Mathematics Class IX

Question 1 of 15

1. Question
1 point(s)
A chord AB of a circle lies at 9 cm from the centre 0. The radius of the circle is 15 cm. What is the length of the chord?
- 24 cm
- 12 cm
- 18 cm
- 27 cm
Correct

Incorrect

Hint

Let the mid-point of AB be C. Use Pythagoras theorem to calculate the length of CB.

(1) $\begin{equation*}\begin{split} CB^{2} + OC^{2} &= OB^{2} \\ CB^{2} + 9^{2} &= 12^{2} \\ CB^{2}+81 &= 225 \\ CB^{2} &= 144 \\ CB &= 12 \\ AB &= 2 \times 12 \\ AB &= 24 \end{split}\end{equation*}$

Length of the chord AB is 24 cm.
Question 2 of 15

2. Question
1 point(s)
Given the PQ, QR and RS are the chords of a circle. PQ and RS both are perpendicular to QR. Which of the following is correct?
- PQ = QR
- PQ = RS
- RS = QR
- PR = QR
Correct

Incorrect

Hint

If we join PR and QS, both will be the diameters of the circle and they will same length and therefore PQ and RS are also equal in length.
Question 3 of 15

3. Question
1 point(s)
In the following figure PQ and PR are the chords of a circle such that PQ = PR = 8 cm. The circle has a radius of 5 cm and centre O. Point S is the mid-point of QR. What is the length of OS?
- 1.5 cm
- 1.6 cm
- 1.4 cm
- 1.3 cm
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} (OS+5)^{2} + SR^{2} &= 8^{2} \\ OS^{2} + SR^{2} &= 5^{2} \end{split}\end{equation*}$

Subtract the equations

(2) $\begin{equation*}\begin{split} 10 OS + 25 &= 39 \\ 10 OS &= 14 \\ OS &= 1.4 \end{split}\end{equation*}$
Question 4 of 15

4. Question
1 point(s)
Angles subtended by the chords PR and PQ at the centre of a circle O are 58° and 122° respectively. What is the measure of the angle PRQ?
- 96°
- 92°
- 88°
- 90°
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} \angle OPR + \angle ORP + 58^{\circ} &= 180^{\circ} \\ 2 \angle ORP &= 122^{\circ} \\ \angle ORP &= 61^{\circ} \\ \angle ORQ + \angle OQR + 122^{\circ} &= 180^{\circ} \\ 2 \angle ORQ &= 58^{\circ} \\ \angle ORQ &= 29^{\circ} \\ \angle PRQ &= \angle ORP + \angle ORQ \\ \angle PRQ &= 61^{\circ} + 29^{\circ} \\ \angle PRQ &= 90^{\circ} \end{split}\end{equation*}$
Question 5 of 15

5. Question
1 point(s)
ABCD is a cyclic quadrilateral with angle DCB as 40° and angle BAC as 60°. What is the measure of the angle CAD?
- 60°
- 80°
- 90°
- 70°
Correct

Incorrect

Hint

Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} \angle BAD + \angle DCB &= 180^{\circ} \\ \angle BAD + 40^{\circ} &= 180^{\circ} \\ \angle BAD &= 140^{\circ} \\ \angle CAD &= \angle BAD - \angle BAC \\ \angle CAD &= 140^{\circ} - 60^{\circ} \\ \angle CAD &= 80^{\circ} \end{split}\end{equation*}$
Question 6 of 15

6. Question
1 point(s)
AB is a chord of the circle with centre O. What is the measure of $b + c$?
- 100°
- 60°
- 120°
- 90°
Correct

Incorrect

Hint

Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} 2c &= a \\ a+b+b &= 180^{\circ} \\ a+2b &= 180^{\circ} \\ 2c+2b &= 180^{\circ} \\ 2(b+c) &= 180^{\circ} \\ b+c &= 90^{\circ} \end{split}\end{equation*}$
Question 7 of 15

7. Question
1 point(s)
What is the value of $k$ in the following figure?
- 115°
- 130°
- 90°
- 100°
Correct

Incorrect

Hint

Angle in a semicircle is a right angle. Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} \angle PRQ + \angle RQP + \angle QPR &= 180^{\circ} \\ 90^{\circ} + \angle RQP + 25^{\circ} &= 180^{\circ} \\ \angle RQP &= 65^{\circ} \\ \angle RQP + \angle PSR &= 180^{\circ} \\ 65^{\circ} + k &= 180^{\circ} \\ k &= 115^{\circ} \end{split}\end{equation*}$
Question 8 of 15

8. Question
1 point(s)
PQRS and TQRS are cyclic quadrilaterals in the following figure. What is the measure of the angle STQ?
- 83°
- 90°
- 93°
- 87°
Correct

Incorrect

Hint

Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} \angle P + \angle R &= 180^{\circ} \\ 87^{\circ} + \angle R &= 180^{\circ} \\ \angle R &= 93^{\circ} \\ \angle R + \angle STQ &= 180^{\circ} \\ 93^{\circ} + \angle STQ &= 180^{\circ} \\ \angle STQ &= 87^{\circ} \end{split}\end{equation*}$
Question 9 of 15

9. Question
1 point(s)
What is the value of $k$ in the following figure?
- 144°
- 72°
- 156°
- 78°
Correct

Incorrect

Hint

Plot a point S anywhere on the major arc PQ. Join PS and QS. The angle PSR will be half of the angle POR. Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} \angle PSR + \angle PQR &= 180^{\circ} \\ \dfrac{k}{2} + 108^{\circ} &= 180^{\circ} \\ \dfrac{k}{2} &= 72^{\circ} \\ k &= 144^{\circ} \end{split}\end{equation*}$
Question 10 of 15

10. Question
1 point(s)
The length of a chord of a circle is equal to the radius. What is the measure of the angle subtended by this chord at the centre of the circle?
- 90°
- 60°
- 45°
- None of these
Correct

Incorrect

Hint

The triangle formed will be an equilateral triangle. Therefore, the angle subtended by this chord at the centre of the circle is 60°.
Question 11 of 15

11. Question
1 point(s)
What is the value of $k$ in the following figure?
- 62°
- 112°
- 124°
- 56°
Correct

Incorrect

Hint

Plot a point S anywhere on the major arc PQ. Join PS and QS. The angle PSR will be half of the angle POR. Sum of opposite angles of a cyclic quadrilateral is 180°.

(1) $\begin{equation*}\begin{split} \angle PSR &= 56^{\circ} \\ \angle PQR + \angle PSR &= 180^{\circ} \\ k + 56^{\circ} &= 180^{\circ} \\ k &= 124^{\circ} \end{split}\end{equation*}$
Question 12 of 15

12. Question
1 point(s)
The length of PR and PQ in the following diagram are 20 cm and $10\sqrt{3}$ cm. Find the length of PQ.
- 8 cm
- 12 cm
- 10 cm
- None of these
Correct

Incorrect

Hint

Triangle drawn in a semicircle is a right triangle.

(1) $\begin{equation*}\begin{split} PQ^{2} + QR^{2} &= PR^{2} \\ PQ^{2} + (10\sqrt{3})^{2} &= 20^{2} \\ PQ^{2} +300 &= 400 \\ PQ^{2} &= 100 \\ PQ &= 10\end{split}\end{equation*}$

Length of PQ is 10 cm.
Question 13 of 15

13. Question
1 point(s)
The lengths of two chords PQ and RS of a circle are 12 cm and 16 cm respectively. They lie on the opposite sides of the centre of the circle. If PQ lies at a distance of 8 cm from the centre then find the distance between the chord RS and centre of the circle.
- 6 cm
- 8 cm
- 10 cm
- 5 cm
Correct

Incorrect

Hint

Let the midpoints of PQ and RS be A and B respectively.

(1) $\begin{equation*}\begin{split} OA^{2} + AQ^{2} &= OQ^{2} \\ 8^{2} + 6^{2} &= OQ^{2} \\ 100 &= OQ^{2} \\ 10 &= OQ \\ OB^{2} + BS^{2} &= OS^{2} \\ OB^{2} + 8^{2} &= 10^{2} \\ OB^{2} &= 36 \\ OB &= 6 \end{split}\end{equation*}$

The distance between the chord RS and the centre of circle O is 6 cm.
Question 14 of 15

14. Question
1 point(s)
The lengths of two chords PQ and RS of a circle are 20 cm and 44 cm respectively. They lie on the opposite sides of the centre of the circle. If the distance between the chords is 24 cm then find the radius of the circle.
- $10\sqrt{6}$ cm
- $10\sqrt{5}$ cm
- $6\sqrt{10}$ cm
- $5\sqrt{10}$ cm
Correct

Incorrect

Hint

Let the midpoints of PQ and RS be A and B respectively. Let $OA = x$ , therefore $OB=24-x$

(1) $\begin{equation*}\begin{split} OA^{2} + AQ^{2} &= OQ^{2} \\ x^{2} + 10^{2} &= r^{2} \\ x^{2} + 100 = r^{2} \\ OB^{2} + BS^{2} &= OS^{2} \\ (24-x)^{2} + 22^{2} &= r^{2} \\ 576-48x+x^{2}+484 &= r^{2} \end{split}\end{equation*}$

Subtracting the equations, we get

(2) $\begin{equation*}\begin{split} 576-48x+484-100 &= 0 \\ 960 &= 48x \\ 20 &= x \\ 20^{2} + 10^{2} &= r^{2} \\ 500 &= r^{2} \\ 10\sqrt{5} &= r \end{split}\end{equation*}$
Question 15 of 15

15. Question
1 point(s)
ADBO and ABCO are parallelograms and DBC is a straight line. What is the measure of the angle DAC?
- 120°
- 60°
- 72°
- 90°
Correct

Incorrect

Hint

The sides of both the parallelograms are equal to the radius. The triangle DAB and BOA are equilateral triangles. Therefore, $\angle DAB = 60^{\circ}$ . The triangles BAC and OAC are congruent to each other. Therefore, $\angle BAC = 30^{\circ}$ .

(1) $\begin{equation*}\begin{split} \angle DAC &= \angle DAB + \angle BAC \\ \angle DAC &= 60^{\circ} + 30^{\circ} \\ \angle DAC &= 90^{\circ} \end{split}\end{equation*}$