All practice sets solution of Circle lesson 10th std . Maths part 2.
please Please please :(
Answers
Answer:
Step-by-step explanation:
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:
Let the two circles having centres P and Q touch each other internally at point R.
Here, QR = 3.5 cm, PR = 4.8 cm
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 1
The two circles touch each other internally.
∴ By theorem of touching circles,
P – Q – R
PQ = PR – QR
= 4.8 – 3.5
= 1.3 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]
Question 2.
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:
Let the two circles having centres P and R touch each other externally at point Q.
Here, PQ = 5.5 cm, QR = 4.2 cm
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 2
The two circles touch each other externally.
∴ By theorem of touching circles,
P – Q – R
PR = PQ + QR
= 5.5 + 4.2
= 9.7 cm
[The distance between the centres of the circles touching externally is equal to the sum of their radii]
Question 3.
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other
i. externally
ii. internally.
Solution:
i. Circles touching externally:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 3
ii. Circles touching internally:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 4
Question 4.
In the adjoining figure, the circles with centres P and Q touch each other at R A line passing through R meets the circles at A and B respectively. Prove that –
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 5
i. seg AP || seg BQ,
ii. ∆APR ~ ∆RQB, and
iii. Find ∠RQB if ∠PAR = 35°.
Solution:
The circles with centres P and Q touch each other at R.
∴ By theorem of touching circles,
P – R – Q
i. In ∆PAR,
seg PA = seg PR [Radii of the same circle]
∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem]
Similarly, in ∆QBR,
seg QR = seg QB [Radii of the same circle]
∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem]
But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles]
∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.
∴ seg AP || seg BQ [Alternate angles test]
ii. In ∆APR and ∆RQB,
∠PAR ≅ ∠QRB [From (i) and (iii)]
∠APR ≅ ∠RQB [Alternate angles]
∴ ∆APR – ∆RQB [AA test of similarity]
iii. ∠PAR = 35° [Given]
∴ ∠RBQ = ∠PAR= 35° [From (iv)]
In ∆RQB,
∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)]
∴ ∠RQB + 2 ∠RBQ = 180°
∴ ∠RQB + 2 × 35° = 180°
∴ ∠RQB + 70° = 180°
∴ ∠RQB = 110°
Question 5.
In the adjoining figure, the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 6
Construction : Draw seg AF ⊥ seg BD.
Solution:
i. The circles with centres A and B touch each other at E. [Given]
∴ By theorem of touching circles,
A – E – B
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2 7
∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
∠AFD = 90° [Construction]
∴ ∠CAF = 90° [Remaining angle of ꠸AFDC]
∴ ꠸AFDC is a rectangle. [Each angle is of measure 900]
∴ AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B – F – C]
∴ 6 = BF + 4 BF = 2 cm
Also, AB = AE + EB
= 4 + 6 = 10 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]