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A square, ABCD has a side of length 12 cm each. An equilateral triangle, MAB exists such that M is outside the square ABCD. If a circle is drawn in such a way that it passes through the points M, C and D, then what will be the radius of that circle?
- a)12√3
- b)12√2
- c)12 cm
- d)4√3 cm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A square, ABCD has a side of length 12 cm each. An equilateral triangl...
The problem figure in the question can be drawn as:
If we construct an equilateral triangle with base CD, we will get:
If we construct an equilateral triangle with base CD, we will get:
AM and DK are parallel to each other (makes 60 degrees with the base) and AD and MK are parallel to each other.
∴AMKD is a parallelogram and hence AM= DK and AD= MK.
But since AM=AB=AD.
So, AMKD is a rhombus.
Also, since MK= KD and M and D are the points on the circle, K must be centre of the circle and so, the radius = KD = CD = 12 cm.
∴AMKD is a parallelogram and hence AM= DK and AD= MK.
But since AM=AB=AD.
So, AMKD is a rhombus.
Also, since MK= KD and M and D are the points on the circle, K must be centre of the circle and so, the radius = KD = CD = 12 cm.
Most Upvoted Answer
A square, ABCD has a side of length 12 cm each. An equilateral triangl...
Since triangle MAB is equilateral, we know that angle MBA is 60 degrees. Therefore, angle MBC is 120 degrees (since BC is perpendicular to AB), and angle MBD is 60 degrees (since BD is perpendicular to AB).
Since MC and MD are radii of the circle, we know that they have the same length, which we can call r. Let's call the center of the circle O.
Since angle MBC is 120 degrees, we know that angle OBC is half of that, or 60 degrees. Therefore, triangle OBC is a 30-60-90 triangle, and we can use the ratios of the sides to find that BC has length 2r.
Similarly, since angle MBD is 60 degrees, we know that angle OBD is 30 degrees. Therefore, triangle OBD is also a 30-60-90 triangle, and we can use the ratios of the sides to find that BD has length 2r*sqrt(3).
Since BD has length 12 (since it is a side of the square), we can set up the equation 2r*sqrt(3) = 12 and solve for r to get r = 6/sqrt(3) = 2sqrt(3) cm.
Therefore, the radius of the circle is 2sqrt(3) cm.
Since MC and MD are radii of the circle, we know that they have the same length, which we can call r. Let's call the center of the circle O.
Since angle MBC is 120 degrees, we know that angle OBC is half of that, or 60 degrees. Therefore, triangle OBC is a 30-60-90 triangle, and we can use the ratios of the sides to find that BC has length 2r.
Similarly, since angle MBD is 60 degrees, we know that angle OBD is 30 degrees. Therefore, triangle OBD is also a 30-60-90 triangle, and we can use the ratios of the sides to find that BD has length 2r*sqrt(3).
Since BD has length 12 (since it is a side of the square), we can set up the equation 2r*sqrt(3) = 12 and solve for r to get r = 6/sqrt(3) = 2sqrt(3) cm.
Therefore, the radius of the circle is 2sqrt(3) cm.
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A square, ABCD has a side of length 12 cm each. An equilateral triangle, MAB exists such that M is outside the square ABCD. If a circle is drawn in such a way that it passes through the points M, C and D, then what will be the radius of that circle?a)12√3b)12√2c)12 cmd)4√3 cmCorrect answer is option 'C'. Can you explain this answer?
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