**The Arc of a Circle – Maharashtra Board Class 8 Solutions for Mathematics**

MathematicsGeneral ScienceMaharashtra Board Solutions

**Exercise – 56**

**Solution 1:**

The angles which intercept the arc LXT are ∠LNT and ∠LMT.

**Solution 2:**

The angles which intercept the arcs are:

∠COB intercepts arc CYB.

∠AOC intercepts arc AZC.

∠AOB intercepts arc BXA.

**Exercise – 57**

**Solution 1:**

Minor arcs: arc TXS, arc SYN, arc NMD, and arc DZT

Major arcs: arc TYD, arc NXD, arc TMS, and arc SZN

Semicircular arcs: arc TXN, arc TYN, arc SYD, arc SMD, arc SZD, arc SXT, arc NZT, and arc NMT

**Exercise – 58**

**Solution 1:**

Measure of the central angle = 120°

The measure of the central angle corresponding to a minor arc is also the measure of the minor arc.

Hence, the measure of the corresponding minor arc = 120°.

Measure of the major arc

= 360° – Measure of the corresponding minor arc

= 360° – 120°

= 240°

Hence, the measure of the major arc = 240°.

Thus, the measure of the corresponding minor arc is 120° and the measure of the major arc corresponding to this minor arc is 240°.

**Solution 2:**

The measure of the central angle corresponding to the minor arc is also the measure of the minor arc.

Measure of the minor arc = 160° …(given)

Hence, the measure of the central angle = 160°

Measure of the major arc

= 360° – Measure of the corresponding minor arc

= 360° – 160°

= 200°

Hence, the measure of the major arc = 200°.

Thus, the measure of the corresponding central angle is 160° and the measure of the corresponding major arc is 200°.

**Solution 3:**

The minor arcs formed by the endpoints of the diameters PN and TC are PXT, TYN, NZC and CWP.

The major arcs formed by the endpoints of the above diameters are PZT, CYP, CXN and TWN.

m∠POT = 100° …(given)

m∠NOC = 100° …(vertically opposite angles).

m∠POC = 180° – m∠POT = 180° – 100° = 80°

m∠TON = 80° …(vertically opposite angles).

Measures of minor arcs:

m(arc PXT) = m∠POT = 100°

m(arc TYN) = m∠TON = 80°

m(arc NZC) = m∠NOC = 100°

m(arc CWP) = m∠POC = 80°

Measures of major arcs:

m(arc PZT) = 360° – m(corresponding minor arc PXT)

= 360° – 100°

= 260°

m(arc CYP) = 360° – m(corresponding minor arc CWP)

= 360° – 80°

= 280°

m(arc CXN) = 360° – m(corresponding minor arc NZC)

= 360° – 100°

= 260°

m(arc TWN) = 360° – m(corresponding minor arc TYN)

= 360° – 80°

= 280°

**Exercise – 59**

**Solution 1:**

The ∠PML is inscribed in the arc PTL and the ∠TNL in the arc TPL.

**Solution 2:**

An infinite number of angles can be inscribed in the arc LMN.

**Exercise – 60**

**Solution 1:**

Measure of the central angle = 120°

Thus, the measure of the arc intercepted by a central angle of 120° is 120°.

**Solution 2:**

**Exercise – 61**

**Solution 1:**

□ABCD is a cyclic quadrilateral since all its vertices lie on the same circle.

According to the property of cyclic quadrilateral, the opposite angles of a cyclic quadrilateral are supplementary.

Thus, in □ABCD

m∠B + m∠D = 180°

∴ 85° + m∠D = 180°

∴ m∠D = 180° – 85° = 95°

Also, m∠C + m∠A = 180°

∴ 105° + m∠A = 180°

∴ m∠A = 180° – 105° = 75°

Thus, m∠A = 75° and m∠D = 95°.

**Solution 2:**

According to the property of cyclic quadrilaterals, the opposite angles of a cyclic quadrilateral are supplementary.

Now, it is given that the opposite angles of the cyclic quadrilateral are x and 3x.

∴ x + 3x = 180°

∴ 4x = 180°

∴ x = 45°

∴ 3x = 3 × 45 = 135°

Hence, the measures of the angles of the given quadrilateral are 45° and 135° respectively.