Chapter 10 Circles Ex 10.4 Solutions
Question - 1 : - Two circles of radii 5 cm and 3 cm intersect at two points and thedistance between their centres is 4 cm. Find the length of the common chord.
Answer - 1 : -
Given parameters are:
OP = 5cm
OS = 4cm and
PS = 3cm
Also, PQ = 2PR
Now, suppose RS = x.The diagram for the same is shown below.
Consider the ΔPOR,
OP2 =OR2+PR2
⇒ 52 =(4-x)2+PR2
⇒ 25 = 16+x2-8x+PR2
∴ PR2 =9-x2+8x — (i)
Now consider ΔPRS,
PS2 =PR2+RS2
⇒ 32 =PR2+x2
∴ PR2 =9-x2 — (ii)
By equating equation(i) and equation (ii) we get,
9 -x2+8x =9-x2
⇒ 8x = 0
⇒ x = 0
Now, put the value ofx in equation (i)
PR2 =9-02
⇒ PR = 3cm
∴ The length of thecord i.e. PQ = 2PR
So, PQ = 2×3 = 6cm
Question - 2 : - If two equal chords of a circle intersect within the circle, provethat the segments of one chord are equal to corresponding segments of the otherchord.
Answer - 2 : -
Let AB and CD be twoequal cords (i.e. AB = CD). In the above question, it is given that AB and CDintersect at a point, say, E.
It is now to be proventhat the line segments AE = DE and CE = BE
Construction Steps:
Step 1: From the center of thecircle, draw a perpendicular to AB i.e. OM ⊥ AB
Step 2: Similarly, draw ON ⊥ CD.
Step 3: Join OE.
Now, the diagram is asfollows-
Proof:
From the diagram, itis seen that OM bisects AB and so, OM ⊥ AB
Similarly, ON bisectsCD and so, ON ⊥ CD
It is known that AB =CD. So,
AM = ND — (i)
and MB = CN — (ii)
Now, triangles ΔOMEand ΔONE are similar by RHS congruency since
OME = ONE (They areperpendiculars)
OE = OE (It is thecommon side)
OM = ON (AB and CD areequal and so, they are equidistant from the centre)
∴ ΔOME ΔONE
ME = EN (by CPCT) —(iii)
Now, from equations(i) and (ii) we get,
AM+ME = ND+EN
So, AE = ED
Now from equations(ii) and (iii) we get,
MB-ME = CN-EN
So, EB = CE (Henceproved).
Question - 3 : - If two equal chords of a circle intersect within the circle, provethat the line joining the point of intersection to the centre makes equalangles with the chords.
Answer - 3 : -
From the question weknow the following:
(i) AB and CD are 2chords which are intersecting at point E.
(ii) PQ is thediameter of the circle.
(iii) AB = CD.
Now, we will have toprove that BEQ = CEQ
For this, thefollowing construction has to be done:
Construction:
Draw two perpendiculars aredrawn as OM ⊥ AB and ON ⊥ D. Now, join OE. Theconstructed diagram will look as follows:Now, consider thetriangles ΔOEM and ΔOEN.
Here,
(i) OM = ON [Since theequal chords are always equidistant from the centre]
(ii) OE = OE [It isthe common side]
(iii) OME = ONE [Theseare the perpendiculars]
So, by RHS congruencycriterion, ΔOEM ΔOEN.
Hence, by CPCT rule,MEO = NEO
∴ BEQ = CEQ (Hence proved).4. If a line intersects two concentric circles (circles with the samecentre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).
Answer - 4 : -
The given image is asfollows:
First, draw a linesegment from O to AD such that OM ⊥ AD.
So, now OM isbisecting AD since OM ⊥ AD.
Therefore, AM = MD —(i)
Also, since OM ⊥ BC, OM bisects BC.
Therefore, BM = MC —(ii)
From equation (i) andequation (ii),
AM-BM = MD-MC
∴ AB = CD
Question - 5 : - Three girls Reshma, Salma and Mandip are playing a game by standing ona circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma toMandip, Mandip to Reshma. If the distance between Reshma and Salma and betweenSalma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Answer - 5 : - 
Let the positions ofReshma, Salma and Mandip be represented as A, B and C respectively.
From the question, weknow that AB = BC = 6cm.
So, the radius of thecircle i.e. OA = 5cm
Now, draw aperpendicular BM ⊥ AC.
Since AB = BC, ABC canbe considered as an isosceles triangle. M is mid-point of AC. BM is theperpendicular bisector of AC and thus it passes through the centre of thecircle.
Now,
let AM = y and
OM = x
So, BM will be =(5-x).
By applying Pythagoreantheorem in ΔOAM we get,
OA2 =OM2 +AM2
⇒ 52 =x2 +y2 — (i)
Again, by applyingPythagorean theorem in ΔAMB,
AB2 =BM2 +AM2
⇒ 62 =(5-x)2+y2 — (ii)
Subtracting equation(i) from equation (ii), we get
36-25 = (5-x)2 +y2 -x2-y2
Now, solving thisequation we get the value of x as
x = 7/5
Substituting the valueof x in equation (i), we get
y2 +(49/25)= 25
⇒ y2 =25 – (49/25)
Solving it we get thevalue of y as
y = 24/5
Thus,
AC = 2×AM
= 2×y
= 2×(24/5) m
AC = 9.6 m
So, the distancebetween Reshma and Mandip is 9.6 m.
A circular park of radius 20m is situated in a colony. Three boysAnkur, Syed and David are sitting at equal distance on its boundary each havinga toy telephone in his hands to talk each other. Find the length of the stringof each phone.
Answer - 6 : -
First, draw a diagramaccording to the given statements. The diagram will look as follows.
Here the positions ofAnkur, Syed and David are represented as A, B and C respectively. Since theyare sitting at equal distances, the triangle ABC will form an equilateraltriangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and itpasses through the centre O.
Also, O is thecentroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of atriangle a metres then BD = a/2 m.
Applying Pythagorastheorem in ΔABD,
AB2 =BD2+AD2
⇒ AD2 =AB2 -BD2
⇒ AD2 =a2 -(a/2)2
⇒ AD2 =3a2/4
⇒ AD = √3a/2
OA = 2/3 AD
20 m = 2/3 × √3a/2
a = 20√3 m
So, the length of thestring of the toy is 20√3 m.