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In this problem, we can use the property :

The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle

So Angle BCD = 2 * Angle BAD = 2 * 30 = 60

Let's see the options

I. The length of minor arc BD is greater than 10.

length of minor arc BD = 30/360 * 2 * 22/7 *9 = 11/7 * 3 = 33/7 = 4.7

This is not true.

II. Angle ABD measures 90 degrees.

This is correct, as the angle subtended by diameter of the circle on any point on the circumference of the circle is 90.

III. Triangle BCD is equilateral.

As BC = BD = Radius of the circle in Triangle BCD and Angle BCD = 60. Triangle

Answer is D

Posted from my mobile device


How do you know that BD = radius?

Posted from my mobile device

We know Angle BCD = 60

We know BC = BD = Radius of the circle.

As BCD is now Isoceles, therefore Angle CBD = Angle CDB

therefore, Angle BCD + Angle CBD + Angle CDB = 180

60 + Angle CBD + Angle CDB = 180

Angle CBD + Angle CDB = 120

therefore Angle CBD = Angle CDB = 60

As all three angles of Triangle BCD are equal, so BD = Radius of the circle and Triangle BCD is an equilateral Triangle.
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Attachment:
FullSizeRender.jpg
FullSizeRender.jpg [ 76.77 KiB | Viewed 4719 times ]

Ans:D
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Here ! We know that angle ABD is 90 degrees (subtended by the diameter). Therefore, the triangle ABD is a 30-60-90 right angled triangle.
The sides of a 30-60-90 triangle are in the ratio 1:3^(1/2):2

Since the hypotenuse (diameter here) is 18, therefore side BD is the smallest side (opp to the smallest angle i.e. 30 degrees).
Hence, BD = 1/2 * 18 = 9

Therefore, only II is true.


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Here ! We know that angle ABD is 90 degrees (subtended by the diameter). Therefore, the triangle ABD is a 30-60-90 right angled triangle.
The sides of a 30-60-90 triangle are in the ratio 1:3^(1/2):2

Since the hypotenuse (diameter here) is 18, therefore side BD is the smallest side (opp to the smallest angle i.e. 30 degrees).
Hence, BD = 1/2 * 18 = 9

Therefore, only II is true.

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Since ABD = 90°, and C is the center of the circle.
CA = CB(because both are the radius of the circle)
ABC is also 30°(angles opposite to equal sides are equal)
Therefore, CBD = 90°-30°(60°)

Since CB and CD are both the radius of the circle, BDC is also 60°
Since the triangle has two angles which are 60°, the third angle is also 60°

So triangle BCD is equilateral. Hence III is also correct!
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pushpitkc
Anirudh1527
Here ! We know that angle ABD is 90 degrees (subtended by the diameter). Therefore, the triangle ABD is a 30-60-90 right angled triangle.
The sides of a 30-60-90 triangle are in the ratio 1:3^(1/2):2

Since the hypotenuse (diameter here) is 18, therefore side BD is the smallest side (opp to the smallest angle i.e. 30 degrees).
Hence, BD = 1/2 * 18 = 9

Therefore, only II is true.

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Since ABD = 90°, and C is the center of the circle.
CA = CB(because both are the radius of the circle)
ABC is also 30°(angles opposite to equal sides are equal)
Therefore, CBD = 90°-30°(60°)

Since CB and CD are both the radius of the circle, BDC is also 60°
Since the triangle has two angles which are 60°, the third angle is also 60°

So triangle BCD is equilateral. Hence III is also correct!

Ah, yes ! Sorry, that too. Thanks :)



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How to derive the length of an arc (formula)?

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How to derive the length of an arc (formula)?

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Length of arc is \(\frac{x}{360}*2*pi*\)radius of circle

given x is the angle made by the arc in the circle.
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Imo D
Calculating arc length we have arc BD =60/360*2πr
BD 6.28 which is less than 10
Rest two are easy

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Bunuel

In the circle above with center C, diameter AD = 18. Which of the following must be true?

I. The length of minor arc BD is greater than 10.

II. Angle ABD measures 90 degrees.

III. Triangle BCD is equilateral.

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III


Attachment:
CircleC.png

II. Angle ABD measures 90 degrees.
From the diagram. \(\triangle\)ABD is inside semicircle with one side AD as diameter. Hence its a right triangle and Angle ABD is \(90^{\circ}\).


III. Triangle BCD is equilateral.
Given from the diagram, \(\angle\)A is \(30^{\circ}\), \(\angle\)B is \(90^{\circ}\) (from II) hence \(\triangle\)ABD 30:60:90 triangle. Therefore \(\angle\)D is \(60^{\circ}\).
In \(\triangle\)BCD; BC = CD = radius of circle. \(\angle\)D is \(60^{\circ}\) hence \(\angle\)B is \(60^{\circ}\). Therefore \(\angle\)C is also \(60^{\circ}\). Therefore Triangle BCD is equilateral.

I. The length of minor arc BD is greater than 10.
Length of arc = 2\(\pi\)r\(\frac{C}{360}\) -----------------(C is the central angle, r is the radius)
In \(\triangle\)BCD; BC = CD = radius of circle. \(\angle\) D is \(60^{\circ}\) hence \(\angle\)B is \(60^{\circ}\). and therefore \(\angle\)C is also \(60^{\circ}\). (from III)
Given AD = 18 which is the diameter. Therefore radius = 9.
Length of minor arc BD = 2\(\pi\)\(9*\frac{60}{360}\) = 2\(\pi\)\(9*\frac{1}{6}\) = 3\(\pi\) = \(3 * 3.14\) = \(9.42\) (approximately) . ------------ I is not true.

II and III only. Answer D....
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Bunuel

In the circle above with center C, diameter AD = 18. Which of the following must be true?

I. The length of minor arc BD is greater than 10.

II. Angle ABD measures 90 degrees.

III. Triangle BCD is equilateral.

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III


Attachment:
CircleC.png

I saw too much calculation to solve this problem, once you know 3 basic rules of triangles you can solve it in 30 sec:
- this is a 30-60-90 degree triangle, triangle ABD is inside semicircle with one side AD as diameter.

1 - 30-60-90 degrees triangle is a triple K - K*sqt(2) - K*sqt(3) where K= 6*sqt(3) - FALSE
2 - TRUE - see above
3 - BCD triangle is isosceles --> so 2 angles are 60, the third angle comes by itself. - TRUE



Thanks,
Matteo
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