AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.4

Question 1.

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:

’O’ is the centre

∠AOB = 100°

Thus ∠ACB = \(\frac{1}{2}\) ∠AOB

[∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]]

= \(\frac{1}{2}\) x 100° = 50°

∠ACB and ∠ADB are supplementary

[ ∵ Opp. angles of a cyclic quadrilateral]

∴ ∠ADB = 180°-50° = 130°

[OR]

∠ADB is the angle made by the major arc \(\widehat{\mathrm{ACB}}\) at D.

∴ ∠ADB = \(\frac{1}{2}\)∠AOB [where ∠AOB is the angle; made by \(\widehat{\mathrm{ACB}}\) at the centre]

= \(\frac{1}{2}\) [360° – 100°] [from the figure]

= \(\frac{1}{2}\) x 260° = 130°

Question 2.

In the figure, ∠BAD = 40° then find ∠BCD.

Solution:

‘O’ is the centre of the circle.

∴ In ΔOAB; OA = OB (radii)

∴ ∠OAB = ∠OBA = 40°

(∵ angles opp. to equal sides)

Now ∠AOB = 180° – (40° + 40°)

(∵ angle sum property of ΔOAB)

= 180°-80° = 100°

But ∠AOB = ∠COD = 100°

Also ∠OCD = ∠ODC [OC = OD]

= 40° as in ΔOAB

∴ ∠BCD = 40°

(OR)

In ΔOAB and ΔOCD

OA = OD (radii)

OB = OC (radii)

∠AOB = ∠COD (vertically opp. angles)

∴ ΔOAB ≅ ΔOCD

∴ ∠BCD = ∠OBA = 40°

[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

Question 3.

In the figure, ‘O’ is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.

Solution:

‘O’ is the centre; ∠POR = 120°

∠PQR = \(\frac{1}{2}\)∠POR [∵ angle made by an arc at the centre is, twice the angle made by it on the remaining part]

∠PSR = \(\frac{1}{2}\) [Angle made by \(\widehat{\mathrm{PQR}}\) at the centre]

∠PSR = \(\frac{1}{2}\) [360° – 120°] from the fig.

= \(\frac{1}{2}\) x 240 = 120°

Question 4.

If a parallelogram is cyclic, then it is a rectangle. Justify.

Solution:

Let □ABCD be a parallelogram such

that A, B, C and D lie on the circle.

∴∠A + ∠C = 180° and ∠B + ∠D = 180°

[Opp. angles of a cyclic quadri lateral are supplementary]

But ∠A = ∠C and ∠B = ∠D

[∵ Opp. angles of a ||gm are equal]

∴∠A = ∠C =∠B =∠D = \(\frac{180}{2}\) = 90°

Hence □ABCD is a rectangle

Question 5.

In the figure, ‘O’ is the centr of the circle. 0M = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.

OM bisects AB.

∴ AM = \(\frac{\mathrm{AB}}{2}=\frac{8}{2}\) = 4 cm

OA^{2} = OM^{2} + AM^{2} [ ∵ Pythagoras theorem]

OA \(\begin{array}{l}

=\sqrt{3^{2}+4^{2}} \\

=\sqrt{9+16}=\sqrt{25}

\end{array}\)

= 5cm

Question 6.

In the figure, ‘O’ is the centre of the circle and OM, ON are the perpen-diculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6 cm. Find RS.

Solution:

‘O’ is the centre of the circle.

OM = ON and 0M ⊥ PQ; ON ⊥ RS

Thus the chords FQ and RS are equal.

[ ∵ chords which are equidistant from the centre are equal in length]

∴ RS = PQ = 6cm

Question 7.

A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the

radius of the circle.

Solution:

A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle.

∴ Radius = 4 cm.

Question 8.

Draw a circle with any radius and then draw two chords equidistant

from the centre.

Solution:

- Draw a circle with centre P.
- Draw any two radii.
- Mark off two points M and N oh these radii. Such that PM = PN.
- Draw perpendicular through M and N to these radii.

Question 9.

In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD.

Solution:

‘O’ is the centre of the circle.

AB, CD are equal chords

⇒ They subtend equal angles at the centre.

∴ ∠AOB =∠COD = 70°

Now in ΔOCD

∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]

∴ ∠OCD + ∠ODC + 70° = 180°

= ∠OCD +∠ODC = 180° – 70° = 110°

∴ ∠OCD + ∠ODC = 110° = 55°