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Class 10 Most Important Questions on Chapter 10 -Circle
Class 10 Most Important Questions on Chapter 10 -Circle are helpful for boosting the preparation of class 10 CBSE Board exam Term 2.Class 10 Most Important Questions on Chapter 10 -Circle will give you an idea about the type of questions which are going to ask in class 10 CBSE Board exam Term 2. Along NCERT Solutions of Chapter 10 -Circle the study of Class 10 Most Important Questions on Chapter 10 -Circle are also important for the preparation of class 10 CBSE Board exam Term 2.
You can study NCERT Solutions For Class 10 Maths Chapter 10 Circle
Click here-NCERT Solutions For Class 10 Maths Chapter 10 Circle
Class 10 Most Important Questions on Chapter 10 -Circle
Q1.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 such that OQ = 12 cm. Length PQ is
Ans.
OP ⊥ PQ (radius of the circle is perpendicular to the tangent of the circle)
Applying Pythagoras theorem
PQ = √(OQ² – OP²)
PQ = √(12² – 5²)
PQ = √(144 – 25) = √119
Q2.If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
Ans.
It is given to us
∠APB = 80°
∠AOB + ∠APB = 180° ( since OA⊥AP and OB⊥BP)
OP is the bisector of ∠AOB and ∠APB
Therefore
∠APO = ∠APB/2 = 80°/2 = 40°
∠POA = 180° – (∠PAO + ∠APO) = 180° – (90°+40°)= 180°- 130° = 50°
Q3.In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then calculate ∠OAB.
Ans.
PA =PB (tangent drawn from an external point to the circle are equal)
∠PAB = ∠PBA(angles opposite to equal sides of a trianngle)
Since ∠APB = 60° is given
∠APB + ∠PAB +∠PBA = 180°
∠APB + 2∠PAB = 180°
60° + 2∠PAB = 180°
2∠PAB = 120°
∠PAB = 60°
∠PAO = 90° (the tangent is perpendicular to radius of the circle)
∠OAB = ∠PAO – ∠PAB = 90° – 60° = 30°
Class 10 Most Important Questions on Chapter 10 -Circle
Q4.In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K&M respectively. If EK = 9 cm, calculate the perimeter of EDF
Ans. Since tangents to the circle drawn from an external point are equal
EK = EM =9 cm
Parimeter of ΔEDF
ED +DF +EF
ED +DH +FH +EF
ED +DK +FM +EF
EK + EM = 9 + 9 =18 cm
Q5.In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then calculate ∠BAT
Ans.AT is given tangent on the circle
OA is the radius of the circle
Since radius of the circle is perpendicular to the tangent
OA ⊥ AT
∴∠OAT =90°
It is given that ∠AOB =100°
OA = OB (radii of the circle)
∴∠OAB =∠OBA
2∠OAB +∠AOB=180°
2∠OAB +100°=180°
2∠OAB =80°
∠OAB =40°
∠BAT = ∠OAT – ∠OAB = 90°- 40°=50°
Class 10 Most Important Questions on Chapter 10 -Circle
Q6.In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.
Ans.OC is the radius of the circle and PCQ is the tangent on the circle
∠OCP = 90° (OC ⊥PQ)
∠CAB =∠ACO =30°(angles opposite to equal sides of a trianngle)
∠ACP =∠OCP – ∠ACO=90° -30°=60°
Q7.In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP
Ans.Let D is a point of contact between the tangent BC and the circle,then
BP =BD …(i)and DC =CQ….(ii)
Perimeter of ΔABC =AB +BC +AC=5+6+4 =15 cm
AB +BC +AC=15 cm
AB +BD +DC+AC=15
From the equation (i) and the equation (ii)
AB +BP +CQ+DC=15
AP + AQ = 15
2AP = 15(since AP=AQ)
AP = 15/2 =7.5 cm
Q8.In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre O and radius 4 cm. If PA ⊥ PB, then find the length of each tangent.
Ans. It is given that two tagents PA and PB are such that PA ⊥ PB
Drawing radius OA and OB
OA ⊥ PA and OB ⊥PB (radius of the circle is perpendicular to the tangent of the circle)
Therefore
∠APB =∠PBO = ∠PAO =90°
Hence PAOB is a square
So,AP =OA =BO =PB = 4 cm
Length of each tangent is 4 cm
Q9.In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC
Ans.The given line segments are
PA = 4 cm, BP = 3 cm and AC = 11 cm
PA = AR = 4 cm (tangent drawn from an external point on the circle are equal)
CR = AC – AR = 11 – 4 = 7 cm
CR = CQ = 7 cm (tangent drawn from an external point on the circle are equal)
BP = BQ = 3 cm(tangent drawn from an external point on the circle are equal)
BC = BQ + CQ = 3 + 7 = 10 cm
Q10.In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. Calculate the radius of the circle inscribed in the triangle (in cm).
Ans.
AC = √(12²+ 5²) = 13 cm
Area of ΔABC = (AB×r)/2 + (BC×r)/2 +(AC×r)/2
(BC×AB)/2 = (AB×r)/2 + (BC×r)/2 +(AC×r)/2
(12 ×5)/2 = 5r/2 + 12r/2 +13r/2
30r/2 = 30
r = 2 cm
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| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
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| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
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