(A) R1<\/sub> + R2<\/sub> = R Explanation: Question 2.<\/p>\n (A) R1<\/sub> + R2<\/sub> = R Explanation: Question 3.<\/p>\n (A) Area of the circle = Area of the square Explanation: Question 4.<\/p>\n (A) r2<\/sup> sq. units Explanation: Question 5.<\/p>\n (A) 22 : 7 Explanation: Question 6.<\/p>\n (A) 10 m Explanation: Question 7.<\/p>\n (A) 36\u03c0 cm2<\/sup> Explanation: Question 8.<\/p>\n (A) 256 cm2<\/sup> Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) 225 m2<\/sup> Explanation: Question 2.<\/p>\n (A) 19.625 m2<\/sup> Explanation: Question 3.<\/p>\n (A) 7.85 m2<\/sup> Explanation: Question 4.<\/p>\n (A) 58.758 m2<\/sup> Explanation: Question 5.<\/p>\n (A) Coordinate geometry II. Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) \\(\\frac{35}{2} \\mathrm{~mm}\\) Explanation: Question 2.<\/p>\n (A) 100 mm Explanation: Question 3.<\/p>\n (A) 528 mm Explanation: Question 4.<\/p>\n (A) \\(\\frac{385}{2} \\mathrm{~mm}^{2}\\) Explanation: Question 5.<\/p>\n (A) Areas Related to circles III. Read the following text and answer the following question on the basis of the same: Design I: This design is made with a circle of radius 32 cm leaving equilateral triangle ABC in the middle as shown in the given figure. Question 1.<\/p>\n (A) 12\u221a3 cm Question 2.<\/p>\n (A) 8 cm Refer Design II:<\/p>\n Question 3.<\/p>\n (A) 1264 cm2<\/sup> Explanation: Question 4.<\/p>\n (A) 124 cm2<\/sup> Explanation: Question 5.<\/p>\n (A) 378 cm2<\/sup> Explanation: IV. Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) 180 mm Explanation: Question 2.<\/p>\n (A) 44 mm2 Explanation: Refer to Design B<\/p>\n Question 3.<\/p>\n (A) 48.49 mm Question 4.<\/p>\n (A) 18\u03c0 Question 5.<\/p>\n (A) 2 Explanation: <\/p>\n","protected":false},"excerpt":{"rendered":" Areas Related to Circles Class 10 MCQ Questions with Answers Question 1. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then: (A) R1 + R2 = R (B) (C) R1 + R2 < R (D) Answer: (B) Explanation: …<\/p>\n
\n(B) \\(R_{1}^{2}+R_{2}^{2}=R^{2}\\)
\n(C) R1 + R2 < R
\n(D) \\(R_{1}^{2}+R_{2}^{2}=R^{2}\\)
\nAnswer:
\n(B) \\(R_{1}^{2}+R_{2}^{2}=R^{2}\\)<\/p>\n
\nAccording to the given condition, Area of circle =Area of first circle + Area of second circle
\n\u03c0R2<\/sup> = \u03c0\u00a0\\(R_{1}{ }^{2}\\) + \u03c0\\(R_{2}{ }^{2}\\)
\nR2<\/sup> =\\(R_{1}{ }^{2}\\) + \\(R_{2}{ }^{2}\\).<\/p>\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nIf the sum of the circumferences of two circles with radii R1<\/sub> and R2<\/sub> is equal to the circumference of a circle of radius R, then:<\/h2>\n
\n(B) R1<\/sub> + R2<\/sub> > R
\n(C) R1<\/sub> + R2<\/sub> < R
\n(D) Nothing definite can be said about the relation among R1<\/sub>, R2<\/sub> and R
\nAnswer:
\n(A) R1<\/sub> + R2<\/sub> = R<\/p>\n
\nAccording to question, Circumference of circle
\n= Circumference of first circle + Circumference of second circle
\n2\u03c0R = 2\u03c0R1<\/sub> + 2\u03c0R2<\/sub>
\nR = R1<\/sub> + R2<\/sub><\/p>\nIf the circumference of a circle and the perimeter of a square are equal, then:<\/h2>\n
\n(B) Area of the circle > Area of the square
\n(C) Area of the circle < Area of the square
\n(D) Nothing definite can be said about the relation between the areas of the circle and square.
\nAnswer:
\n(B) Area of the circle > Area of the square<\/p>\n
\nAccording to question, Circumference of a circle = Perimeter of square Let ‘r’ and ‘a’ be the radius of circle and side of square.
\n2\u03c0r = 4 a
\n\\(\\frac{22}{7}\\)r = 2a
\n11r = 7a
\nr = \\(\\frac{7 a}{11}\\) ….(i)
\nA1<\/sub> = \u03c0R2<\/sup>
\nFrom equation (i), we have
\nA1<\/sub> = \u03c0\\(\\left(\\frac{7 a}{11}\\right)^{2}\\)
\n= \\(\\frac{22}{7}\\left(\\frac{49 a^{2}}{121}\\right)\\)
\n= \\(\\frac{14 a^{2}}{11}\\)
\nA2<\/sub> = a2<\/sup>
\nFrom equation (ii) and (iii), we haveA1 = \\(\\frac{14}{11}\\)A2
\nA1<\/sub> > A2<\/sub>
\nArea equation is greater than the area of square.<\/p>\nArea of the largest triangle that can be inscribed in a semi-circle of radius ‘r’ units is:<\/h2>\n
\n(B) \\(\\frac{1}{2}\\) r2<\/sup> sq. units
\n(C) 2r2<\/sup> sq units
\n(D) \u221a2r2<\/sup> sq. units
\nAnswer:
\n(A) r2<\/sup> sq. units<\/p>\n
\nTake a point C on the circumference of the semi-circle and join it by the end points of diameter AB.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-1a.png?resize=185%2C131&ssl=1\")
\n\u2220C = 90 [Angle in a semi – circle is right angle]
\nSo ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 CD
\n= \\(\\frac{1}{2}\\) \u00d7 2r \u00d7 r = r2<\/sup> sQuestion units<\/p>\n<\/p>\nIf the perimeter of a circle is equal to that of a square, then the ratio of their areas is:<\/h2>\n
\n(B) 14 : 11
\n(C) 7 : 22
\n(D) 11 : 14
\nAnswer:
\n(B) 14 : 11<\/p>\n
\nLet the radius of circle be ‘r’ and side of square be ‘a’.
\nAccording to given question,
\nPerimeter of circle = Perimeter of square
\n2\u03c0r = 4a
\n\u2234 a = \\(\\frac{\\pi r}{2}\\)
\nSo, \\(\\frac{ Area of circle}{Area of square}\\) = \\(\\frac{\\pi r^{2}}{\\left(\\frac{\\pi r}{2}\\right)^{2}}\\) [From equation (i)]
\nSolving equation (i), we get result as \\(\\frac{14}{11}\\).<\/p>\nIt is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be:<\/h2>\n
\n(B) 15 m
\n(C) 20 m
\n(D) 24 m
\nAnswer:
\n(A) 10 m<\/p>\n
\nArea of first circular park whose diameter is 16 m,
\n= \u03c0\\(\\left(\\frac{16}{2}\\right)^{2}\\)
\n= \u03c0\\((8)^{2}\\)
\n= 64\u03c0 m2<\/sup>
\nArea of second circle park whose dimeter is 12 m,
\n= \u03c0\\(\\left(\\frac{12}{2}\\right)^{2}\\)
\n= \u03c0(6)2<\/sup>
\n= 36\u03c0 m2<\/sup>
\nAccording to question,
\nArea of single circlar park = Area of first circular park + Area of second circular park
\n\u03c0r2<\/sup> = 64\u03c0 + 36\u03c0
\n\u03c0r2<\/sup> = 100\u03c0
\nr =10 m<\/p>\nThe area of the circle that can be inscribed in a square of side 6 cm is:<\/h2>\n
\n(B) 18\u03c0 cm2<\/sup>
\n(C) 12\u03c0 cm2<\/sup>
\n(D) 9\u03c0 cm2<\/sup>
\nAnswer:
\n(D) 9\u03c0 cm2<\/sup><\/p>\n
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-2a.png?resize=119%2C145&ssl=1\")
\nGiven, side of square = 6 cm
\nDiameter of a circle, (d) = Side of square = 6 cm
\nRadius of a circle (r) = \\(\\frac{d}{2}\\) = 6 = 3 cm.
\nArea of circle = \u03c0r2<\/sup>
\n\u03c0(3)2<\/sup> = 9\u03c0 cm2<\/sup><\/p>\nThe area of the square that can be inscribed in a circle of radius 8 cm is:<\/h2>\n
\n(B) 128 cm2<\/sup>
\n(C) 64 cm2<\/sup>
\n(D) 64 cm2<\/sup>
\nAnswer:
\n(B) 128 cm2<\/sup><\/p>\n
\nGiven, radius of circle, r =OC = 8 cm
\nDiameter of the circle
\n= AC = 2 \u00d7 OC
\n= 2 \u00d7 8 = 16 cm
\nWhich is equal to the diagonal of a square.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-1.png?resize=145%2C137&ssl=1\")
\nLet side of square be ‘a’a.
\nUsing Pythagoras theorem,
\nAB2<\/sup> + BC2<\/sup> = AC2
\na2<\/sup> + a2<\/sup> = 162<\/sup>
\n2a2<\/sup> = 256
\na2<\/sup> = 128 cm 2<\/sup><\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is 50 cm.
\nReason (R): If the perimeter and the area of a circle are numerically equal, then the radius of the circle is 2 units.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn the case of assertion:
\nLet r1<\/sub> = 24 cm and r2<\/sub> = 7 cm Area of first circle = \u03c0\\(r_{1}^{2}\\) = \u03c0(24)2<\/sup> = 576\u03c0 cm2<\/sup>
\nArea of second circle = \u03c0\\(r_{2}^{2}\\) = \u03c0(7)2<\/sup> = 49\u03c0 cm2<\/sup> cm2<\/sup>
\nAccording to question,
\nArea of circle = Area of first circle + Area of second circle
\n\u03c0R2<\/sup> = 576\u03c0 + 49\u03c0
\n[where, R be radius of circle]
\nR2<\/sup> =625 = 25 cm
\nDiameter of a circle = 2R = 2 \u00d7 25 = 50 cm.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nLet the radius of the circle be r.
\nCircumference of circle = 2\u03c0r
\nArea of circle = \u03c0r2<\/sup>
\nGiven that, the circumference of the circle and the area of the circle are equal. This implies,
\n2\u03c0r = \u03c0r2<\/sup>
\nr = 2
\nTherefore, the radius of the circle is 2 units.
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not correct explanation for assertion.<\/p>\n<\/p>\nAssertion (A): In covering a distance s meter, a circular wheel of radius r meter makes \\(\\frac{s}{2 \\pi r}\\) revolution.
\nReason (R): The distance travelled by a circular wheel of diameter d cm in one revolution is 2\u03c0d cm.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nThe distance covered in one revolution is 2\u03c0r, that is, its circumference.
\n\u2234 Assertion is correct In case of reason:
\nBecause the distance travelled by the wheel in one revolution is equal to its circumference Circumference of wheel = \u03c0 \u00d7 diameter ,
\n= \u03c0 \u00d7 d
\n= \u03c0d
\nHence the given answer in the question is incorrect.
\n\u2234Reason is incorrect.
\nHence, assertion is correct and reason is incorrect.<\/p>\nAssertion (A): If circumferences of two circles are equal, then their areas will be equal.
\nReason (R): If the areas of two circles are equal, then their circumferences are equal.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nIf circumferences of two circles are equal, then their corresponding radii are equal. So, their areas will be equal.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nIf areas of two circles are equal, then their corresponding radii are equal. So, their circumference will be equal.
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
\nI. Read the following text and answer the following question on the basis of the same:
\nA horse is tied to a peg at one comer of a square shaped grass field of sides 15 m by means of a 5 m long rope (see the given figure)
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-2.png?resize=254%2C211&ssl=1\")
<\/p>\nWhat is the area of the grass field?<\/h2>\n
\n(B) 225 m
\n(C) 255 m2<\/sup>
\n(D) 15 m
\nAnswer:
\n(A) 225 m2<\/sup><\/p>\n
\nArea of square = (side)2<\/sup>
\n= 15 \u00d7 15
\n= 225 m2<\/sup><\/p>\nThe area of that part of the field in which the horse can graze.<\/h2>\n
\n(B) 19.265 m2<\/sup>
\n(C) 19 m2<\/sup>
\n(D) 78.5 m2<\/sup>
\nAnswer:
\n(A) 19.625 m2<\/sup><\/p>\n
\nFrom the figure, it can be observe that the horse can graze a sector of 90\u00b0 in a circle of 5 m radius.
\nArea that can be grazed by horse = Area of sector
\n= \\(\\frac{90^{\\circ}}{360^{\\circ}}\\) \u00d7 \u03c0r2<\/sup>
\n= \\(\\frac{1}{4}\\) \u00d7 3 . 14 \u00d7 5 \u00d7 5
\n= 19 . 625 m2<\/sup><\/p>\n<\/sup><\/p>\nThe grazing area if the rope were 10 m long instead of 5 m.<\/h2>\n
\n(B) 785 m2<\/sup>
\n(C) 225 m2<\/sup>
\n(D) 78.5 m2<\/sup>
\nAnswer:
\n(D) 78.5 m2<\/sup><\/p>\n
\nArea that can be grazed by the horse when length of rope is 10 m long
\n= \\(\\frac{90^{\\circ}}{360^{\\circ}}\\) \u00d7 \u03c0r2<\/sup>
\n= \\(\\frac{1}{4}\\) \u00d7 3 . 14 \u00d7 10 \u00d7 10
\n= 78.5 m2<\/sup><\/p>\nThe increase in the grazing area if the rope were 10 m long instead of 5 m.<\/h2>\n
\n(B) 58.875 m2<\/sup>
\n(C) 58 m2<\/sup>
\n(D) 78.5 m2<\/sup>
\nAnswer:
\n(B) 58.875 m2<\/sup><\/p>\n
\nIncrease in grazing area
\n= (78.5 – 19.625) m2<\/sup>
\n= 58.875 m2<\/sup><\/p>\nThe given problem is based on which concept?<\/h2>\n
\n(B) Area related to circles
\n(C) Circle
\n(D) None of these
\nAnswer:
\n(B) Area related to circles<\/p>\n
\nIn a workshop, brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the given figure.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-3.png?resize=201%2C211&ssl=1\")
<\/p>\nWhat is the radius of the circle?<\/h2>\n
\n(B) \\(\\frac{5}{2} \\mathrm{~mm}\\)
\n(C) 35 mm
\n(D) 10 mm
\nAnswer:
\n(A) \\(\\frac{35}{2} \\mathrm{~mm}\\)<\/p>\n
\nRadius of circle = \\(\\frac{\\text { Diameter }}{2}\\)
\n= \\(\\frac{35}{2} \\mathrm{~mm}\\)<\/p>\nWhat is the circumference of the brooch?<\/h2>\n
\n(B) 110 mm
\n(C) 50 mm
\n(D) 10 mm
\nAnswer:
\n(B) 110 mm<\/p>\n
\nCircumference of brooch = 2\u03c0r
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{35}{2}\\)
\n= 110 mm<\/p>\n<\/p>\nWhat is the total length of silver wire required?<\/h2>\n
\n(B) 825 mm
\n(C) 285 mm
\n(D) 852 mm
\nAnswer:
\n(C) 285 mm<\/p>\n
\nLength of wire required
\n= 110 + 5 x 35
\n= 110 + 175
\n= 285 mm.<\/p>\nWhat is the area of the each sector of the brooch?<\/h2>\n
\n(B) \\(\\frac{358}{4} \\mathrm{~mm}^{2}\\)
\n(C) \\(\\frac{585}{4} \\mathrm{~mm}^{2}\\)
\n(D) \\(\\frac{385}{4} \\mathrm{~mm}^{2}\\)
\nAnswer:
\n(D) \\(\\frac{385}{4} \\mathrm{~mm}^{2}\\)<\/p>\n
\nIt can be observed from the figure that an angle of each 10 sectors of the circle is subtending at the centre of the circle.
\n\u2234Area of each sector = \\(\\frac{36}{360^{\\circ}}\\) \u00d7 \u03c0r2<\/sup>
\n= \\(\\frac{1}{10}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{35}{2}\\) \u00d7 \\(\\frac{35}{2}\\)
\n= \\(\\frac{385}{4}\\) mm2<\/p>\nThe given problem is based on which mathematical concept ?<\/h2>\n
\n(B) Circles
\n(C) Construction
\n(D) none of these
\nAnswer:
\n(A) Areas Related to circles<\/p>\n
\nAREAS RELATED TO CIRCLES Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam, your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below.
\nObserve these carefully.
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-4.png?resize=262%2C283&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-5.png?resize=217%2C192&ssl=1\")
<\/p>\n
\nDesign II: This Pookalam is made with 9 circular design each of radius 7 cm.
\nRefer Design I:<\/p>\nThe side of equilateral triangle is<\/h2>\n
\n(B) 32\u221a3 cm
\n(C) 48 cm
\n(D) 64 cm
\nAnswer:
\n(B) 32\u221a3 cm<\/p>\nThe altitude of the equilateral triangle is<\/h2>\n
\n(B) 12 cm
\n(C) 48 cm
\n(D) 52 cm
\nAnswer:
\n(C) 48 cm<\/p>\nThe area of square is<\/h2>\n
\n(B) 1764 cm2<\/sup>
\n(C) 1830 cm2<\/sup>
\n(D) 1944 cm2<\/sup>
\nAnswer:
\n(B) 1764 cm2<\/sup><\/p>\n
\nradius = 7 cm
\ndiameter = 2 x 7 cm = 14 cm
\nside of square = 14 cm + 14 cm + 14 cm
\n= 42 cm.
\nArea of square = side2<\/sup>
\n= (42 cm)2<\/sup>
\n1764 cm2<\/sup><\/p>\n<\/sup><\/p>\nArea of each circular design is<\/h2>\n
\n(B) 132 cm2<\/sup>
\n(C) 144 cm2<\/sup>
\n(D) 154 cm2<\/sup>
\nAnswer:
\n(D) 154 cm2<\/sup><\/p>\n
\nradius = 7 cm
\nArea of each circular design = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 154 cm2<\/sup><\/p>\nArea of the remaining portion of the square ABCD<\/h2>\n
\n(B) 260 cm2<\/sup>
\n(C) 340 cm2<\/sup>
\n(D) 278 cm2<\/sup>
\nAnswer:
\n(A) 378 cm2<\/sup><\/p>\n
\nArea of 9 circular design = 9 \u00d7 \u03c0r2<\/sup>
\n= 9 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 1386 cm2<\/sup>
\nArea of square = 1764 cm2<\/sup>
\nArea of remaining portion of square – Area of 9 circular design
\n= 1764 cm 2<\/sup> – 1386 cm2<\/sup>
\n= 378 cm2<\/sup><\/p>\n
\nA Brooch A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-6.png?resize=185%2C185&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-10.png?resize=244%2C201&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-8.png?resize=207%2C192&ssl=1\")
\nDesign A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts.
\nDesign B: Brooch b is made two colours i.e. Gold and silver. Outer part is made with Gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.
\nRefer to Design A<\/p>\nThe total length of silver wire required is<\/h2>\n
\n(B) 200 mm
\n(C) 250 mm
\n(D) 280 mm
\nAnswer:
\n(B) 200 mm<\/p>\n
\nDiameter = 28 mm
\nradius = 14 mm
\nTotal length of wire = length of 4 diameter + circumference of circle.
\n= 4 \u00d7 28 + 2\u03c0r2<\/sup>
\n= 112 + 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 14
\n= 112 + 88
\n= 200 mm<\/p>\nThe area of each sector of the brooch is<\/h2>\n
\n(B) 52 mm2
\n(C) 77 mm2
\n(D) 68 mm2
\nAnswer:
\n(C) 77 mm2<\/p>\n
\nArea of each sector of Brooch
\n= \\(\\frac{1}{8}\\) \u00d7 Area of Brooch
\n= \\(\\frac{1}{8}\\) \u00d7 \u03c0r2<\/sup>
\n= \\(\\frac{1}{8}\\) x \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 14
\n= 77 mm2<\/sup><\/p>\nThe circumference of outer part (golden) is<\/h2>\n
\n(B) 82.2 mm
\n(C) 72.50 mm
\n(D) 62.86 mm
\nAnswer:
\n(D) 62.86 mm<\/p>\nThe difference of areas of golden and silver parts is<\/h2>\n
\n(B) 44\u03c0
\n(C) 51\u03c0
\n(D) 64\u03c0
\nAnswer:
\n(C) 51\u03c0<\/p>\n<\/p>\nA boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm ?<\/h2>\n
\n(B) 3
\n(C) 4
\n(D) 5
\nAnswer:
\n(C) 4<\/p>\n
\nCircumference of silver part of Brooch
\n= 44 cm
\n2\u03c0r = 44 mm
\n2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r = 44
\nr = 7 mm.
\nradius of whole Brooch
\n= 7 mm + 8 mm
\n= 10 mm.
\nCircumference of outer edge
\n= 2\u03c0r
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 10
\n= \\(\\frac{440}{7}\\) mm
\nlet the number of revolutions = n
\nNow, According to question,
\nn . 2\u03c0r = 80\u03c0
\nn . \\(\\frac{440}{7}\\) = 80\u03c0
\nn . \\(\\frac{440}{7}\\) = 80 x \\(\\frac{22}{7}\\)
\nn = 4<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n