{"id":10869,"date":"2023-05-02T11:21:34","date_gmt":"2023-05-02T05:51:34","guid":{"rendered":"https:\/\/wbbsesolutions.net\/?p=10869"},"modified":"2024-04-16T12:08:24","modified_gmt":"2024-04-16T06:38:24","slug":"wbbse-solutions-for-class-9-maths-mensuration-chapter-3","status":"publish","type":"post","link":"https:\/\/wbbsesolutions.net\/wbbse-solutions-for-class-9-maths-mensuration-chapter-3\/","title":{"rendered":"WBBSE Solutions For Class 9 Maths Mensuration Chapter 3 Area Of Circles"},"content":{"rendered":"

Mensuration Chapter 3 Area Of Circles<\/h2>\n

Chapter 3 Area Of Circles Area of a circle<\/strong><\/p>\n

Area of circles circumscribed in Determination of in-radius and circum-radius<\/span><\/p>\n

Area of a circle = x (radius)\u00b2 = \u03c0r\u00b2 [ r = radius of circle]<\/span><\/p>\n

The area of the region closed by two concentric circles= \\(\\pi\\left(\\mathrm{R}^2-r^2\\right)=\\pi(\\mathrm{R}+r)(\\mathrm{R}-r)\\) \u00a0 <\/span>[R and r are the radii of the circles ]<\/span><\/p>\n

Let the sides AB, BC, CD, \u2026\u2026\u2026\u2026 <\/span>of a regular polygon touch the circle with the center at O and the number <\/span>of sides of it be n.\u00a0<\/span><\/p>\n

Let us draw the radii OP, OQ, OR,………………<\/span>contacts. <\/span><\/p>\n

\u2234 these are<\/span> perpendiculars to AB, BC, CD,… etc. respectively.<\/span><\/p>\n

Read and Learn More WBBSE Solutions For Class 9 Maths<\/a><\/strong><\/span><\/p>\n

Let the radius = r,\u00a0<\/span><\/p>\n

\u2234 r = OP = OQ = OR = ………..<\/span><\/p>\n

Now, 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>\n

Now, 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>\n

If 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>\n

Now, 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

\"WBBSE<\/p>\n

 <\/p>\n

Chapter 3 Area Of Circles Area Of Circles Circumscribing A Square<\/h2>\n

Area Of Circles Circumscribing A Square:-<\/strong><\/p>\n

If a circle is circumscribed in a square, then the diameter of the circle is equal to the diagonal of the square.<\/span><\/p>\n

Now, 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

\"WBBSE<\/p>\n

 <\/p>\n

Chapter 3 Area Of Circles Area Of Circles Circumscribing A Rectangle<\/h2>\n

Area Of Circles Circumscribing A Rectangle:-<\/strong><\/p>\n

If a circle is circumscribed in a rectangle, then the diameter of the circle is equal to the diagonal of the rectangle.<\/span><\/p>\n

Now, if the length and breadth of the rectangle be a units and b units respectively,<\/span><\/p>\n

then 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

\"WBBSE<\/p>\n

 <\/p>\n

Chapter 3 Area Of Circles To Find The In-Radius Of A Triangle<\/h2>\n

The In-Radius Of A Triangle:-<\/strong><\/p>\n

Let O be the in-radius of the \u0394ABC.<\/span><\/p>\n

Let us draw perpendiculars OD, OE, and OF to the sides BC, CA, and AB respectively.<\/span><\/p>\n

Let the in-radius of \u0394ABC be r.<\/span><\/p>\n

\u2234 OD = OE = OF = r units<\/span><\/p>\n

Now, 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

 <\/p>\n

\"WBBSE<\/p>\n

 <\/p>\n

Chapter 3 Area Of Circles\u00a0To Find The In Radius And Circum Radius Of An Equilateral Triangle<\/h2>\n

 <\/p>\n

The In Radius And Circum Radius Of An Equilateral Triangle:-<\/strong><\/p>\n

Let 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>\n

Since, 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>\n

Now, 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>\n

and circum-radius = \\(\\frac{2}{3} \\times \\frac{\\sqrt{3} a}{2} \\text { units }=\\frac{a}{\\sqrt{3}} \\text { units }\\)<\/span><\/p>\n

If 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>\n

and 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

 <\/p>\n

Chapter 3 Area Of Circles Select The Correct Answer (MCQ)<\/h2>\n

 <\/p>\n

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
  1. \\(\\frac{1}{2}\\)<\/li>\n
  2. \\(\\frac{1}{4}\\)<\/li>\n
  3. 1<\/span><\/li>\n
  4. \\(\\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>\n

    and 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

     <\/p>\n

    Question 2. The ratio of the areas of two squares circumscribed and inscribed in a circle is.<\/b><\/p>\n

      \n
    1. 4:1<\/li>\n
    2. 1:4<\/li>\n
    3. 2:1<\/li>\n
    4. 1:2<\/li>\n<\/ol>\n

      Solutio<\/strong>n:<\/b><\/p>\n

      Let 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>\n

      The 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>\n

      The required ratio = (2r units)\u00b2: (\u221a2r units)\u00b2 = 4r\u00b2 sq-units: 2r\u00b2 sq-units = 2: 1.<\/p>\n

       <\/p>\n

      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
      1. 4 units<\/li>\n
      2. 2 units<\/li>\n
      3. 4\u221a2 units<\/li>\n
      4. 2\u221a2 units<\/li>\n<\/ol>\n

        Solution:\u00a0<\/strong><\/p>\n

        Let 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
        1. 4:1<\/li>\n
        2. 1:4<\/li>\n
        3. 2:1<\/li>\n
        4. 1:2<\/li>\n<\/ol>\n

          Solution:\u00a0<\/strong><\/p>\n

          The 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>\n

          The 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

           <\/p>\n

          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>\n

          The 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>\n

          Radius 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>\n

          The 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>\n

          Now, 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>\n

          Given<\/strong><\/p>\n

          Jaya 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>\n

          Since 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>\n

          The diagonal of the square = 8\u221a2 cm.<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          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>\n

          Given<\/strong><\/p>\n

          Sumit 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>\n

          Also, 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>\n

          Again, 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>\n

          As 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

           <\/p>\n

          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>\n

          Palas 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

          The ratio of the areas of the circles = \\(\\pi(4 r)^2: \\pi(5 r)^2=16 \\pi r^2: 25 \\pi r^2=16: 25\\)<\/span><\/p>\n

          The required ratio = 16: 25.<\/span><\/p>\n

           <\/p>\n

          Question 2. There is a road of the breadth of 7 m of a circular park along its outer circumference. If the circumference of the circle is 352 m, then find the area of the road. To concrete the road at a rate of Rs. 20 per sq-m, what will be the required expenditure?<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          There is a road of the breadth of 7 m of a circular park along its outer circumference.<\/b><\/p>\n

          The circumference of the circle is 352 m.<\/b><\/p>\n

          To concrete the road at a rate of Rs. 20 per sq-m<\/b><\/p>\n

          Let the radius of the circular park be r m.<\/span><\/p>\n

          \u2234 its circumference = 2\u03c0r m<\/span><\/p>\n

          As per question, 2\u043br = 352\u00a0\u21d2 <\/span>\\(2 \\times \\frac{22}{7} \\times r=352\\)<\/span><\/p>\n

          \u21d2 \\(r=\\frac{352 \\times 7}{2 \\times 22}=56\\)<\/span><\/p>\n

          The radius of the park = 56 m.<\/span><\/p>\n

          \u2234 Including the road the<\/span> radius of the park = (56+7) m = 63 m\u00a0<\/span><\/p>\n

          \u2234 Including the road the area of the circle = \u03c0 x 63\u00b2 sq-cm<\/span><\/p>\n

          Again, the area of the circle = <\/span>\u03c0 x 56\u00b2<\/span><\/p>\n

          \u2234 The area of the road only = \\(\\left(\\pi \\times 63^2-\\pi \\times 56^2\\right) \\mathrm{sq}-\\mathrm{m}=\\pi\\left(63^2-56^2\\right) \\mathrm{sq}-\\mathrm{m}\\)<\/span><\/p>\n

          = \u03c0 x (63 + 56) x (63 – 56) sqm<\/p>\n

          = \\(\\frac{22}{7} \\times 119 \\times 7 \\mathrm{sq}-\\mathrm{m}=2618 \\mathrm{sq}-\\mathrm{m}\\)<\/span><\/p>\n

          Now the expenditure to concrete the road= is Rs. 20 x 2618\u00a0<\/span><\/p>\n

          = Rs. 52360.<\/span><\/p>\n

          \u2234 The required area is 2618 sqm and the required expenditure = is Rs. 52360.<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Question 3. There is a road of equal breadth around the circular field of Bakultala. The outer circumference of the road is 132 m more than its inner circumference. If the area of the road is 14190 sqm, then find the area of the field.<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          There is a road of equal breadth around the circular field of Bakultala.<\/b><\/p>\n

          The outer circumference of the road is 132 m more than its inner circumference.<\/b><\/p>\n

          the area of the road is 14190 sqm,<\/b><\/p>\n

          Let the inner radius and outer radius of the circular road be \\(r_1 \\mathrm{~m} \\text { and } r_2\\) respectively.<\/span><\/p>\n

          \u2234 inner circumference = \\(2 \\pi r_1 \\mathrm{~m} \\text { and outer circumference }=2 \\pi r_2 \\mathrm{~m} \\text {. }\\)<\/span><\/p>\n

          As per the question, \\(2 \\pi r_2-2 \\pi r_1=132 \\Rightarrow 2 \\times \\frac{22}{7}\\left(r_2-r_1\\right)=132\\)<\/span><\/p>\n

          \\(r_2-r_1=\\frac{132 \\times 7}{2 \\times 22} \\Rightarrow\\left(r_2-r_1\\right)=21\\) \u2026\u2026\u2026\u2026\u2026(1)<\/span><\/p>\n

          Given that the area of the road = 14190 sqm<\/span><\/p>\n

          \u2234 \\(\\pi\\left(r_2^2-r_1^2\\right)=14190 \\Rightarrow\\left(r_2+r_1\\right)\\left(r_2-r_1\\right)=14190 \\times \\frac{7}{22} \\Rightarrow\\left(r_2+r_1\\right) \\times 21=14190 \\times \\frac{7}{22}\\) \u2026\u2026\u2026\u2026\u2026\u2026(2)<\/span><\/p>\n

          Adding (1) + (2) we get, \\(r_2-r_1+r_2+r_1=21+215 \\Rightarrow 2 r_2=236 \\Rightarrow r_2=118\\)<\/span><\/p>\n

          \\(r_1=215-118=97\\)\u00a0 \u00a0 \u2234 the radius of the field = 97 m<\/span><\/p>\n

          \u2234 <\/span>The area of the circle = <\/span>\\(\\pi r^2 \\mathrm{sq}-\\mathrm{m}=\\frac{22}{7} \\times(97)^2 \\mathrm{sq}-\\mathrm{m}=\\frac{206998}{7} \\mathrm{sq}-\\mathrm{m}=29571 \\frac{1}{7} \\mathrm{sq}-\\mathrm{m} .\\)<\/span><\/p>\n

           <\/p>\n

          Question 4. The area of a circular field is 154 sq-cm. Find the perimeter of the square circumscribing the circle and also find the area of the square. If the square would be inscribed in the circle, then what would be the perimeter and area of the square?<\/b><\/p>\n

          Solution:\u00a0<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          The area of a circular field is 154 sq-cm.<\/b><\/p>\n

          The square would be inscribed in the circle.<\/b><\/p>\n

          Let the radius of the field = r cm,<\/span><\/p>\n

          \u2234 Area of the field = r\u00b2 sq-cm.<\/span><\/p>\n

          As per the question, \\(\\pi r^2=154 \\Rightarrow \\frac{22}{7} \\times r^2=154 \\Rightarrow r^2=\\frac{154 \\times 7}{22} \\Rightarrow r^2=49 \\Rightarrow r=7\\)<\/span><\/p>\n

          The radius of the field = 7 cm.<\/span><\/p>\n

          \u2234 Diameter of the field 2r= 2 x 7 cm = 14 cm.<\/span><\/p>\n

          The side of the square circumscribing the circular field is equal to the diameter of the field.\u00a0<\/span><\/p>\n

          \u2234 The perimeter of the square = 4 x side = 4 x 14 cm = 56 cm.<\/span><\/p>\n

          \u2234 Area of the square = (side)\u00b2 = (14)\u00b2 sq-cm = 196 sq-cm.<\/span><\/p>\n

          Again, when the square is inscribed in the circular field, the diagonal of the square is equal to the diameter of the field.<\/span><\/p>\n

          \u2234 Now, if the side of the square be x cm, then \\(\\sqrt{2} x=14 \\Rightarrow x=\\frac{14}{\\sqrt{2}}=7 \\sqrt{2}\\)<\/span><\/p>\n

          \u2234 Then the perimeter of the square = 4 x side = 4 x 7\u221a2 cm = 28 \u221a2 cm.\u00a0<\/span><\/p>\n

          and the area of it (side)\u00b2= (7\u221a2)\u00b2 sq-cm = 98 sq-cm<\/span><\/p>\n

          The perimeter and area of the square 28 \u221a2 cm and 98 sq-cm.<\/span><\/p>\n

           <\/p>\n

          Question 5. Lina bought a bracelet from a fair. There is 269-5 sq-cm metal in the bracelet. If the outer diameter of the bracelet is 28 cm, find its inner radius.<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given\u00a0<\/strong><\/p>\n

          Lina bought a bracelet from a fair. There is 269-5 sq-cm metal in the bracelet.<\/b><\/p>\n

          The outer diameter of the bracelet is 28 cm.<\/b><\/p>\n

          Let the inner radius be r cm<\/span><\/p>\n

          Given that the outer-diameter = 28 cm,\u00a0<\/span><\/p>\n

          \u2234 outer-radius = 14 cm<\/span><\/p>\n

          There is 269.5 sq-cm metal in the bracelet.<\/span><\/p>\n

          \u2234 \u03c0 (14\u00b2 – r\u00b2) = 269.5<\/span><\/p>\n

          (196 – r\u00b2) = 269.5 x \\(\\frac{7}{22}\\)<\/span><\/p>\n

          \u2234 <\/span>Inner-radius = 2 x \\(\\frac{21}{2}\\) <\/span>cm = 21 cm.<\/span><\/p>\n

           <\/p>\n

          Question 6. The radius of the circumcircle of a right-angled triangle is 5 cm and the perpendicular distance of the hypotenuse from its opposite vertex is 4 cm. Find the area of the triangle.\u00a0<\/b><\/p>\n

          Solution:\u00a0<\/b><\/p>\n

          Given\u00a0<\/strong><\/p>\n

          The radius of the circumcircle of a right-angled triangle is 5 cm and the perpendicular distance of the hypotenuse from its opposite vertex is 4 cm.<\/b><\/p>\n

          We know that the circumcentre of the circumcircle of a right-angled triangle is the midpoint of the hypotenuse and its circum-radius is half of the hypotenuse.<\/p>\n

          In the \u2220B is the right angle of the \u0394ABC.<\/p>\n

          The midpoint of the hypotenuse AC is O and BD is the perpendicular drawn from B to AC.<\/span><\/p>\n

          Now, AC = 2 x OA = 2 x 5 cm = 10 cm and BD = 4 cm.<\/span><\/p>\n

          \u2234 Area of \u0394ABC = \\(\\frac{1}{2}\\) <\/span>\u00d7 AC x BD sq-cm<\/span><\/p>\n

          \u00a0= \\(\\frac{1}{2}\\) <\/span>x 10 x 4 sq-cm 20 sq-cm<\/span><\/p>\n

          \u2234 The required area = <\/span>20 sq-cm.<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Question 7. In the given figure, OPQR is a rhombus three of whose vertices are on the circle with the center at O. If the area of the rhombus is 32\u221a3 sq-cm, find the radius of the circle.<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          OPQR is a rhombus three of whose vertices are on the circle with the center at O. <\/b><\/p>\n

          The area of the rhombus is 32\u221a3 sq-cm.<\/b><\/p>\n

          Let the radius of the circle = r cm.<\/span><\/p>\n

          OPQR is a rhombus,<\/span><\/p>\n

          \u2234 OP = PQ = QR = RO<\/span><\/p>\n

          \u2234 OP = OR radius. <\/span><\/p>\n

          OP = OQ = PQ = r cm\u00a0<\/span><\/p>\n

          \u2234 \u0394OPQ is an equilateral triangle.<\/span><\/p>\n

          Now area of the rhombus = 2 x (area of \u0394OPQ)<\/span><\/p>\n

          = \\(2 \\times \\frac{\\sqrt{3}}{4} \\times r^2 \\mathrm{sq}-\\mathrm{cm}=\\frac{\\sqrt{3}}{2} r^2 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          As per the question, \\(\\frac{\\sqrt{3}}{2} r^2=32 \\sqrt{3} \\Rightarrow r^2=64 \\Rightarrow r=8\\)<\/span><\/p>\n

          \u2234 The required radius = 8 cm.<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Question 8. The circle inscribed in an equilateral triangle of sides 24 cm just touches the sides <\/strong>of the triangle. Find the area of the rest part of the triangle (let \u221a3 = 1-732)<\/b><\/p>\n

          Solution:\u00a0<\/strong><\/p>\n

          Given\u00a0<\/strong><\/p>\n

          The circle inscribed in an equilateral triangle of sides 24 cm just touches the sides <\/strong>of the triangle.<\/b><\/p>\n

          The \u0394ABC is an equilateral triangle, each side of which is 24 cm.\u00a0<\/span><\/p>\n

          AD is the perpendicular drawn from A to BC.\u00a0<\/span><\/p>\n

          Since the triangle is equilateral,<\/span><\/p>\n

          \u2234 D bisects BC.<\/span><\/p>\n

          \u2234 BD = <\/span>CD = \\(\\frac{24}{2}\\) cm = 12 cm.\u00a0<\/span><\/p>\n

          The incentre and centroid of the \u0394ABC are the same points O.<\/span><\/p>\n

          \u2234 OD = \\(\\frac{1}{3}\\) AD…………(1)<\/span><\/p>\n

          Height of the triangle ABC= AD = \\(\\mathrm{ABC}=\\mathrm{AD}=\\frac{\\sqrt{3}}{2} \\times 24 \\mathrm{~cm}=12 \\sqrt{3} \\mathrm{~cm}\\)<\/span><\/p>\n

          \u2234 from (1) we get, OD = \\(\\mathrm{OD}=\\frac{1}{3} \\times \\mathrm{AD}=\\frac{1}{3} \\times 12 \\sqrt{3} \\mathrm{~cm}=4 \\sqrt{3} \\mathrm{~cm}\\)<\/span><\/p>\n

          \u2234 Area of the circle = \\(\\pi \\times(\\mathrm{OD})^2=\\left\\{\\frac{22}{7} \\times(4 \\sqrt{3})^2\\right\\} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = \\(\\left(\\frac{22}{7} \\times 48\\right) \\mathrm{sq}-\\mathrm{cm}=150 \\cdot 86 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          Area of the \u0394ABC = \\(\\frac{\\sqrt{3}}{4} \\times(\\text { side })^2=\\frac{\\sqrt{3}}{4} \\times(24)^2 \\mathrm{sq}-\\mathrm{cm}=144 \\sqrt{3} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = 144 x 1.732 sq. cm<\/span><\/p>\n

          = 249.40 sq. cm<\/span><\/p>\n

          \u2234 Area of the \u0394ABC = (249.40 – 150.86) sq. cm<\/span><\/p>\n

          = 98.54 sq. cm<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

          The area of the rest part of the triangle = 98.54 sq. cm<\/span><\/p>\n

           <\/p>\n

           <\/p>\n

          Question 9. ABCD is a square of side 12 cm. Four arcs, each of radius 6 cm have been drawn with centers at A, B, C, and D respectively, and in the remaining part, some sketches are also drawn. Find the perimeter and area of the sketched region.<\/b><\/p>\n

          Solution:\u00a0<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          ABCD is a square of side 12 cm.<\/b><\/p>\n

          Four arcs, each of radius 6 cm have been drawn with centers at A, B, C, and D respectively, and in the remaining part, some sketches are also drawn.<\/b><\/p>\n

          Let the four circles drawn with centers A, B, C, and D respectively, and a radius of 6 cm intersect each other at P, Q, R, and S.<\/span><\/p>\n

          We have to find the perimeter and area of the curved region PQRS<\/span><\/p>\n

          Now, area of the square ABCD = (12)\u00b2 sq-cm = 144 sq.cm<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

           <\/p>\n

          Each of the arcs PS, PQ, QR, and RS have produced an angle of 90\u00b0 at each center.\u00a0<\/span><\/p>\n

          Since each of the radii of the circles is equal.\u00a0<\/span><\/p>\n

          \u2234 arc PS arc PQ = arc QR arc arc RS<\/span><\/p>\n

          = \\(\\frac{\\text { angle produced at the centre }}{360^{\\circ}} \\times(\\text { circumference of circle })\\)<\/span><\/p>\n

          = \\(\\left(\\frac{90^{\\circ}}{360^{\\circ}} \\times 2 \\pi \\times 6\\right) \\mathrm{cm}=3 \\pi \\mathrm{cm}=3 \\times \\frac{22}{7} \\mathrm{~cm}=\\frac{66}{7} \\mathrm{~cm}\\)<\/span><\/p>\n

          \u2234 Perimeter of the curved region\u00a0<\/span><\/p>\n

          PQRS = 4 x arc Ps = (4 x \\(\\frac{66}{7}\\)) cm<\/span><\/p>\n

          = 37 \\(\\frac{5}{7}\\)<\/p>\n

          Again, the region APS = <\/span>region BPQ = region CQR = region DRS<\/span><\/p>\n

          = \\(\\frac{\\text { angle produced at the centre }}{360^{\\circ}} \\times \\text { area of the circle }\\)<\/span><\/p>\n

          = \\(\\left[\\frac{90^{\\circ}}{360^{\\circ}} \\times \\pi \\times(6)^2\\right] \\mathrm{sq}-\\mathrm{cm}=\\frac{198}{7} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          \u2234 Area of the 4 regions = \\(\\left(4 \\times \\frac{198}{7}\\right) \\mathrm{sq}-\\mathrm{cm}=\\frac{792}{7} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          Area of the curved region PQRS = (Area of the square) (Area of 4 regions)<\/span><\/p>\n

          = \\(\\left(144-\\frac{792}{7}\\right) \\mathrm{sq}-\\mathrm{cm}=30 \\frac{6}{7} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Alternative Method:<\/strong><\/p>\n

          \u00a0<\/span>The radius of each of the circles with centers at A, B, C, and D is 6 cm and the length of the arc cut by each of the four circles is part \\(\\frac{1}{4}\\) of the circumference of the circles, and the area of each part <\/span>is also \\(\\frac{1}{4}\\) part of the area of the circle.<\/span><\/p>\n

          \u2234 The required perimeter = \\(4 \\times \\frac{1}{4} \\times 2 \\times \\frac{22}{7} \\times 6 \\mathrm{~cm}=37 \\frac{5}{7} \\mathrm{~cm}\\)<\/span><\/p>\n

          and required area = \\(\\left\\{(6+6)^2-4 \\times \\frac{1}{4} \\times \\pi \\times 6^2\\right\\} \\mathrm{sq}-\\mathrm{cm}=30 \\frac{6}{7} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

           <\/p>\n

          Question 10. In the given find the area of the lined region.<\/b><\/p>\n

          Solution<\/strong><\/p>\n

          Area of each circle = \\(\\pi \\times(r)^2[r=\\text { radius }]=\\left[\\frac{22}{7} \\times(3 \\cdot 5)^2\\right] \\mathrm{sq}-\\mathrm{cm}=38 \\cdot 5 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          The arc PS has produced an angle <\/span>of 90\u00b0 at the center A. \u00a0 <\/span>[<\/span>ABCD is a square]<\/span><\/p>\n

          \u2234 Area of the region APS<\/span><\/p>\n

          = \\(=\\frac{\\text { angle produced at the centre }}{360^{\\circ}} \\times \\text { area of circle }\\)<\/span><\/p>\n

          = \\(\\left[\\frac{90^{\\circ}}{360^{\\circ}} \\times 38 \\cdot 5\\right] \\mathrm{sq}-\\mathrm{cm}=9.625 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          \u2234 The area of the lined region of the circle with the center at A<\/span><\/p>\n

          = (38.5 – 9.625) sq-cm\u00a0<\/span><\/p>\n

          = 28.875 sq-cm\u00a0<\/span><\/p>\n

          \u2234 The required area = (4 x 28.875) sq-cm<\/span><\/p>\n

          = <\/span>115.5 sq-cm<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

          The area of the lined region = <\/span>115.5 sq-cm<\/span><\/p>\n

           <\/p>\n

          Alternative Method:<\/strong><\/p>\n

          The radius of each of the circles with centers at A, B, C, and D is 3.5 cm.<\/span><\/p>\n

          Now in the lined region, \\(\\frac{1}{4}\\) <\/span>part of the area of each of the circles must be subtracted and \\(\\frac{3}{4}\\) part <\/span>of the area of each of the circles must be added.<\/span><\/p>\n

          The area of the lined region = \\(4 \\times \\frac{3}{4} \\times \\frac{22}{7} \\times(3 \\cdot 5)^2 \\mathrm{sq}-\\mathrm{cm}=115 \\cdot 5 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

           <\/p>\n

          Question 11. In the given find the area and perimeter of the shaded region.\u00a0<\/b><\/p>\n

          Solution:<\/b><\/p>\n

          The perimeter of the shaded region = line segment AB + arc AB.<\/p>\n

          = \\(\\sqrt{\\mathrm{OA}^2+\\mathrm{OB}^2}\\left [ \\angle \\mathrm{AOB}=90^{\\circ}\\right]+\\frac{1}{4} \\times 2 \\times \\frac{22}{7} \\times 12 \\mathrm{~cm}\\)(radius 12 cm)<\/span><\/p>\n

          = \\(\\sqrt{(12)^2+(12)^2} \\mathrm{~cm}+\\frac{132}{7} \\mathrm{~cm}\\)<\/span><\/p>\n

          = \\(\\left(12 \\sqrt{2}+\\frac{132}{7}\\right) \\mathrm{cm}\\)<\/span><\/p>\n

          = 35.83 cm<\/span><\/p>\n

          \u2234 Area of the shaded region<\/span><\/p>\n

          = \\(\\left\\{\\frac{1}{4} \\times \\frac{22}{7} \\times(12)^2-\\frac{1}{2} \\times 12 \\times 12\\right\\} \\text { sq-cm }\\)<\/span><\/p>\n

          = \\(\\frac{288}{7} \\mathrm{sq}-\\mathrm{cm}=41 \\frac{1}{7} \\mathrm{sq}-\\mathrm{cm} .\\)<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Question 12. Protul has drawn an equilateral triangle ABC the side of which is 10 cm. Sumita has drawn three arcs each of length 5 cm with centers at A, B, and C, and have colored some region in the middle position. Find the area of the colored region. (Take \u221a3 = 1.732)<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          Protul has drawn an equilateral triangle ABC the side of which is 10 cm. <\/b><\/p>\n

          Sumita has drawn three arcs each of length 5 cm with centers at A, B, and C, and have colored some region in the middle position.<\/b><\/p>\n

          The area of the equilateral \u0394ABC = \\(\\frac{\\sqrt{3}}{4} \\times(\\text { side })^2=\\left[\\frac{\\sqrt{3}}{4} \\times(10)^2\\right] \\text { sq-cm }\\)<\/span><\/p>\n

          = \\(25 \\sqrt{3} \\mathrm{sq}-\\mathrm{cm}=25 \\times 1.732 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = <\/span>43.3 sq-cm (approx.)<\/span><\/p>\n

          Since the arc PR has produced an angle of 60\u00b0. at A,<\/span><\/p>\n

          \u2234 <\/span>Area of the region APR = \\(=\\frac{\\text { angle produced at the centre }}{360^{\\circ}}\\) x (area of the circle)<\/span><\/p>\n

          = \\(\\left[\\frac{60^{\\circ}}{360^{\\circ}} \\times \\pi \\times(5)^2\\right] \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = \\(\\left[\\frac{1}{6} \\times \\frac{22}{7} \\times 25\\right] \\mathrm{sq}-\\mathrm{cm}=\\frac{275}{21} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          Similarly, the areas of the regions BPQ and CQR are \\(\\frac{275}{21}\\)<\/span>sq-cm.<\/span><\/p>\n

          . The area of the colored region= (area of the triangle) – (area of the three equal regions)<\/span><\/p>\n

          = \\(\\left(43 \\cdot 3-3 \\times \\frac{275}{21}\\right) \\mathrm{sq}-\\mathrm{cm}=(43 \\cdot 3-39 \\cdot 29) \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = 4.01 sq-cm (approx.)<\/span><\/p>\n

           <\/p>\n

          \"WBBSE<\/p>\n

           <\/p>\n

          Question 13. Rabeya has drawn an equilateral triangle of side 21 cm on a large piece of paper. Find the area of the incircle of the triangle.<\/b><\/p>\n

          Solution:<\/strong> <\/span><\/p>\n

          Given<\/strong><\/p>\n

          Rabeya has drawn an equilateral triangle of side 21 cm on a large piece of paper.<\/b><\/p>\n

          The side of the equilateral triangle is 21 cm.<\/span><\/p>\n

          \u2234 height = \\(\\frac{\\sqrt{3}}{2} \\times 21 \\mathrm{~cm}=\\frac{21}{2} \\sqrt{3} \\mathrm{~cm}\\)<\/span><\/p>\n

          The median of the equilateral triangle = its height.<\/span><\/p>\n

          Now, inradius of the incircle = \\(\\frac{1}{3}\\) x median<\/span><\/p>\n

          = \\(\\frac{1}{3} \\times \\frac{21 \\sqrt{3}}{2} \\mathrm{~cm}=\\frac{7 \\sqrt{3}}{2} \\mathrm{~cm}\\)<\/span><\/p>\n

          \u2234 Area of the incircle = \\(\\frac{22}{7} \\times\\left(\\frac{7 \\sqrt{3}}{2}\\right)^2 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = \\(\\frac{22}{7} \\times \\frac{7 \\times \\sqrt{3}}{2} \\times \\frac{7 \\times \\sqrt{3}}{2} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          = 115.5 sq. cm<\/span><\/p>\n

          The area of the incircle of the triangle = 115.5 sq. cm<\/span><\/p>\n

           <\/p>\n

          Question 14. The perimeter of a triangle is 32 cm and the area of its incircle is 38-5 sq-cm. Find the area of the triangle.<\/strong><\/p>\n

          Solution:<\/strong><\/p>\n

          Given\u00a0<\/strong><\/p>\n

          The perimeter of a triangle is 32 cm and the area of its incircle is 38-5 sq-cm.<\/strong><\/p>\n

          Let the radius of the incircle be r cm.<\/span><\/p>\n

          \u2234 Area of the incle = \u03c0r\u00b2 sq. cm<\/span><\/p>\n

          \u2234 As per the question, \\(\\pi r^2=\\frac{385}{10} \\Rightarrow \\frac{22}{7} r^2=\\frac{385}{10} \\Rightarrow r^2=\\frac{385 \\times 7}{10 \\times 22}=\\frac{49}{4} \\Rightarrow r=\\frac{7}{2}=3 \\cdot 5\\)<\/span><\/p>\n

          \u2234 The radius\u00a0of the incircle = 3.5 cm.<\/span><\/p>\n

          Now, area of the triangle = \\(\\frac{1}{2} \\times \\text { perimeter } \\times \\text { radius }=\\frac{1}{2} \\times 32 \\times 3.5 \\mathrm{sq}-\\mathrm{cm}=56 \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

           <\/p>\n

          Question 15. Find the radii of the incircle and circumcircle of a triangle of sides 20 cm, 15 cm and 25 cm. Also, find their areas.<\/b><\/p>\n

          Solution:<\/strong><\/p>\n

          Given<\/strong><\/p>\n

          A triangle of sides 20 cm, 15 cm, and 25 cm.<\/b><\/p>\n

          \u00a020\u00b2 + 15\u00b2 = 400+ 225 = 625 = 25\u00b2<\/span><\/p>\n

          \u2234 The given triangle is a right-angled triangle, the hypotenuse of which is 25 cm.<\/span><\/p>\n

          Now, the radius of the circumcircle = \\(\\frac{\\text { hypotenuse }}{2}=\\frac{25}{2} \\mathrm{~cm}=12 \\frac{1}{2} \\mathrm{~cm}\\)<\/span><\/p>\n

          \u2234 Area of the circumcircle = \\(\\frac{22}{7} \\times\\left(\\frac{25}{2}\\right)^2 \\mathrm{sq}-\\mathrm{cm}=\\frac{22}{7} \\times \\frac{625}{4} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          Again, area of the triangle = \\(\\frac{6875}{14} \\mathrm{sq}-\\mathrm{cm}=491 \\frac{1}{14} \\mathrm{sq}-\\mathrm{cm}\\)<\/span><\/p>\n

          Let the radius of the incircle be r cm,<\/span><\/p>\n

          \u2234 \\(\\frac{1}{2}\\) <\/span>perimeter of the triangle x r = 150<\/span><\/p>\n

          \\(\\frac{1}{2} \\times(15+20+25) \\times r=150 \\Rightarrow \\frac{1}{2} \\times 60 \\times r=150 \\Rightarrow r^{\\prime}=\\frac{150}{30}=5\\)<\/span><\/p>\n

          \u2234 The radius of the incircle = 5 cm<\/p>\n

          \u2234 Area of the incircle = \\(\\frac{22}{7} \\times 5^2 \\mathrm{sq}-\\mathrm{cm}=\\frac{550}{7} \\mathrm{sq}-\\mathrm{cm}=78 \\frac{4}{7} \\mathrm{sq}-\\mathrm{cm}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

          Mensuration Chapter 3 Area Of Circles Chapter 3 Area Of Circles Area of a circle Area of circles circumscribed in Determination of in-radius and circum-radius Area of a circle = x (radius)\u00b2 = \u03c0r\u00b2 [ r = radius of circle] The area of the region closed by two concentric circles= \u00a0 [R and r are … Read more<\/a><\/p>\n","protected":false},"author":12,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[34],"tags":[],"yoast_head":"\nWBBSE Solutions For Class 9 Maths Mensuration Chapter 3 Area Of Circles - WBBSE Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/wbbsesolutions.net\/wbbse-solutions-for-class-9-maths-mensuration-chapter-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"WBBSE Solutions For Class 9 Maths Mensuration Chapter 3 Area Of Circles - WBBSE Solutions\" \/>\n<meta property=\"og:description\" content=\"Mensuration Chapter 3 Area Of Circles Chapter 3 Area Of Circles Area of a circle Area of circles circumscribed in Determination of in-radius and circum-radius Area of a circle = x (radius)\u00b2 = \u03c0r\u00b2 [ r = radius of circle] The area of the region closed by two concentric circles= \u00a0 [R and r are ... 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