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Andhra Pradesh SSC Class 10 Solutions For Maths – Tangent and Secants to a Circle (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 9 Tangent and Secants to a Circle
Exercise 9.1:
Question 1:
Fill in the blanks
- A tangent to a circle intersects it in …… point (s).
- A line intersecting a circle in two points is called a ……..
- A circle can have …… parallel tangents at the most.
- The common point of a tangent to a circle and the circle is called ………
- We can draw ……. tangents to a given circle.
Solution :
- One
- Secant of a circle
- Two
- Point of contact
- Infinite
AP SSC 10th Class Textbook Solutions
Question 2:
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Solution :
Question 3:
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution :
Steps of construction:
- Draw a circle with centre ‘O’ and radius ‘r’
- Mark the point ‘P’ anywhere on the circles. Join OP.
- Draw a perpendicular line through the point P and name it as AB, as shown in figure.
- AB is the required tangent to the given circle.
- Draw another line parallel to the tangent AB and intersect circle at two points and name it as CD
- CD is the required secant.
Question 4:
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Solution :
Question 5:
Prove that the tangents to a circle at the end points of a diameter are parallel.
Solution :
Two lines are parallel to each other only if a line that cuts through them makes the same alternate angles with both of them.
In the case, since the diameter is perpendiculer to the tangent, the two lines are perpendiculer to the diameter on the either side. Each line makes an angle of 90 degrees with diameter.
Hence, the alternate anglers are 90°
∴ Tangents drawn at both ends of diameter are parallel to each other
Exercise 9.2:
Question 1:
- Choose the correct answer and give justification for each. The angle between a tangent to a circle and the radius drawn at the point of contact is
- From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
- If AP and AQ are the two tangents a circle with centre O so that ∠POQ = 1100, then ∠PAQ is equal to
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then ∠POA is equal to- In the figure XY and X1 Y1 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X1 Y1 at B then ∠AOB =
Solution :
i. d. Since the tangent at any point is perpendicular to the radius through the point of contact.
ii. a.
Tangent is perpendicular to the radius at point of contact here PQ is tangent segment and OP is radius of circle
OP ⊥ PQ ⇒ ∠OPQ = 90°
Now in ∆OPQ,
OQ2 = OP2 + PQ2 (Pythagoras theorem)
242 = 242 + 252
625 = 576 + PQ2
49 = PQ2
PQ = 7 cm
iii. From figure
Here; ∠OPT = ∠OQT = 90° (Since radius is perpendicular to tangent)
Hence, ∠POQ + ∠PTQ = 180°
Or, 110° + ∠PTQ = 180°
Or, ∠PTQ = 180° – 110°
Or, ∠PTQ = 70°
Ans: b.
iv. a ∠AOB=360°-180°-80°=100°, ∠POA=∠AOB/2=50°
v. c If we join O and C we get ∠OCA=90°
∠OPA = 90°, we get OPAC is a square. Similarly OCBQ is square. ⇒ ∠AOB=90°
Question 2:
Two concentric circles of radii 5 cm and 3cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Solution :
Given AO=5cm (radii of larger circle)
OD=3cm(radii of smaller circle)
So, AO2=OD2+AD2 (Pythagoras theorem)
52=32+AD2
25-9=AD2
AD=4
Since AD=DB
AB=8cm.
Question 3:
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution :
Construction:
Draw a circle with centre O. Draw parallelogram ABCD which touches the circle P, Q, R and S
Given, AB ∥ DC
AD ∥ BC
To Prove: ABCD is a rhombus
In ∆AOB and ∆DOC;
AB = DC (given that AD ∥ BC)
∠AOB = ∠DOC (Vertically opposite angles)
∠BAO = ∠DCO (Alternate angles)
Hence; ∆AOB ≈ ∆DOC
Hence; AO = CO and BO = DO
Since diagonals are bisecting each other, so given parallelogram is a rhombus.
Question 4:
A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See adjacent figure). Find the sides AB and AC.
Solution :
Since the external tangent are equal
DC=CE=3 cm
BD=BF=9 cm
Let AF=AE= x cm
AB2=BC2+AC2 (using Pythagoras theorem)
(AF+FB)2=(BD+DC)2+(AE+EC)2
(x+9)2=122+(3+x)2
X2+18x+81=144+9+6x+x2
12x=72
X=6
So, AB=AF+FB=15cm
AC=EC+AE= 9cm.
Question 5:
Draw a circle of radius 6cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythogoras Theorem.
Solution :
AB2+AO2=OB2 (using pythogaros theorem)
AB2+62=102
AB2=100-36
AB=8 cm.
Question 6:
Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6cm and measure its length. Also verify the measurement by actual calculation.
Solution :
Steps of Construction:
Step 1: Draw two concentric circles with centre O and radii
4cm and 6 cm such that OP = 6 cm, OQ = 4 cm.
Step 2: Join OP and bisect it at M. i.e. M is the mid-point of OP
i.e. OM = PM = 3 cm.
Step 3: Taking M as centre with OM as radius draw a circle
intersecting the smaller circle in two points namely T and S.
Step 4: Join PT and PS.
PT and PS are the required tangents from a point P to the smaller circle, whose radius is 4 cm. By measurement: PT = 4.5 cm.
Verification. OTP is right Δ at T
OP2=OT2+PT2
PT2=OP2-OT2
PT2=26-16=20
PT=2 √5=4.48cm.
Question 7:
Draw a circle with the help of a bangle, Take a point outside the circle. Construct the pair of tangents from this point to the circle measure them. Write conclusion.
Solution :
Steps of construction:
- Draw a circle, take a point outside the circle anywhere let it be R.
- Join RO and draw a perpendicular bisector of it. Let M be the midpoint of RO
- Make an arc across the circle at two places such as A and B.
- Join RA and RB. Then RA and RB are the required two tangents.
Diagram:
Question 8:
In a right triangle ABC, a circle with a side AB as diameter is to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Solution :
Question 9:
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? Hint : The distance of two points to the point of contact is the same.
Solution :
Given: We are given a circle with centre ‘O’ and o point R outside it. We have to construct tangent from point R to the circle.
Steps of construction:
- Join RO and draw a perpendicular bisector of it. Let M be the midpoint of RO
- Make an arc across the circle at two places such as A and B.
- Join RA and RB. Then RA and RB are the required two tangents.
Diagram:
Exercise 9.3:
Question 1:
A chord of a circle of radius 10 cm. subtends aright angle at the centre. Find the area of the corresponding: (use π = 3.14)
- Minor segment
- Major segment
Solution :
Question 2:
A chord of a circle of radius 12 cm. subtends an angle of 1200 at the centre. Find the area of the corresponding minor segment of the circle (use π 3.14 and \(\sqrt{3}\) = 1.732)
Solution :
Question 3:
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 1150 Find the total area cleaned at each sweep of the blades. (use π = \(\frac{22}{7}\) )
Solution :
Question 4:
Find the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are with each side of the square as diameter (use π = 3.14)
Solution :
Question 5:
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use π = \(\frac{22}{7}\))
Solution :
Area of square=72=49cm2
Area of 2 semi-circle=πr2=3.14×3.5×3.5=38.465cm2
Area of the shaded region
=area of square-area of 2 semicircle
=49-38.465
=10.5cm2.
Question 6:
In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm. , find the area of the shaded region. (use π = \(\frac{22}{7}\))
Solution :
Question 7:
AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 300, find the area of the shaded region. (use π = \(\frac{22}{7}\))
Solution :
Question 8:
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm. each. (use π = 3.14)
Solution :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 9 Tangent and Secants to a Circle are helpful to complete your math homework.
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