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These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 6.

CBSE Sample Papers for Class 10 Maths Paper 6

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 6 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections A, B, C andD.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each,
•  There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.

Section-A

Question 1.
Explain why 3 × 5 × 7 + 7 is composite number.

Question 2.
If sin A = [latex]\frac { 1 }{ \sqrt { 2 } } [/latex] and tan B = 1. Prove that sin (A+ B) = 1, where A, B are acute angles.

Question 3.
In Fig. DE || BC. HAD = x, DB = x – 2,AE=x +2 and EC = x – 1, find the value ofx.

Question 4.
Find the distance of origin from the point P(3, -2). (NTSE2012)

Question 5.
Find the 9th term from the end (towards the first term) ofthe AP. 5,9,13,……, 185

Question 6.
Find the discriminant ofthe quadratic equation 3[latex]\sqrt { 3 } [/latex]x² +10x + [latex]\sqrt { 3 } [/latex] = 0

Section-B

Question 7.
Show that 5- [latex]\sqrt { 3 } [/latex] is not irrational.

Question 8.
Cards marked with numbers 13,14,15,…….., 60 are placed in a box and mixed thoroughly. One card is
drawn at random from the box. Find the probability that the number on drawn card is
(i) divisible by 5
(ii) a number which is a perfect square

Question 9.
If p, q, r are in A.P. then find the value of p³ +r³ – 8q³.

Question 10.
For all real values of c, the pair of equations
x – 2y = 8
5x- 10y = c
have a unique solution. Justify whether it is true or false.

Question 11.
If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP= [latex s=2]\frac { 3 }{ 7 } [/latex]AB, where Plies on the line segment AB.

Question 12.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹ 5 coin?

Section-C

Question 13.
Show that any positive odd integer is of the form 6q – 1 or 6q + 5, where q is some integer.

Question 14.
What must be subtracted from x³ – 6x² – 15x + 80, so that the result is exactly divisible by x² + x— 12.

Question 15.
If a cos θ + b sinθ = m and a sin θ – b cosθ = n, prove that a² + b² = m² + n²
OR
If x = r sin A cos C, y = r sin A sin C, z = r cos A, Prove that : r² = x²+ y²+ z²

Question 16.
Calculate the area of the shaded portion.
The quadrants shown in the figure are each of radius 7 cm. [Take π = [latex s=2]\frac { 22 }{ 7 } [/latex]]

Question 17.
If mean of the following frequency distribution is 7.5, then find the value of p.

Question 18.
An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the Pillar if one cubic cm. of iron weights 7.8 gms.
OR
A cylinder whose height is two-third of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Question 19.
The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. The fraction is [latex s=2]\frac { a }{ b } [/latex] . Find [latex s=2]\frac { a+b }{ 2 } [/latex]

Question 20.
If the diagonal BD of a quadrilateral ABCD bisects both ∠B and ∠D then show that [latex s=2]\frac { AB }{ BC } [/latex] = [latex s=2]\frac { AD }{ CD } [/latex]
OR
In an equilateral triangle ABC, AD is drawn perpendicular to BC. Prove that 4 AD² = 3BD²

Question 21.
Find the values ofk so that the area of the triangle with vertices (1, -1), (-4,2k) and (-k, -5) is 24 sq. units.
OR
If the point P (x, y) is equidistant from the points A (a + b, b – a) and B (a – b, a + b). Prove that bx = ay.

Question 22.
If from an external point P of a circle with centre 0, two tangents PQ and PR are drawn such that ∠QPR =120°, prove that 2PQ = PO.

Section-D

Question 23.
The ratio of the sums of first m and first n terms of an A.P. is m²: n². Show that the ratio of its m^th and n^th terms is (2m -1): (2n – 1).

Question 24.
If the areas of two similar triangles are equal, prove that they are congruent.
OR
If the corresponding sides of two triangles are proportional (i.e., in the same ratio), then prove that their corresponding angles are equal and hence the two triangles are similar.

Question 25.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.
OR
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.
(Take [latex]\sqrt { 3 } [/latex] =1.732)

Question 26.
A semicircular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of the cup.

Question 27.
If θ be an acute angle such that tan² θ = 1 – a², then find the value of {sec θ + tan³ θ cosec θ}²

Question 28.
Solve the quadratic equation (x² – 5x)² – 7(x² – 5x) + 6 = 0.

Question 29.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are [latex s=2]\frac { 3 }{ 5 } [/latex] times the corresponding sides of the given triangle.

Question 30.
Following table gives the ages in years of militants operating in a certain area of a country:

If mean of the above distribution is 47.2, find how many militants in the age groups 49-52 are active in the area?
OR
One hundred students took an examination. Compute values ofx and y from the following data if mean is 42.

Solutions
Section-A

Solution 1.
3 × 5 ×7+7 = 7 (3 × 5 + 1) = 7 (16), which has more than two factors.
∴ It is a composite number. (1)

Solution 2.
sinA= [latex s=2]\frac { 1 }{ \sqrt { 2 } } [/latex] ⇒ A=45°

and tan B = 1 ⇒ B = 45° (1/2)
∴ sin (A + B) = sin (45° + 45°) ⇒ sin 90° = 1 Hence Proved. (1/2)

Solution 3.
In ∆ABC, we have DE || BC
∴ [latex s=2]\frac { AD }{ DB } [/latex] = [latex s=2]\frac { AE }{ EC } [/latex] By Thale’s Theorem] (1/2)

⇒ [latex s=2]\frac { x }{ x-2 } [/latex] = [latex s=2]\frac { x+2 }{ x-1 } [/latex]
⇒x(x-1) = (x-2)(x +2) ⇒ x² – x = x² – 4 ⇒x = 4 (1/2)

Solution 4.
Distance = [latex]\sqrt { { (0-3) }^{ 2 }+{ (0+2) }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } [/latex] (1)

Solution 5.
Common difference of the AP = d = 9 – 5 = 4
Last term of the AP = l= 185 (1/2)
Since the n^th term from the end of an AP is given by l – (n – 1 )d.
So, the 9^th term from the end is = 185 – (9 – 1)4 = 185 – 32= 153 (1/2)

Solution 6.
D = b² – 4ac = (10)² – 4(3[latex]\sqrt { 3 } [/latex] )([latex]\sqrt { 3 } [/latex]) = 100 – 36 = 64 (1)

Section-B

Solution 7.
If 5 – [latex]\sqrt { 3 } [/latex] is not irrational.
Let 5 – [latex]\sqrt { 3 } [/latex] = [latex s=2]\frac { p }{ q } [/latex] [p and q are integers, q ≠ 0] (1/2)
⇒[latex s=2]\frac { 5q-p }{ q } [/latex] = [latex]\sqrt { 3 } [/latex] ⇒ [latex s=2]\frac { 5q-p }{ 3 } [/latex] is rational number  (1)
So, [latex]\sqrt { 3 } [/latex] should be rational.
But [latex]\sqrt { 3 } [/latex] is irrational, it cannot be equal to rational number, Hence 5 -[latex]\sqrt { 3 } [/latex] is irrational. (1/2)

Solution 8.
Total no. of cards = 60-12 = 48
Numbers are 13,14,15,16,……….,60.
(i) Numbers divisible by 5 are 15, 20,25, 30, 35, 40, 45, 50, 55, 60. (1/2)
∴ Favourable outcomes = 10
∴ P(no. is divisible by 5) = [latex s=2]\frac { 10 }{ 48 } [/latex] = [latex s=2]\frac { 5 }{ 24 } [/latex] (1/2)
(ii) Perfect square numbers are 16,25,36,49 (1/2)
∴ Ptperfect square) = [latex s=2]\frac { 4 }{ 48 } [/latex] = [latex s=2]\frac { 1 }{ 12 } [/latex] (1/2)

Solution 9.

Solution 10.

Solution 11.

Solution 12.
Number of 50 p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
Total number of coins = 180 (1/2)
Then (i) P (50 p coin) = [latex s=2]\frac { 100 }{ 180 } [/latex] = [latex s=2]\frac { 5 }{ 9 } [/latex]
(ii) P (not a ₹ 5 coin) = 1 – P (a ₹ 5 coin) (1/2)
= 1- [latex s=2]\frac { 10 }{ 180 } [/latex] = 1- [latex s=2]\frac { 1 }{ 18 } [/latex] = [latex s=2]\frac { 17 }{ 18 } [/latex] (1)

Section-C

Solution 13.
Let a be any positive integer and b = 6
Then by Euclid’s algorithm, a = bq + r, o < r < b (1/2)
Wehavea = 6q+r   …(1)
Since 0 ≤ r < 6, So, r = 0,1,2,3,4,5
∴ From(1), forr = 0, a = 6q
forr = 1, a = 6q + 1
forr = 2, a = 6q + 2
forr = 3, a = 6q + 3
forr = 4, a = 6q + 4
forr = 5, a = 6q + 5   (1/2)
Since 6q is divisible by 2,
∴ 6q is even. (1)
6q + 1 is not divisible by 2. So, 6q + 1 l is odd.
6q + 2 is divisible by 2
∴ 6q + 2 is even. (1)
6q + 3 is not divisible by 2. So, 6q + 3 is odd.
6q + 4 is divisible by 2.
∴ 6q + 4 is even.
6q + 5 is not divisible by 2. So, 6q + 5 is odd. (1/2)
So, we see that 6q, 6q + 2, 6q + 4 are even. Since the number which are not divisible by 2 are odd integer.
∴ 6q +1, 6q + 3, 6q + 5 are odd integer. Hence any positive odd integer is ofthe form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. (1/2)

Solution 14.
Let (ax + b) be subtracted from given polynomial p(x), so that it is exactly divisible by x² + x – 12 Then, .
q (x) = x³-6x² – 15x + 80 – (ax + b)
= x³ -6x² – (15 + a)x + (80 -b) (1/2)
∵ Dividend = Divisor × quotient + remainder (A)
But remainder will be zero.
∴ Dividend = Divisor x quotient
q (x) = (x² + x – 12) x quotient
∴q(x) = x³-6x²-(15+a)x + (80 – b) (1/2)
Also, x(x² +x- 12)-7(x² +x- 12)
= x³ + x² – 7x² – 12x – 7x + 84
= x³ – 6x² – 19x + 84 (1)
Hence, x³ – 6x² – 19x + 84 = x³ – 6x² – (15 + a)x + (80 – b)
– 15 – a = – 19 ⇒ a = + 4 and 80 – b = 84 ⇒ b = – 4 (1)
Hence, ifinp (x) we subtracted 4x – 4 = (ax + b), then it is exactly divisible by x² + x – 12

Solution 15.
m = a cos θ + b sin θ
(m)²= (a cos θ + b sin θ)²
m² = a² cos² θ + bs² sin²θ + 2ab cos θ sin θ …(i) (1)
also, n = a sin θ – b cos θ ;
(n)² = (a sin θ – b cos θ)²
n² = a²sin² θ + b² cos² θ – 2ab cos θ sin θ …(ii) (1)
On adding equation (i) and (ii), we get
m² + n² = a² cos² θ + b² sin2 θ + a² sin² θ + b² cos² θ
= ²(cos² θ + sin² θ) + b²(sin²θ + cos² θ) (1)
m² + n² = a² + b²
Hence proved.
OR
x = r sin A cos C, y = r sin A sin C and z = r cos A
x²+ y²+ z² = r²sin²A cos²C + =(r² sin²A ) (cos²C + sin²C) +r² cos²A  (1)
= r² sin² A(1) + r² cos² A  (1/2)
= r²(sin² A + cos² A) =r² Hence Proved. (1/2)

Solution 16.
Draw square PQRS as shown in the figure given below.
Here PQ= QR = PS= RS = 14 cm (1)
Area of the shaded portion = Area of the square PQRS – Area of four equal quadrants (1)
= 14 × 14 – 4 × [latex s=2]\frac { 1 }{ 4 } [/latex]π × (7)² =196 – [latex s=2]\frac { 22 }{ 7 } [/latex] 7 ×7
= 196-154 =42 cm²

Solution 17.

Solution 18.
V of cylindrical part = π(8)² (240) = 64π × 240 (1/2)
V of conical part = [latex s=2]\frac { 1 }{ 3 } [/latex]π × 64 × 36 =64π(12) (1/2)

Total volume of iron = 64π(240 + 12) = 64×[latex s=2]\frac { 22 }{ 7 } [/latex] ×252 = 64 × 22× 36 cm³ (1)
∴ Total weight ofpillar = (64) (22) (36) × 7.8 = 395366.4 gms = 395.3664 kg (1)
OR

Solution 19.

Solution 20.


OR

Solution 21.
∵ Area of the triangle = 24 sq. units (1)
∴[latex s=2]\frac { 1 }{ 2 } [/latex][1(2k + 5)-4(-5 + 1)-k(-1-2k)] = 24
So, 2k² + 3k- 27 = 0 ⇒ (2k + 9) (k – 3) = 0
After solving we get k = 3, -[latex s=2]\frac { 9 }{ 2 } [/latex] (1)
OR
The point P (x, y) is equidistant form the points A (a + b, b – a) and B (a – b, a + b)
∴ PA= PB
⇒ [latex]\sqrt { { (a+b-x) }^{ 2 }+{ (b-a-y) }^{ 2 } } =\sqrt { { (a-b-x) }^{ 2 }+{ (a+b-y) }^{ 2 } } [/latex] (1/2)
⇒ (a + b – x)² + (b – a – y)² = (a – b – x)² + (a + b – y)² (1/2)
⇒ (a + b – x)² – (a – b – x)² = (a + b – y)² – (b – a – y)² (1/2)
⇒ (a + b-x + a-b-x) (a + b-x-a + b + x) = (a + b-y + b-a-y) (a + b-y-b+ a + y) (1/2)
⇒ (2a-2x)(2b) = (2b-2y)(2a)
⇒ (a – x) b = (b – y) a
⇒ ab – bx = ab – ay ⇒ bx = ay (1)

Solution 22.

Section-D

Solution 23.

Solution 24.
Let ∆ABC ~ ∆PQR and .
area (∆ABC) = area (∆PQR) (Given) (1/2)

OR

Solution 25.

OR

Solution 26.

Solution 27.

Solution 28.

Solution 29.

Steps of construction
1. Draw a line segment BC = 6.5 cm.
2. Take B as centre and mark an arc of length 5.5 cm above BC.
3. Similarly, take C as centre and cut an arc above BC of length 5 cm which cuts previous arc at point A.
4. Join AB and AC. Therequired triangle is formed i.e., ∆ABC
5. Draw a ray BX below BC.
6. Mark equal arcs B₁, B₂, B₃, B₄ and B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
7. Join B₅C and draw a line || to B₅C from point B₃ which cuts BC at C’.
8. Drawaline||toACfromC’whichcutsAB at A’. (1)
9. Now, ∆A’BC’ is the required triangle.

Solution 30.


OR

We hope the CBSE Sample Papers for Class 10 Maths paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 6, drop a comment below and we will get back to you at the earliest.

UP Board Solutions for Class 11 English Translation Chapter 15 Agreement of Subject and Verb

These Solutions are part of UP Board Solutions for Class 11 English. Here we have given UP Board Solutions for Class 11 English Translation Chapter 15 Agreement of Subject and Verb.

Exercise 33

1. My brother as well as friends was invited to the feast.
2. None of the news is good.
3. Every boy and girl was eager to sing.
4. Pity as well as justice make a man great.
5. You as well as I have performed our duty.
6. The army was marching towards the enemy.
7. This is the shop which was highly damaged by fire.
8. Everyone of you is suitable for this post.
9. Where are your spectacles ?
10.Ten thousand rupees is a good salary.
11. Either he or his friends have deceived you.
12. All the thieves with their chief have run away.
13. The committee has elected its president.
14. Not only the doctor but also the compounder went on trying to save the patient.
15. Honey and ghee is a good medicine.

We hope the UP Board Solutions for Class 11 English Translation Chapter 15 Agreement of Subject and Verb help you. If you have any query regarding UP Board Solutions for Class 11 English Translation Chapter 15 Agreement of Subject and Verb, drop a comment below and we will get back to you at the earliest.

UP Board Solutions for Class 7 Maths Chapter 11 वृत्त

These Solutions are part of UP Board Solutions for Class 7 Maths. Here we have given UP Board Solutions for Class 7 Maths Chapter 11 वृत्त.

अभ्यास 11 (a)

प्रश्न 1.
पाश्र्व चित्र में Oवृत्त का केन्द्र है। निम्नलिखित कथनों (UPBoardSolutions.com) में सत्य/असत्य कथनों को बताइएः

(i) रेखाखंड AB जीवा है। (✓)
(ii) QF त्रिज्या है। (✗)
(iii) OD त्रिज्या है। (✓)
(iv) PC जीवा है। (✗)

प्रश्न 2.
अर्धवृत्त में बने कोण की माप होती है।
(i) 30°
(ii) 60°
(iii) 180°
(iv) 90° (✓)

प्रश्न 3.
आकृति 11.5 के अनुसार अपनी अभ्यास पुस्तिका पर एक (UPBoardSolutions.com) आकृति खींचकर उसके दीर्घ वृत्तखंड को छायांकित कीजिए।

उत्तर :
चित्र में दीर्घ वृत्तखंड ADB है।

प्रश्न 4.
आकृति 11.6 में 0 वृत्त का केन्द्र है। आकृति में निर्मित किन्हीं दी त्रिज्यखंडों के नाम लिखिए।

उत्तर :
चित्र में निर्मित्त दो त्रिज्यखंड OAB तथा OBC है।

प्रश्न 5.
2.5 सेमी त्रिज्या का एक वृत्त खींचिए जिसका केन्द्र 0 (UPBoardSolutions.com) है। इस वृत्त को दो अर्द्धवृत्तों में विभक्त कीजिए।

रचना – सर्वप्रथम 0 को केन्द्र मानकर 2.5 सेमी त्रिज्या का एक वृत्त खींचा। AP वृत्त का व्यास AB खींचा, जिससे वृत्त दो भागों में विभक्त होगा।

प्रश्न 6.
आकृति 11.7 में 0 वृत्त का केन्द्र है। ∠ACB कितने अंश का है? अपने उत्तर के पक्ष में कारण बताइए।

उत्तर :
चूँकि अर्द्धवृत्त में बना कोण समकोण होता है,
अतः ∠ACB = 90°

अभ्यास 11 (b)

प्रश्न 1.
अर्द्धवृत्त का अंशमाप होता है। (UPBoardSolutions.com)
(i) 45°
(ii) 90°
(iii) 180°
(iv) 360°
उत्तर :
(iii) 180°

प्रश्न 2.
किसी वृत्त में यदि उसके किसी लघुचाप का (UPBoardSolutions.com) अंशमाप 70° है, तो उसके दीर्घचाप का अंशमाप कितना होगा?
हल :
लघुचाप का अंशमाप = 70°
दीर्घचाप का अंशमाप = 360° – 70° = 290°

प्रश्न 3.
किसी चाप द्वारा केन्द्र पर अन्तरित कोण तथा उसके द्वारा वृत्त के शेष भाग पर स्थित किसी बिन्दु पर अन्तरित कोण में क्या सम्बन्ध होता है?
उत्तर :
किसी चाप द्वारा केन्द्र पर अन्तरित कोण उसके द्वारा वृत्त के (UPBoardSolutions.com) शेष भाग पर स्थित किसी बिन्दु पर अन्तरित कोण का दो गुना होता है।

प्रश्न 4.
आकृति 11.17 में 0 वृत्त का केन्द्र है। चाप AXB को अंशमाप बताइए।

हल :
∠APB = 35°
चाप AXB का अंशमाप = ∠ AOB
= 2 ∠ APB
= 2 × 35° = 70°

प्रश्न 5.
आकृति 11.18 में लधुचाप BCD एवं दीर्घचाप BAD (UPBoardSolutions.com) के अन्तर्गत कोणों के नाम बताइए।

हल :
लघुचाप BCD के अन्तर्गत कोण = ∠ BCD
दीर्घचाप BAD के अन्तर्गत कोण = ∠ BAD

प्रश्न 6.
आकृति 11.19 में o वृत्त का केन्द्र है। A, C और B वृत (UPBoardSolutions.com) पर तीन बिन्दु है, तथा ∠AOB को प्रतिवर्ती कोण = 240° है तो ACB का मान ज्ञात कीजिए।
हल :

∠ACB = [latex]\frac { 1 }{ 2 }[/latex] ∠ AOB
∴ ∠ACB = [latex]\frac { 1 }{ 2 }[/latex] × 240 = 120°

अभ्यास 11 (c)

प्रश्न 1.
आकृति 11.25 में एक ही वृत्तखंड में बने कोणों (UPBoardSolutions.com) के नाम लिखिए।”
हल :

∠ ACB और ∠ AEB एक ही वृत्तखंड में बने दीर्घ कोण हैं तथा ∠ADB और ∠AFB एक ही वृत्तखंड में बने लघुकोण हैं।

प्रश्न 2.
आकृति 11.26 मे बने कोण PRQ=45°, तो ∠PSQ का मान बताइए।
हल :

चूँकि एक ही वृत्तखंड में बने (UPBoardSolutions.com) कोण बराबर होते हैं।
अतः ∠ PSQ = ∠ PRQ = 45°

प्रश्न 3.
आकृति 11.27 में यदि ∠ACB=x, तो ∠ADB एवं ∠AEB के मान बताइए।
हल :

चूंकि एक ही वृत्तखंड में बने कोण बराबर होते हैं। अतः
∠ ADB = ∠ ACB = x°
∠AEB = ∠ ACB = x°

प्रश्न 4.
आकृति 11.28 में बने कोणों के सम्बन्ध में निम्नलिखित (UPBoardSolutions.com) कथनों में सत्य/असत्य कथनों को छाँटिएः
उत्तर :

(i) ∠BDC   = ∠BAC   – सत्य
(ii) ∠BDC  = ∠BCA  – असत्य
(iii) ∠ACB = ∠ADB  – सत्य
(iv) ∠BDA = ∠CDB  – असत्य
(v) ∠ACD   = ∠DBA  – सत्य

प्रश्न 5.
आकृति 11.29 में ∠ACB के बराबर निम्नलिखित (UPBoardSolutions.com) में से कौन-सा कोण है?
(i) ∠ABD
(ii) ∠ADB
(iii) ∠DBC
(iv) ∠BAD

उत्तर :
(ii) ∠ADB

दक्षता अभ्यास – 11

प्रश्न 1.
आकृति 11.32 में वृत्त को केन्द्र 0 है। रेखा BOD, ∠AOC की समद्विभाजक है, (UPBoardSolutions.com) तथा ∠COD=50°, तो ∠ABC की माप होगीः
(i) 50°
(ii) 25°
(iii) 100°
(iv) 120°

हल :
∠ AOC = 2 ∠ COD= 2 × 50° = 100°
∴ ∠ ABC = [latex]\frac { x }{ 12 }[/latex] ∠ AOC = [latex]\frac { x }{ 12 }[/latex] × 100° = 50° (i)

प्रश्न 2.
आकृति 11.33 में AB वृत्त की जीवा है बिन्दु C तथा D (UPBoardSolutions.com) वृत्त पर हैं। यदि ∠ADB = 45° तो ∠ACB की माप होगीः
(i) 90°
(ii) 135°
(iii) 45०
(iv) 223°

हल :
∠ ACB =∠ ADB = 45° (iii)

प्रश्न 3.
आकृति 11.34 में बिन्दु 0 वृत्त का केन्द्र है और ∠AOB = 60°, (UPBoardSolutions.com) तो ∠ADB की माप होगीः
(i) 120°
(ii) 150°
(iii) 140°
(iv) 30°

हल :
वृहत्त कोण AOB = 360° -∠ AOB = 360° – 60° = 300°
∴ ∠ ADB = [latex]\frac { 1 }{ 2 }[/latex] × बृहत AOB = [latex]\frac { 1 }{ 2 }[/latex] × 300° = 150° (ii)

प्रश्न 4.
आकृति 11.35 में बिन्दू 0 वृत्त का एक केन्द्र है। इस पर तीन बिन्दु A, B तथा C हैं। (UPBoardSolutions.com) यदि ∠ACB = 40° तो ∠AOB की माप होगीः
(i) 20°
(ii) 40°
(iii) 60°
(iv) 80°

हल :
∠AOB = 2 ∠ ACB = 2 × 40° = 80° (iv)

प्रश्न 5.
आकृति 11.36 में बिन्दु 0 वृत्त का एक केन्द्र है। वृत्त की दो समान जीवाए AC और BC खींची गई हैं। (UPBoardSolutions.com) ∠ABC का मान ज्ञात कीजिए।
हल :

चूँकि अर्द्धवृत्त का कोण समकोण होता है ∠ACB = 90°
प्रश्नानुसार, AC = BC4∠ CAB = ∠ABC
∠CAB + ∠ABC + ∠ ACB = 180°
∠ABC + ∠ABC + 90° = 180°
∠ ABC = 180° – 90° = 90°
∠ABC = [latex]\frac { { 90 }^{ \circ } }{ 2 } [/latex] = 45०

प्रश्न 6.
निम्नांकित वृत्तों में प्रत्येक का केन्द्र 0 है। प्रत्येक में x का (UPBoardSolutions.com) मान ज्ञात कीजिए।

हलः
(i) x° = ∠AOB = 2 ∠ACB = 2 × 40° = 80°
(ii) x° = ∠ACB = 1/2 ∠AOB = [latex]\frac { 1 }{ 2 }[/latex] × 105° = 52.5°
(iii) x° = ∠ACB = 1/2 ∠AOB = [latex]\frac { 1 }{ 2 }[/latex] × 70° = 35°

प्रश्न 7.
वृत्त की एक जीवा की लम्बाई उसकी त्रिज्या के बराबर है। (UPBoardSolutions.com) इस जीवा द्वारा लघु वृत्तखंड पर अन्तरित कोण ज्ञात कीजिए।
हल :
चूंकि वृत्त की जीवा AB वृत्त की त्रिज्या के बराबर है। अत: ∆OAB समबाहु त्रिभुज होगा तथा तीनों कोण बराबर होंगे।

∠OAB +∠OBA+∠AOB = 180°
∠AOB +∠AOB +∠AOB = 180°
3∠AOB = 180°
∠AOB = [latex]\frac { { 180 }^{ \circ } }{ 3 } [/latex] = 60°
बृहत्त कोण AOB = 360° – ∠ AOB = 360° – 60° = 300° (UPBoardSolutions.com)
∠ACB = 3 बृहत्त ∠AOB = [latex]\frac { 1 }{ 2 }[/latex] × 300° = 150°
अतः लघु वृत्तखंड पर अन्तरित कोण = 150°

प्रश्न 8.
3.0 सेमी त्रिज्या का एक वृत्त खींचिए। इस वृत्त की:एक जीवा । खींचकर वृत्त को दो वृतखंडों में विभक्त कीजिए।
हल :

सर्वप्रथम सेमी त्रिज्या लेकर परकार की सहायता से एक (UPBoardSolutions.com) वृत्त खींचा। किसी भी लम्बाई की एक जीवा AB खींची। अत: वृत्त, दीर्घ वृत्तखंड AYB तथा लघु वृत्तखंडे AXB में विभाजित हो गया।

प्रश्न 9.
अर्द्धवृत्त किसे कहते हैं? चित्र बनाकर स्पष्ट कीजिए।
उत्तर :

वृत्त के आधे भाग को अर्द्धवृत्त कहते हैं।

प्रश्न 10.
आकृति 11.38 में बिन्दु 0 वृत्त का केन्द्र हैं। AOB (UPBoardSolutions.com) वृत्त का व्यास है और ∠COB=40°। ज्ञात कीजिए:

(i) दीर्घचाप BC का अंशमाप
(ii) दीर्घचाप AC को अंशमाप
(iii) लघुचाप AC का अंशमाप
(iv) अर्द्धवृत्त ACB का अंशमाप।

हल :

(i) दीर्घचाप BC का अंशमाप = 360° – 40° = 320°
(ii) दीर्घचाप AC का अंशमाप = 360° – 140° = 220°
(iii) लघुचाप AC का अंशमाप = 360°-220° = 140°
(iv) अर्धवृत्त ACB का अंशमाप = 180°

प्रश्न 11.
आकृति 11.39 में 0 वृत्त का केन्द्र है। इसके अन्तर्गत एक (UPBoardSolutions.com) AABC बना है यदि ∠ACB= 30° तो ∠A और ∠B ज्ञात कीजिए।

हल :
चूँकि अर्द्धवृत्त का (UPBoardSolutions.com) कोण समकोण होता है।
अतः ∠A = 90°
∠ACB + ∠ BAC + ∠ABC = 180°
30° + 90° + ∠ABC = 180°
120° + ∠ABC = 180°
∠ABC = 180° – 120° = 60°

प्रश्न 12.
वृत्त की एक जीवा की लम्बाई उसकी त्रिज्या के बराबर है। (UPBoardSolutions.com) इस जीवा द्वारा दीर्घ वृत्तखंड पर अन्तरित कोण ज्ञात कीजिए।
हल :

चूंकि वृत्त की जीवा AB वृत्त की त्रिज्या के बराबर है। (UPBoardSolutions.com) अत: ∆OAB समबाहु त्रिभुज होगा तथा तीनों कोण बराबर होंगे।
∠AOB = 60°
अतः दीर्घ वृतखंड पर अन्तरित कोण = [latex]\frac { 1 }{ 2 }[/latex] x 60° = 30°

प्रश्न 13.
आकृति 11.40 में 0 वृत्त का केन्द्र है। ∠AEB= 130° और ∠EBC=20°, तो ∠BDA का मान ज्ञात कीजिए।
हल :

∠ AEB +∠ BEC = 180°
130° +∠ BEC = 180°
∠ BEC = 180° – 130° = 50°
∆ BEC में, 2 ECB +∠ BEC +∠ EBC = 180°
∠ ECB + 50° +20° = 180° (UPBoardSolutions.com)
∠ACB =∠ ECB = 180° – 70° = 110°
∠ BDA =∠ ACB = 110°

प्रश्न 14.
आकृति 11.41 में 0 वृत्त को केन्द्र है। (UPBoardSolutions.com) ∠ABC=40° और ∠CAB= 80°, तो ∠ADB का मान ज्ञात कीजिए।

हल :
∆ABC में,
∠ ABC + ∠CAB +∠ ACB = 180°
40° + 80° + 2 ACB = 180°
120° + 2 ACB = 180°
∠ ACB = 180° – 120° = 60°

प्रश्न 15.
आकृति 11.42 में 0 वृत्त का केन्द्र है तथा AABC एक समबाहु त्रिभुज है ∠BOC का मान ज्ञात कीजिए।
हल :

चूँकि A ABC एक समबाहु त्रिभुज है। (UPBoardSolutions.com) अतः त्रिभुज के तीनों कोण बराबर होंगे।
∴ ∠ BAC = [latex]\frac { { 180 }^{ \circ } }{ 3 } [/latex] = 60°
∠ BOC = 2 ∠ BAC = 2 × 60° = 120°

प्रश्न 16.
आकृति 11.43 में 0 वृत्त का केन्द्र है और ∠BOD=130°, ∠BCD का माप ज्ञात कीजिए।
हल :

अधिक कोण ∠ BOD = 130°
वृहत कोण ∠ BOD = 360° – 130° = 230°
∠ BCD = 1/2 ∠ BOD (UPBoardSolutions.com)
= [latex]\frac { 1 }{ 2 }[/latex] × 230° = 115°

प्रश्न 17.
आकृति 11.44 में 0 वृत्त का केन्द्र है। (UPBoardSolutions.com) ∠OBC=40°, तो ∠BAC का माप ज्ञात कीजिए।
हल :

A OBC में ∠OCB = ∠OBC= 40°
∠ OBC +∠OCB +∠ BOC = 180°
40° + 40° +∠ BOC = 180°
80° +∠ BOC = 180° (UPBoardSolutions.com)
∠ BOC = 180° – 80° = 100°
∠ BAC = [latex]\frac { 1 }{ 2 }[/latex] ∠ BOC
= [latex]\frac { 1 }{ 2 }[/latex] × 100° = 50°

We hope the UP Board Solutions for Class 7 Maths Chapter 11 वृत्त help you. If you have any query regarding UP Board Solutions for Class 7 Maths Chapter 11 वृत्त, drop a comment below and we will get back to you at the earliest.