WBBSE Solutions For Class 9 Maths<\/a><\/strong><\/span><\/p>\nLet the radius = r,\u00a0<\/span><\/p>\n\u2234 r = OP = OQ = OR = ………..<\/span><\/p>\nNow, the area of the polygon ABCD \u2026\u2026\u2026\u2026.<\/span><\/p>\n= \u0394OAB + \u0394OBC + \u0394OCD + ………..<\/span><\/p>\n= \\(\\frac{1}{2} \\mathrm{AB} \\cdot \\mathrm{OP}+\\frac{1}{2} \\mathrm{BC} \\cdot \\mathrm{OQ}+\\frac{1}{2} \\mathrm{CD} \\cdot \\mathrm{OR}+\\cdots \\cdots\\)<\/span><\/p>\n= \\(\\frac{1}{2} \\mathrm{AB} \\cdot r+\\frac{1}{2} \\mathrm{BC} \\cdot r+\\frac{1}{2} \\mathrm{CD} \\cdot r+\\cdots \\cdots\\)<\/span><\/p>\n= \\(\\frac{1}{2} r(\\mathrm{AB}+\\mathrm{BC}+\\mathrm{CD}+\\cdots \\cdots)\\)<\/span><\/p>\n= \\(\\frac{1}{2} r \\times \\text { (Perimeter of the polygon) }\\)<\/span><\/p>\nNow, if the number of sides is a large number, i.e., if n tends to infinity, then AB, BC, CD,….. etc. reduces smaller and smaller and ultimately at the limiting position they coincide with the circumference of the circle.<\/span><\/p>\n\u2234 Area of the circle = \\(\\frac{1}{2} r \\times(\\text { circumference of the circle) }\\)<\/span><\/p>\n= \\(\\frac{1}{2} r \\times 2 \\pi r=\\pi r^2\\)<\/span><\/p>\n\u2234 <\/span>Area of a circle = \u03c0r\u00b2 (r= radius). (Proved)<\/span><\/p>\n <\/p>\n
Chapter 3 Area Of Circles\u00a0 Area Of Circles Inscribed In A Square<\/h2>\n
Area Of Circles Inscribed In A Square:-<\/strong><\/p>\nIf a circle is inscribed in a square, then the diameter of the circle is equal to the equal sides of the square.<\/span><\/p>\n\u2234 Radius of the circle = \\(\\frac{\\text { side of the square }}{2}\\)<\/span><\/p>\nNow, if the equal sides of the square be units, then the radius of the circle = \\(\\frac{a}{2}\\) <\/span>units,\u00a0<\/span><\/p>\n\u2234 Area of the circle = \\(\\pi\\left(\\frac{a}{2}\\right)^2 \\text { sq-units }=\\frac{\\pi a^2}{4} \\text { sq-units }\\)<\/span><\/p>\n <\/p>\n
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Chapter 3 Area Of Circles Area Of Circles Circumscribing A Square<\/h2>\n
Area Of Circles Circumscribing A Square:-<\/strong><\/p>\nIf a circle is circumscribed in a square, then the diameter of the circle is equal to the diagonal of the square.<\/span><\/p>\nNow, if the equal sides of the square be a unit, then its diagonal is \u221a2a units.<\/span><\/p>\n\u2234 Diameter = \u221a2a units<\/span><\/p>\n\u2234 Radius = \\(\\frac{\\sqrt{ } 2 a}{2} \\text { units }=\\frac{a}{\\sqrt{ } 2} \\text { units. }\\)<\/span><\/p>\n\u2234 Area of the circle = \\(\\pi \\times\\left(\\frac{a}{\\sqrt{ } 2}\\right)^2 \\text { sq-units }=\\frac{\\pi a^2}{2} \\text { sq-units. }\\)<\/span><\/p>\n <\/p>\n
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Chapter 3 Area Of Circles Area Of Circles Circumscribing A Rectangle<\/h2>\n
Area Of Circles Circumscribing A Rectangle:-<\/strong><\/p>\nIf a circle is circumscribed in a rectangle, then the diameter of the circle is equal to the diagonal of the rectangle.<\/span><\/p>\nNow, if the length and breadth of the rectangle be a units and b units respectively,<\/span><\/p>\nthen the diagonal of the rectangle is \\(\\sqrt{a^2+b^2}\\) units.<\/span><\/p>\n\u2234 Diameter of the circle = \\(\\sqrt{a^2+b^2}\\)<\/span><\/p>\n\u2234 Radius of the circle = \\(\\frac{\\sqrt{a^2+b^2}}{2}\\)<\/span><\/p>\n\u2234 Area of the circle = \\(=\\pi\\left(\\frac{\\sqrt{a^2+b^2}}{2}\\right)^2 \\text { sq-units }=\\frac{\\pi}{4}\\left(a^2+b^2\\right) \\text { sq-units }\\)<\/span><\/p>\n <\/p>\n
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Chapter 3 Area Of Circles To Find The In-Radius Of A Triangle<\/h2>\n
The In-Radius Of A Triangle:-<\/strong><\/p>\nLet O be the in-radius of the \u0394ABC.<\/span><\/p>\nLet us draw perpendiculars OD, OE, and OF to the sides BC, CA, and AB respectively.<\/span><\/p>\nLet the in-radius of \u0394ABC be r.<\/span><\/p>\n\u2234 OD = OE = OF = r units<\/span><\/p>\nNow, the area of \u0394ABC = area of\u0394OBC + area of \u0394OAC + area of \u0394OAB.<\/span><\/p>\n\u21d2 \\(\\frac{1}{2} \\times \\mathrm{BC} \\times \\mathrm{OD}+\\frac{1}{2} \\times \\mathrm{AC} \\times \\mathrm{OE}+\\frac{1}{2} \\times \\mathrm{AB} \\times \\mathrm{OF}\\)<\/span><\/p>\n= \\(\\frac{1}{2} \\times \\mathrm{BC} \\times r+\\frac{1}{2} \\times \\mathrm{AC} \\times r+\\frac{1}{2} \\times \\mathrm{AB} \\times r\\)<\/p>\n
= \\(\\frac{1}{2}(\\mathrm{BC}+\\mathrm{AC}+\\mathrm{AB}) \\times r\\)<\/p>\n
\u2234 Area of the triangle = \\(\\frac{1}{2}\\) x Perimeter of the triangle x in-radius<\/p>\n
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Chapter 3 Area Of Circles\u00a0To Find The In Radius And Circum Radius Of An Equilateral Triangle<\/h2>\n
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The In Radius And Circum Radius Of An Equilateral Triangle:-<\/strong><\/p>\nLet the equal sides of the equilateral triangle be a unit.<\/span><\/p>\n\u2234 Height of the triangle = \\(\\frac{\\sqrt{3}}{2} a \\text { units }\\)<\/span><\/p>\nSince, the height of an equilateral triangle is also a median of it, the <\/span>median\u00a0<\/span><\/p>\n= \\(\\frac{\\sqrt{3}}{2} a \\text { units }\\)<\/span><\/p>\nNow, the median is divided internally at the centroid of the triangle into the ratio 2: 1,<\/span><\/p>\n\u2234\u00a0 in-radius = \\(\\frac{1}{3} \\times\\left(\\frac{\\sqrt{3}}{2} a\\right) \\text { units }=\\frac{a}{2 \\sqrt{3}} \\text { units }\\)<\/span><\/p>\nand circum-radius = \\(\\frac{2}{3} \\times \\frac{\\sqrt{3} a}{2} \\text { units }=\\frac{a}{\\sqrt{3}} \\text { units }\\)<\/span><\/p>\nIf an arc of a circle of radius r units produces an angle e at the center, then the length of that arc = \\(\\frac{\\theta}{360} \\times \\text { circumference of the circle }=\\frac{2 \\pi r \\theta}{360} \\text { units. }\\)<\/span><\/p>\nand the area of the circle = \\(\\frac{\\theta}{360} \\times \\text { area of the circle }=\\frac{\\pi r^2 \\theta}{360} \\text { sq-units }\\)<\/p>\n
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Chapter 3 Area Of Circles Select The Correct Answer (MCQ)<\/h2>\n
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Question 1. The area of a circle is x sq-units. If the circumference is y-units and the diameter be z-units, <\/b>then the value of \\(\\frac{x}{yz}\\)is<\/b><\/p>\n\n- \\(\\frac{1}{2}\\)<\/li>\n
- \\(\\frac{1}{4}\\)<\/li>\n
- 1<\/span><\/li>\n
- \\(\\frac{1}{8}\\)<\/li>\n<\/ol>\n
Solution :<\/b><\/p>\n
The area of the circle = \\(\\pi\\left(\\frac{z}{2}\\right)^2 \\text { sq-units }\\)<\/span><\/p>\nand circumference = \\(y=2 \\pi\\left(\\frac{z}{2}\\right) \\text { units }\\)<\/span><\/p>\n\u2234 \\(\\frac{x}{y z}=\\frac{\\frac{\\pi z^2}{4}}{\\pi z \\cdot z}=\\frac{1}{4}\\)<\/span><\/p>\n <\/p>\n
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Question 2. The ratio of the areas of two squares circumscribed and inscribed in a circle is.<\/b><\/p>\n\n- 4:1<\/li>\n
- 1:4<\/li>\n
- 2:1<\/li>\n
- 1:2<\/li>\n<\/ol>\n
Solutio<\/strong>n:<\/b><\/p>\nLet the radius of the circle = r units.<\/span><\/p>\n\u2234 The side of the circumscribed square = diameter of the circle = 2r units and the diagonal of the inscribed square = diameter of the circle = 2r units.<\/span><\/p>\nThe side of the square inscribed in the circle= \\(\\frac{\\text { diagonal }}{\\sqrt{2}}=\\frac{2 r}{\\sqrt{2}} \\text { units }=\\sqrt{2} r \\text { units }\\)<\/span><\/p>\nThe required ratio = (2r units)\u00b2: (\u221a2r units)\u00b2 = 4r\u00b2 sq-units: 2r\u00b2 sq-units = 2: 1.<\/p>\n
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Question 3. The numerical value of the circumference and area of a circle are equal. The length of the diagonal of the circumscribed circle is<\/b><\/p>\n\n- 4 units<\/li>\n
- 2 units<\/li>\n
- 4\u221a2 units<\/li>\n
- 2\u221a2 units<\/li>\n<\/ol>\n
Solution:\u00a0<\/strong><\/p>\nLet the radius of the circle is r units,\u00a0<\/span><\/p>\n\u2234 \\(2 \\pi r=\\pi r^2 \\Rightarrow r=2\\)<\/span><\/p>\n\u2234 The side of the square circumscribing the circle = <\/span>the diameter = 2r units 2 x 2 units = 4 units<\/span><\/p>\n\u2234 Diagonal = side x \u221a2 = 4\u221a2 units.<\/span><\/p>\n <\/p>\n
Question 4. The ratio of the areas of the circles circumscribing and inscribed in an equilateral triangle <\/b>is<\/b><\/p>\n\n- 4:1<\/li>\n
- 1:4<\/li>\n
- 2:1<\/li>\n
- 1:2<\/li>\n<\/ol>\n
Solution:\u00a0<\/strong><\/p>\nThe circum-radius: in-radius of the equilateral triangle = \\(\\frac{2}{3} \\times \\text { height : } \\frac{1}{3} \\times \\text { height }\\)<\/span><\/p>\n= 2: 1<\/span><\/p>\nThe ratio of the areas is 2\u00b2: 1\u00b2 = 4: 1.<\/span><\/p>\n <\/p>\n
Chapter 3 Area Of Circles Short Answer Type Questions<\/h2>\n
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Question 1. If the radius of a circle is increased by 10%, then what percentage of its area will be increased?<\/b><\/p>\n
Solution:<\/b><\/p>\n
Given<\/strong><\/p>\nThe radius of a circle is increased by 10%<\/b><\/p>\n
Let the initial radius of the circle be 100r units.<\/p>\n
\u200d\u00a0initial area = \\(\\pi \\times(100 r \\text { units })^2=10000 \\pi r^2 \\text { sq-units }\\)<\/span><\/p>\nRadius when increased by 10% = \\(\\left[100 r \\times\\left(1+\\frac{10}{100}\\right)\\right] \\text { units }=110 r \\text { units }\\)<\/span><\/p>\n\u2234 Area after increment = <\/span>\\(\\left(12100 \\pi r^2-10000 \\pi r^2\\right) \\text { sq-units }=2100 \\pi r^2 \\text { sq-units }\\)<\/span><\/p>\n\u2234 Increased area = \\(\\left(12100 \\pi r^2-10000 \\pi r^2\\right) \\text { sq-units }=2100 \\pi r^2 \\text { sq-units }\\)<\/span><\/p>\n\u2234 Percentage of increment = <\/span>\\(\\left(\\frac{2100 \\pi r^2}{10,000 \\pi r^2} \\times 100\\right) \\%=21 \\%\\)<\/span><\/p>\n <\/p>\n
Question 2. If the circumference of a circle is decreased by 50%, then what percentage of its area will be decreased?<\/b><\/p>\n
Solution:\u00a0<\/b><\/p>\n
Given<\/strong><\/p>\nThe circumference of a circle is decreased by 50%<\/b><\/p>\n
Let the initial radius of the circle = r units.<\/p>\n
\u2234 initial circumference = 2\u03c0r units and area = \u03c0<\/span>r\u00b2 units.<\/span><\/p>\nNow, circumference after discriminant = \\(\\frac{<\/span>2\u03c0r}{2}\\) = <\/span>units = \u03c0r unit [ \\(50 \\%=\\frac{50}{100}=\\frac{1}{2}\\) ]<\/span><\/p>\n\u2234 Then the radius = \\(\\frac{\\text { circumference }}{2 \\pi}=\\frac{\\pi r}{2 \\pi} \\text { units }=\\frac{r}{2} \\text { units }\\)<\/span><\/p>\n\u2234 Area after discriminant = \\(\\pi \\times\\left(\\frac{r}{2}\\right)^2 \\text { sq-units }=\\frac{\\pi r^2}{4} \\text { sq-units }\\)<\/span><\/p>\n\u2234 Decreased area = \\(\\left(\\pi r^2-\\frac{\\pi r^2}{4}\\right) \\text { sq-units } \u20b9 \\frac{3 \\pi r^2}{4} \\text { sq-units }\\)<\/span><\/p>\n\u2234 Required percentage = \\(\\left(\\frac{\\frac{3 \\pi r^2}{4}}{\\pi r^2} \\times 100\\right) \\%=\\left(\\frac{3 \\pi r^2}{4} \\times \\frac{1}{\\pi r^2} \\times 100\\right) \\%=75 \\%\\)<\/span><\/p>\n <\/p>\n
Question 3. Jaya inscribed a circle in a square. This circle is also the circumcircle of an equilateral triangle, each of whose sides are 4\u221a3 cm. Find the diagonal of the square.<\/b><\/p>\n
Solution:<\/strong> <\/span><\/p>\nGiven<\/strong><\/p>\nJaya inscribed a circle in a square.<\/b><\/p>\n
This circle is also the circumcircle of an equilateral triangle, each of whose sides are 4\u221a3 cm.<\/b><\/p>\n
The side of the equilateral triangle is 4\u221a3 cm<\/span><\/p>\n\u2234 Height = \\(\\frac{\\sqrt{3}}{2} \\times 4 \\sqrt{3} \\mathrm{~cm}=6 \\mathrm{~cm} .\\)<\/span><\/p>\n\u2234 Radius of the circumcircle= \\(\\frac{2}{3}\\) x 6 cm = 4 cm.<\/span><\/p>\n\u2234 Diameter = 2 x 4 cm = 8 cm<\/span><\/p>\nSince the circle is inscribed in the square, the side of the square is equal <\/span>to the diameter of the circle.<\/span><\/p>\n\u2234 Side of the square = 8 cm.<\/span><\/p>\nThe diagonal of the square = 8\u221a2 cm.<\/span><\/p>\n <\/p>\n
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Question 4. Sumit cut a wire into two equal parts and bent one of them into a square and the other into a circle. If the area of the square be 33 sq-cm more than that of the circle, find the length of the wire.<\/b><\/p>\n
Solution:<\/strong><\/p>\nGiven<\/strong><\/p>\nSumit cut a wire into two equal parts and bent one of them into a square and the other into a circle.<\/b><\/p>\n
The area of the square be 33 sq-cm more than that of the circle.<\/b><\/p>\n
Let the length of the wire be 2a cm.<\/span><\/p>\n\u2234 length of each part =\\(\\frac{2a}{2}\\) cm = a cm\u00a0<\/span><\/p>\nAlso, let the side of the square be x cm,\u00a0<\/span><\/p>\n\u2234 Perimeter = 4x cm.<\/span><\/p>\n\u2234 4x = a \u21d2 x = \\(\\frac{a}{4}\\)<\/span><\/p>\nAgain, let the radius of the circle be r cm.\u00a0<\/span><\/p>\n\u2234 Circumference = 2\u03c0r cm <\/span><\/p>\n\u2234 \\(2 \\pi r=a \\Rightarrow r=\\frac{a}{2 \\pi}\\)<\/span><\/p>\nAs per the question, \\(\\frac{14 a^2-11 a^2}{176}=33 \\Rightarrow 3 a^2=33 \\times 176 \\Rightarrow a^2=\\frac{33 \\times 176}{3}\\)<\/span><\/p>\n\u21d2 \\(\\frac{14 a^2-11 a^2}{176}=33 \\Rightarrow 3 a^2=33 \\times 176 \\Rightarrow a^2=\\frac{33 \\times 176}{3}\\)<\/span><\/p>\n\u21d2 \\(a=\\sqrt{11 \\times 11 \\times 16}=11 \\times 4=44\\)<\/span><\/p>\n\u2234 The length of the wire = 2 x 44 cm = 88 cm.<\/span><\/p>\n <\/p>\n
Chapter 3 Area Of Circles Long Answer Type Questions<\/h2>\n
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Question 1. Palas and Piyali have drawn two circles, the radius of which is 4: 5; then find the ratio of their areas.<\/b><\/p>\n
Solution:<\/b><\/p>\n
Given<\/strong><\/p>\nPalas and Piyali have drawn two circles, the radius of which is 4: 5.<\/b><\/p>\n
Let the radii of the circles be 4r units and 5r units respectively.<\/span><\/p>\n