### Practice Set 3.4 Circle Class 10th Mathematics Part 2 MHB Solution

Practice Set 3.4

1. In figure 3.56, in a circle with centre O, length of chord AB is equal to the radius of…
2. In figure 3.57, is cyclic. side PQ ≅ side RQ. ∠ PSR = 110°, Find-(1) measure of ∠…
3. is cyclic, ∠R = (5x - 13)°, ∠N = (4x + 4)°. Find measures of angle r and triangle…
4. In figure 3.58, seg RS is a diameter of the circle with centre O. Point T lies in the…
5. Prove that, any rectangle is a cyclic quadrilateral.
6. In figure 3.59, altitudes YZ and XT of∆ WXY intersect at P. Prove that, x^{w}sum t (1)…
7. In figure 3.60, m(arc NS) = 125°, m(arc EF) = 37°, find the measure ∠ NMS.…
8. In figure 3.61, chords AC and DE intersect at B. If ∠ ABE = 108°, m(arc AE) = 95°, find…

###### Practice Set 3.4
Question 1.

In figure 3.56, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following. (1) ∠ AOB (2) ∠ ACB

(3) arc AB (4) arc ACB.

(1)In ∆AOB,

AB = OA = OB = radius of circle

⇒ ∆AOB is an equilateral triangle

∠ AOB + ∠ ABO + ∠ BAO = 180° {Angle sum property}

⇒ 3∠ AOB = 180° {All the angles are equal}

∠ AOB = 60°

(2)∠ AOB = 2 × ∠ ACB {The measure of an inscribed angle is half the measure of the arc intercepted by it.}

⇒∠ ACB = 30°

(3)∠ AOB = 60°

⇒arc(AB) = 60° {The measure of a minor arc is the measure of its central angle.}

(4) Using Measure of a major arc = 360°- measure of its corresponding minor arc

⇒arc(ACB) = 360° - arc(AB)

⇒arc(ACB) = 360° - 60° = 300°

Question 2.

In figure 3.57, is cyclic. side PQ ≅ side RQ. ∠ PSR = 110°, Find- (1) measure of ∠ PQR

(2) m(arc PQR)

(3) m(arc QR)

(4) measure of ∠ PRQ

(1) Given PQRS is a cyclic quadrilateral.

∵Opposite angles of a cyclic quadrilateral are supplementary

⇒∠ PSR + ∠ PQR = 180°

⇒∠ PQR = 180° - 110°

⇒∠ PQR = 70°

(2)2 × ∠ PQR = m(arc PR){The measure of an inscribed angle is half the measure of the arc intercepted by it.}

m(arc PR) = 140°

⇒m(arc PQR) = 360° -140° = 220° {Using Measure of a major arc = 360°- measure of its corresponding minor arc}

(3)side PQ ≅ side RQ

∴m(arc PQ) = m(arc RQ){Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent}

⇒m(arc PQR) = m(arc PQ) + m(arc RQ)

⇒m(arc PQR) = 2 × m(arc PQ)

⇒m(arc PQ) = 110°

(4)In ∆ PQR,

∠ PQR + ∠ QRP + ∠ RPQ = 180°{Angle sum property}

⇒∠ PRQ + ∠ RPQ = 180° - ∠ PQR

⇒ 2∠ PRQ = 180° - 70° {∵side PQ ≅ side RQ}

⇒∠ PRQ = 55°

Question 3. is cyclic, ∠R = (5x - 13)°, ∠N = (4x + 4)°. Find measures of and Given MRPN is a cyclic quadrilateral.

⇒∠ R + ∠ N = 180° {Using Opposite angles of a cyclic quadrilateral are supplementary}

⇒ (5x - 13)° + (4x + 4)° = 180°

⇒9x - 9 = 180°

⇒ x – 1 = 20°

⇒ x = 21°

∠R = (5x - 13)° = 5 × 21 – 13 = 105 – 13 = 92°

∠N = (4x + 4)° = 4 × 21 + 4 = 84 + 4 = 88°

Question 4.

In figure 3.58, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠ RTS is an acute angle.  Given RS is the diameter

⇒ ∠ ROS = 180°

m(arc RS) = 180°

Now, ∠ RTS is an external angle.   Hence, ∠ RTS is an acute angle.

Question 5.

Prove that, any rectangle is a cyclic quadrilateral. In ABCD,

∠A = 90°{∵ angle of a rectangle is 90°.}

∠C = 90° {opposite angles are equals}

⇒∠ A + ∠ C = 180°

If opposite angles are supplementary, the quadrilateral is cyclic.

∴ ABCD is cyclic.

Question 6.

In figure 3.59, altitudes YZ and XT of

∆ WXY intersect at P. Prove that, (1) WZPT is cyclic.

(2) Points X, Z, T, Y are concyclic.

(1)In WZPT,

∠ WZP = ∠ WTP = 90° {YZ and XT are the altitudes}

If a pair of opposite angles of a quadrilateral is supplementary, then the

⇒ WZPT is cyclic.

(2)∵ X, Z,T,Y lie on same circle, ∴ they are concyclic.

Question 7.

In figure 3.60, m(arc NS) = 125°, m(arc EF) = 37°, find the measure ∠ NMS. Given m(arc NS) = 125°, m(arc EF) = 37°

Also, ∠ NMS is an external angle, so   Question 8.

In figure 3.61, chords AC and DE intersect at B. If ∠ ABE = 108°, m(arc AE) = 95°, find m(arc DC). Given ∠ ABE = 108°, m(arc AE) = 95°

Using the property of the secant,  ⇒m(arc DC) = 108° × 2 – 95°

⇒m(arc DC) = 121°

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