Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 116:

Question 1:

In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB).

Answer:

P is the centre of the circle.
AB = 13 cm
seg PQ ⊥ chord AB
Perpendicular drawn from the centre of the circle to the chord bisects the chord. 
So, the length of QB = 12AB=132=6.5 cm.
 

Page No 116:

Question 2:

Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Answer:

Perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, PD=12CD=482=24 cm
In OPD,
We apply the Pythagoras theorem,
OP2+PD2=OD2OP2+242=252OP2=252-242=625-576=49OP2=49OP=7 cm
Hence, the distance of the chord CD from the centre O is 7 cm.

Page No 116:

Question 3:

O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the circle.

Answer:

Join OA. 

Let the perpendicular drawn from point O to the chord AB be P.
We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, AP=AB2=242=12 cm
In OPA,
We apply the Pythagoras theorem,
OP2+AP2=OA292+122=OA2OA2=81+144=225OA=225=15 cm
Hence, the radius of the circle is 15 cm.

Page No 116:

Question 4:

C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.

Answer:


C is the centre of the circle.
Radius = 10 cm
Let the chord be AB. 
Distance of the centre to the chord = AB.
CD is perpendiculaar to the chord AB. 
Perpendicular drawn from the centre of the circle to the chord bisects the chord. 
AD=AB2=122=6 cm
In ACD,
We apply the Pythagoras theorem
CD2+AD2=AC2CD2+62=102CD2+36=100CD2=64CD=8 cm
Thus, distance of the chord from the centre is 8 cm. 



Page No 118:

Question 1:

The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. state, why arc PS and arc SQ are congruent. Write the other arcs which are congruent to arc PS

Answer:

PQ perpendicular to RS.
PCS=PCQ=90°arcPS=arcQS           
We know that if the measures of two arcs of circle are same then two arcs are congruent
arcPSarcQS
Similarly, 
arcPSarcPRarcRQ
 

Page No 118:

Question 2:

In the adjoining figure O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. hence find the following

(1) m∠ AOB and m∠ COD
(2) Show that arc AB ≌ arc CD.
(3) Show that chord AB ≌ chord CD

Answer:

(1) MN is the diameter of the circle with centre O. 
MOA+AOB+BON=180°100°+AOB+35°=180°AOB=180°-100°-35°=45°
Similarly, COD=45°
(2) Eqaual arcs subtend equal angles at the centre. 
AOB=COD=45°
So, arc AB ≌ arc CD
(3) Corresponding chords of congruent arcs are congruent.
arc AB ≌ arc CD
So, chord AB ≌ chord CD



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