17. Chapter 17. Secants , Tangents and Their Properties NIOS Class 10 Mathematics Textbook Solutions
Chapter 17. Secants , Tangents and Their Properties NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 17. SECANTS, TANGENTS AND THEIR PROPERTIES
CHECK YOUR PROGRESS 17.1
1. Fill in the blanks:
(i) A tangent is __________ to the radius through the point of contact.
(ii) The lengths of tangents from an external point to a circle are __________
(iii) A tangent is the limiting position of a secant when the two ______ coincide.
(iv) From an external point ________ tangents can be drawn to a circle.
(v) From a point in the interior of the circle, ______ tangent(s) can be drawn to the circle.
Answer: (i) A tangent is perpendicular to the radius through the point of contact.
(ii) The lengths of tangents from an external point to a circle are equal .
(iii) A tangent is the limiting position of a secant when the two points of intersection coincide.
(iv) From an external point two tangents can be drawn to a circle.
(v) From a point in the interior of the circle, no tangent(s) can be drawn to the circle.
2. In Fig. 17.11, ∠POY = 40° , Find the ∠OYP and ∠OYT.
Solution: Given, ∠POY = 40° , ∠OPY = 90° 
In ∆OPY 
, we have
∠OPY+∠POY+∠OYP=180° 
⇒90°+40°+∠OYP=180° 
⇒130°+∠OYP=180° 
⇒∠OYP=180°-130° 
⇒∠OYP=50° 
We join OT .
In ∆OYP 
and ∆OYT 
OP=OT 
[radius]
PY=TY 
[Tangent ]
∠OPY=∠OTY=90° 
∆OYP≅∆OYT 
[RHS]
∠OYP=∠OYT=50° 
∠OYT=50° 
3. In Fig. 17.12, the incircle of ΔPQR is drawn. If PX = 2.5 cm, RZ = 3.5 cm and perimeter of ΔPQR = 18 cm, find the length of QY.
Photo
Solution: Here, PX = 2.5 cm, RZ = 3.5 cm and perimeter of ΔPQR = 18 cm .
We know that, the lengths of two tangents from an external point are equal.
So, PX = PZ = 2.5 cm , RZ = RY = 3.5 cm , QY = QX
The perimeter of ΔPQR = PQ + QR + PR = PX+QX+QY+RY+PZ+RZ
⇒18=2.5+QY+QY+3.5+2.5+3.5 
⇒18=12+2QY 
⇒2QY=18-12=6 
⇒QY=62
⇒QY=3 cm
4. Write an experiment to show that the lengths of tangents from an external point to a circle are equal.
Solution: Given, A circle with centre O. PT and PT′ are two tangents from a point P outside the
circle.
To Prove: PT = PT′
Construction: Join OP, OT and OT′ (see Fig. 17.6)
Photo
Proof: In ΔOPT and ΔOPT′
∠ OTP = ∠ OT′P (Each being right angle)
OT = OT′
OP = OP (Common)
ΔOPT ≅ ΔOPT′ (RHS criterion)
∴ PT = PT′
The lengths of two tangents from an external point are equal .
CHECK YOUR PROGRESS 17.2
1. In Fig. 17.19, if PA = 3 cm, PB = 6 cm and PD = 4 cm then find the length of PC.
Solution: Given, PA = 3 cm, PB = 6 cm and PD = 4 cm
We know that,
PA×PB =PC×PD 
⇒3×6=PC×4 
⇒PC=184=92
⇒PC=4.5 cm 
2. In Fig. 17.19, PA = 4 cm, PB = x + 3, PD = 3 cm and PC = x + 5, find the value of x.
Solution: Given, PA = 4 cm, PB = x + 3, PD = 3 cm and PC = x + 5
We know that, PA ×PB =PC×PD
⇒4×x+3=x+5×3 
⇒4x+12=3x+15 
⇒4x-3x=15-12 
⇒x=3 cm 
3. If Fig. 17.20, if PA = 4cm, PB = 10 cm, PC = 5 cm, find PD.
Solution: Given, PA = 4cm, PB = 10 cm, PC = 5 cm and PD = ?
We know that,
PA×PB=PC×PD 
⇒4×10=5×PD 
⇒PD=4×105
⇒PD=4×2
⇒PD=8 cm 
4. In Fig. 17.20, if PC = 4 cm, PD = (x + 5) cm, PA = 5 cm and PB = (x + 2) cm, find x.
Solution: Given, PC = 4 cm, PD = (x + 5) cm, PA = 5 cm and PB = (x + 2) cm
We know that,
PA×PB=PC×PD
⇒5×x+2=4×x+5
⇒5x+10=4x+20 
⇒5x-4x=20-10 
⇒x=10 
cm
5. In Fig. 17.21, PT = 
cm, OP = 8 cm, find the radius of the circle, if O is the centre of the circle.
Solution: Given, PT = cm, OP = 8 cm
We know that,
PA×PB=PT2
⇒OP-OAOP+OA=PT2
[OA = OB]
⇒OP2-OA2=272 
⇒82-OA2=4×7 
⇒64-28=OA2
⇒OA=36=6 
⇒OA=6 cm 
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