres is a right triangle.
So area of the triangle would be 1/2*15*20 -= 150
If we draw a perpendicular from right angle and say it is X inch
Then area can be written as
1/2*X * 25 = 150
X=12
The chord would be 2X
Hence 24
Answer is E.
Posted from my mobile device
So area of the triangle would be 1/2*15*20 -= 150
If we draw a perpendicular from right angle and say it is X inch
Then area can be written as
1/2*X * 25 = 150
X=12
The chord would be 2X
Hence 24
Answer is E.
Posted from my mobile device
25 Dec 2021, 13:23
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Attachment:
Triangle.JPG [ 52.93 KiB | Viewed 2198 times ]
In the figure, X and Y are two centers of the circle.
AX = 20
AY = 15
XY = 25
We can conclude that
1) XAY is a right angled triangle, with XAY = 90
2) XBY is a right angled triangle, with XBY = 90
3) The triangles are congruent
Area of quadrilateral XAYB = \(\triangle XAY + \triangle XBY \) = 2 * (\(\frac{1}{2}\) * 20 * 15) = 300
We can also see that quadrilateral XAYB is a kite (the diagonals AB and XY are perpendicular to each other)
Let AB = X
Area of kite = \(\frac{1}{2}\) * product of diagonals
= \(\frac{1}{2}\) * 25 * x = 300
x = 24
