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Area of Trapezium and Polygon | Area of Trapezium and Polygon Exercise 20.1

Question 3

A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π= 22/7.)

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  • Solution

  • Transcript

Area of the plot = Area of the Rectangle + 2 × area of one semi-circle

Radius of semi-circle = BC/2 = 24.5/2 = 12.25m

Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m2

Area of the Semi-circular portions = 2 × πr2/2

= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m2

Area of the plot = 882 + 471.625 = 1353.625 m2

"If the sides of the rectangle are 36 meter and 24.5 meters then find the area of the data and take value of pi is 22 by step. First of all, we draw the diagram of the playground. So this is our program we have ABCD is in a shape of the end of your lens is 36 meter and the practice 24.5 meters but in which it is given that the to semicircle on is gonna send it means on its threat. We don't lose any Circle 80 and TCP that person circle on it. We have to find the area of this breakup. First of all, we find the area of rectangle area of rectangle. ABCD is length into breadth so length is given to us 36 meter and practice 24.5 meters. Okay, when we multiply these 2 then we get 882 centimeter Square after that. There are two semicircles hold its breath as we know the area of a circle is pi r squared but has a visitor so area of semi circle area. A circle is pi r squared it is sensible. That's why we divided it by 2. So the formula is pie R square by 2. But here we are two sub surface 1 and 2 so P multiplied by 2 because there are two times and then two and two together. So the value of pi is 22 by 7 and 24 by 5 is the time in the Laboratories 24.1. So it means the radius is half of the diameter. So we put the values of our are is 24.5 divided by 2. So R is 24 Point 5 divided by 2 R square. That's why we write it two times twenty four point five to twenty solve this to on the And that is what is then we get 471 point 6 2 5 centimeter Square. Okay now area of playground. A of playground is equal to LR circuit angle. Plus he loved to San circle area of rectangle is 882 faculty of the semicircle is for seventy one point six two five when we add these two then we can 1353 point 6 and 6 2 5 centimeter Square. So this is our T table data is 1333 1600 five centimeters per I hope you like this video subscribe for more videos till then stay safe, bye-bye."

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Question 3

A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π= 22/7.)

  • Solution

  • Transcript

Area of the plot = Area of the Rectangle + 2 × area of one semi-circle

Radius of semi-circle = BC/2 = 24.5/2 = 12.25m

Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m2

Area of the Semi-circular portions = 2 × πr2/2

= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m2

Area of the plot = 882 + 471.625 = 1353.625 m2

"If the sides of the rectangle are 36 meter and 24.5 meters then find the area of the data and take value of pi is 22 by step. First of all, we draw the diagram of the playground. So this is our program we have ABCD is in a shape of the end of your lens is 36 meter and the practice 24.5 meters but in which it is given that the to semicircle on is gonna send it means on its threat. We don't lose any Circle 80 and TCP that person circle on it. We have to find the area of this breakup. First of all, we find the area of rectangle area of rectangle. ABCD is length into breadth so length is given to us 36 meter and practice 24.5 meters. Okay, when we multiply these 2 then we get 882 centimeter Square after that. There are two semicircles hold its breath as we know the area of a circle is pi r squared but has a visitor so area of semi circle area. A circle is pi r squared it is sensible. That's why we divided it by 2. So the formula is pie R square by 2. But here we are two sub surface 1 and 2 so P multiplied by 2 because there are two times and then two and two together. So the value of pi is 22 by 7 and 24 by 5 is the time in the Laboratories 24.1. So it means the radius is half of the diameter. So we put the values of our are is 24.5 divided by 2. So R is 24 Point 5 divided by 2 R square. That's why we write it two times twenty four point five to twenty solve this to on the And that is what is then we get 471 point 6 2 5 centimeter Square. Okay now area of playground. A of playground is equal to LR circuit angle. Plus he loved to San circle area of rectangle is 882 faculty of the semicircle is for seventy one point six two five when we add these two then we can 1353 point 6 and 6 2 5 centimeter Square. So this is our T table data is 1333 1600 five centimeters per I hope you like this video subscribe for more videos till then stay safe, bye-bye."

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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