CBSE Class 9 Maths Exemplar Chapter 10: Circles NCERT Solutions

Let O be the center of the circle. Draw a perpendicular line segment OP from the center O to the chord AB. A property of circles states that a perpendicular drawn from the center to a chord bisects the chord. Therefore, P is the midpoint of AB.

Given the length of the chord AB = 30 cm, we have AP = PB = $\backslash frac\{1\}\{2\}\; \backslash times\; AB\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 30\; =\; 15$ cm.

The diameter of the circle is AD = 34 cm. The radius of the circle is half of the diameter. So, OA (which is a radius) = $\backslash frac\{1\}\{2\}\; \backslash times\; AD\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 34\; =\; 17$ cm.

Now, consider the right-angled triangle OPA. By the Pythagorean theorem, we have:

$$OA^2\; =\; OP^2\; +\; AP^2$$

We need to find the distance OP. Rearranging the formula:

$$OP^2\; =\; OA^2\; -\; AP^2$$

Substitute the values of OA and AP:

$$OP^2\; =\; (17)^2\; -\; (15)^2$$

$$OP^2\; =\; 289\; -\; 225$$

$$OP^2\; =\; 64$$

Taking the square root of both sides:

$OP\; =\; \backslash sqrt\{64\}\; =\; 8$ cm.

Thus, the distance of the chord AB from the center of the circle is 8 cm. The correct option is (D).

We are given a circle with center O. OA is the radius, and AB is a chord. OD is perpendicular to AB. A key property of circles is that a perpendicular from the center to a chord bisects the chord. In this case, OD is perpendicular to AB, so it bisects AB at point C.

Given AB = 8 cm, we have AC = CB = $\backslash frac\{1\}\{2\}\; \backslash times\; AB\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 8\; =\; 4$ cm.

We are given that the radius OA = 5 cm.

Now, consider the right-angled triangle OAC. By the Pythagorean theorem:

$$OA^2\; =\; AC^2\; +\; OC^2$$

Substitute the known values:

$(5)^2\; =\; (4)^2\; +\; OC^2$

$25\; =\; 16\; +\; OC^2$

Solve for OC:

$OC^2\; =\; 25\; -\; 16$

$OC^2\; =\; 9$

Taking the square root, we get OC = 3 cm (since length must be positive).

OD is also a radius of the same circle, so OD = OA = 5 cm.

The point C lies on the radius OD. The length CD can be found by subtracting OC from OD:

$CD\; =\; OD\; -\; OC$

$CD\; =\; 5\; \backslash text\{\; cm\}\; -\; 3\; \backslash text\{\; cm\}$

$CD\; =\; 2\; \backslash text\{\; cm\}$

Therefore, the length of CD is 2 cm. The correct option is (A).

We are given a situation where three points A, B, and C lie on a circle. We are also given the lengths of two segments, AB = 12 cm and BC = 16 cm, and that AB is perpendicular to BC. This means that the angle $\backslash angle\; ABC$ is a right angle (90 degrees).

A fundamental theorem in geometry states that an angle inscribed in a semicircle is always a right angle. Conversely, if an inscribed angle is a right angle, then the chord subtending it must be a diameter of the circle.

Since $\backslash angle\; ABC\; =\; 90^\{\backslash circ\}$, the chord AC subtends this right angle. Therefore, AC must be the diameter of the circle passing through points A, B, and C.

To find the length of the diameter AC, we can use the Pythagorean theorem in the right-angled triangle ABC:

$$AC^2\; =\; AB^2\; +\; BC^2$$

Substitute the given values:

$$AC^2\; =\; (12\; \backslash text\{\; cm\})^2\; +\; (16\; \backslash text\{\; cm\})^2$$

$$AC^2\; =\; 144\; \backslash text\{\; cm\}^2\; +\; 256\; \backslash text\{\; cm\}^2$$

$$AC^2\; =\; 400\; \backslash text\{\; cm\}^2$$

Taking the square root of both sides to find the length of AC:

$AC\; =\; \backslash sqrt\{400\; \backslash text\{\; cm\}^2\}\; =\; 20\; \backslash text\{\; cm\}$

So, the diameter of the circle is 20 cm.

The radius of a circle is half of its diameter. Therefore, the radius is:

$\backslash text\{Radius\}\; =\; \backslash frac\{AC\}\{2\}\; =\; \backslash frac\{20\; \backslash text\{\; cm\}\}\{2\}\; =\; 10\; \backslash text\{\; cm\}$

Hence, the radius of the circle passing through points A, B, and C is 10 cm. The correct option is (C).