The Arc Of A Circle
Class 8th Mathematics (old) MHB Solution
- In the figure alongside, name the angles which intercept the arc LXT.…
- Observe the figure and name the angles which intercept arcs and name the arc…
- If the measure of a central angle is 120°, find the measure of the…
- The measure of the minor arc of a circle is 160°. Hence, find the measures of…
- In the figure, seg PN and seg TC are the diameters of a circle. Write the…
- Observe the figure given alongside and fill in the blanks. The ∠PML is…
- Draw a circle. Mark an arc LMN on this circle. Inscribe an angle LMN in it. How…
Exercise 56
Question 1.In the figure alongside, name the angles which intercept the arc LXT.
Answer:
According to the diagram,
Arc LXT is intercepted by 2 angles
⇒ one angle is made by point M
As it intersect through point L & T
∴∠LMT isintercept of Arc LXT
⇒ another angle is made by point N
As it also intersect through point L & T
∴∠LMT isanother intercept of Arc LXT
∴∠LMT and∠LNT are the angles which intercept at Arc LXT
Question 2.
Observe the figure and name the angles which intercept arcs and name the arc that each angle intercepts.
Answer:
The angles which intercept arcs are :-
∠ AOB
∠ BOC
∠ COA
In arc AXB
Point A, B are on arms AO & BO
∠ AOB is intercept of arc AXB
In arc BYC
Point B, C are on arms BO & CO
∠ BOC is intercept of arc BYC
In arc CZA
Point C,A are on arms CO & AO
∠ COA is intercept of arc CZA
Exercise 57
Question 1.In the figure, seg SD and seg TN are diameters. Name those minor arcs, major arcs and semicircular arcs whose endpoints are also the endpoints of these diameters.
Answer:
According to the diagram
⇒ Taking point S & T of diameters
Minor arc = SXT
Major arc = SMT
⇒ Taking point S & N of diameters
Minor arc = SYT
Major arc = SZN
⇒ Taking point N & D of diameters
Minor arc = NMD
Major arc = NXD
⇒ Taking point D & T of diameters
Minor arc = TZD
Major arc = TYD
As we know that:-
SD & TN are diameters of circle
⇒ Taking diameter SD
Semicircular arc are STD & SND
⇒ Taking diameter TN
Semicircular arc are TSN & TDN
Exercise 58
Question 1.If the measure of a central angle is 120°, find the measure of the corresponding minor arc and that of the major arc corresponding to this minor arc.
Answer:
If the central angle is 120°
The measure of the corresponding minor arc
m(minor arc) = 120°
Measure of the major arc =
360° - measure of the corresponding minor arc
m(major arc) = 360°-120˚
= 240°
Question 2.
The measure of the minor arc of a circle is 160°. Hence, find the measures of the corresponding central angle and the corresponding major arc.
Answer:
If the m(minor arc) is 160°
The measure of the corresponding central angle
m(minor arc) = central angle
Central angle = 160°
⇒ Measure of the major arc =
360°- measure of the corresponding minor arc
∴ m(major arc) = 360°-160°
= 200°
Question 3.
In the figure, seg PN and seg TC are the diameters of a circle. Write the measures of the arcs formed by the endpoints of these diameters and explain your answers.
Answer:
PN is a straight line
So
∠ POT + ∠ TON = 180°
∠ TON = 180°-100°
∠ POT = ∠ CON = 100°
∠ POC = ∠ TON = 80°
According to the diagram
⇒ Taking point P & C of diameters
Minor arc = m(PWC) = 80°
Major arc = m(PYC) = 360°-80° = 280
⇒ Taking point C & N of diameters
Minor arc = m(CZN) = 100°
Major arc = m(CXN) = 360°-100° = 260°
⇒ Taking point N & T of diameters
Minor arc = m(NYT) = 80°
Major arc = m(NWT) = 360°-80° = 280°
⇒ Taking point P & T of diameters
Minor arc = m(PXT) = 100°
Major arc = m(PZT) = 360° -100° = 260°
PN & TC are diameters of circle
⇒ Taking diameter PN
Semicircular arc are PTN & PCN
m(PTN) = m(PCN) = 180°
⇒ Taking diameter TC
Semicircular arc are TPC & TNC
m(TPC) = m(TNC) = 180°
Exercise 59
Question 1.Observe the figure given alongside and fill in the blanks. The ∠PML is inscribed in the arc …………. and the ∠TNL in the arc ……….
Answer:
⇒ Vertex M is on arc PML
And endpoints P, L of arc PML are on arms PM&ML of ∠ PML
∴ ∠ PML is inscribed on arc PML
⇒ Vertex N is on arc TNL
And endpoints T, L of arc TNL are on arms TN&NL of ∠ TNL
∴ ∠ TNL is inscribed on arc TNL
Question 2.
Draw a circle. Mark an arc LMN on this circle. Inscribe an angle LMN in it.
How many more angles can you inscribe in this arc?
Answer:
In an arc LMN
The vertex M is on the arc
And the endpoints L AND M are on arms of ∠ LMN
Hence ∠ LMNis inscribed on arc LMN
∴ There can be infinite points on arc LMN
And ∴ Infinite number of angles can inscribed in arc LMN
which are passing through end points L & M
Exercise 60
Question 1.The measure of a central angle is 120°. Find the measure of the arc it intercepts.
Answer:
If the central angle is 120°
The arc at which it intercept is the minor arc BDC
The measure of the arc BDC
m(arc BDC) = 120°
Question 2.
In the figure, arc AXB is a semicircle. m∠PAB = 40°. Hence, find the values of
(1) m∠APB,
(2) m(arc PYB),
(3) m(arc AZP),
Answer:
As we know that
Inscribed angle of arc is half the angle intercepted by arc
1. ⇒ ∠ APB = 
*∠ AOB
∠ APB = *180°
∠ APB = 90°
2. ⇒ m(arc PYB) = ∠ POB
∠ POB = 2*∠ PAB
∠ POB = 2*40° = 80°
∴ m(arc PYB) = 80°
3. In Δ APB
∠ A + ∠ P + ∠ B = 180°
∠ B = 180°-∠ P-∠ A
∠ B = 180°-90°-40°
∠ B = 50°
⇒ m(arc AZP) = ∠ POA
∠ POA = 2*∠ PBA
∠ POA = 2*50°
∴ m(arc AZP) = 100°
Exercise 61
Question 1.In the figure, m∠B = 85° and m∠C = 105°, then find the measures of ∠A and ∠D.
Answer:
As we know that
Sum of Opposite angles of cyclic quadrilateral are supplementary
If the opposite angles are ∠A&∠C ; ∠B&∠D
⇒ ∠A + ∠C = 180°
∠A = 180°-105° = 75°
⇒ ∠B + ∠D = 180°
∠ B = 180°-85° = 95°
∴∠A and∠B are 75°and95°
Question 2.
The opposite angles of a cyclic quadrilateral are x and 3x. What are the measures of the angles of this quadrilateral?
Answer:
As we know that
Sum of Opposite angles of cyclic quadrilateral are supplementary
If the opposite angles are x & 3x
∴ x + 3x = 180°
∴ 4x = 180°
∴ x = 
∴ x = 45°
⇒ 3x = 3*45° = 135°
Hence the angles are 