10th Maths 12.1
NCERT Class 10th solution of Exercise 12.3
Exercise 12.1
Unless stated otherwise, use `pi = 22/7`.
Q1. The radii of two circles are `text{19 cm}` and `text{9 cm}` respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
`text{Sol.}`
`text{circumference of big circle}` `text{= circumference of}` (`text{circle}_1`+ `text{circle}_2`)
2`pi``text{R}` = 2`pi``text{r}_1` + 2`pi``text{r}_2`
2`pi``text{R}` = `2``pi`{`19 + 9`}
`text{R}` = `28` `text{cm.}`
`text{Answer :}`
`text{Radius}` = `28` `text{cm.}`
Q2. The radii of two circles are `text{8 cm}` and `text{6 cm}` respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
`text{Sol.}`
`text{area of big circle}`= `text{= area of}`(`text{circle}_1`+`text{circle}_2`)
`pi``text{R}^2` =`pi``text{r}_1`+`pi``text{r}^2`
`pi``text{R}^2` =`pi``{(8)^2+(6)^2}`
`text{R}^2` =`{64+36}`
`text{R}^2` =`100`
`text{R}` =`sqrt{100}`
`text{R}` = `10` `text{ cm.}`
`text{Answer :}`
`text{Radius}` = `10` `text{ cm.}`
Q3. Figure depicts an archery target market with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is `text{21 cm}` and each of the other bands is `text{10.5 cm}` wide. Find the area of each of the five scoring regions.
`text{Sol.}`
`text{The diameter of Gold circle is 21 cm}`
`text{The radius of G circle is r}_1 = ` `21/2 = 10.5` `text{ cm}.`
`text{width of each bands}` `= 10.5 text{ cm}`
`text{area of Gold circle}` = `pi``(text{r}_1)^2`
`text{ }` = `22/7``(text{10.5})^2 = 346.5 ``text{ cm}^2`
`text{r}_2 = text{r}_1 + text{r}_1 = 2text{r}_1 text{ cm}`
`text{area of Red circle}` = `pi``[(text{2r}_1)^2` - `(text{r}_1)^2]`
`text{ }` = `22/7``[4text{r}_1^2` - `text{r}_1^2]`
`text{ }` = `22/7``[3text{r}_1^2]`
`text{ }` = `22/7``[3(text{10.5})^2]`
`text{ }` = `1039.5 text{ cm}^2`
`text{r}_3 = 2text{r}_1 + text{r}_1 = 3text{r}_1 text{ cm}`
`text{area of Blue circle}` = `pi``[(text{3r}_1)^2` - `(text{2r}_1)^2]`
`text{ }` = `pi``[9text{r}_1^2` - `4text{r}_1^2]`
`text{ }` = `pi``[5text{r}_1^2]`
`text{ }` = `22/7``[5(text{10.5})^2]`
`text{ }` = `1732.5 text{ cm}^2`
`text{r}_4 = 3text{r}_1 + text{r}_1 = 4text{r}_1 text{ cm}`
`text{area of Black circle}` = `pi``[(text{4r}_1)^2`- `(text{3r}_1)^2]`
`text{ }` = `22/7``[16text{r}_1^2` - `9text{r}_1^2]`
`text{ }` = `22/7``[7text{r}_1^2]`
`text{ }` = `22/7``[7(text{10.5})^2]`
`text{ }` = `2425.5 text{ cm}^2`
`text{r}_5 = 4text{r}_1 + text{r}_1 = 5text{r}_1 text{ cm}`
`text{area of White circle}` = `pi``[(text{5r}_1)^2` - `(text{4r}_1)^2]`
`text{ }` = `22/7``[25text{r}_1^2` - `16text{r}_1^2]`
`text{ }` = `22/7``[9text{r}_1^2]`
`text{ }` = `22/7``[9(text{10.5})^2]`
`text{ }` = `3118.5 text{ cm}^2`
`text{Answer :}`
`text{Gold } 346.5 ``text{ cm}^2`, `text{Red }``1039.5 text{ cm}^2`, `text{Blue }``1732.5 text{ cm}^2`, `text{Black }``2425.5 text{ cm}^2`, `text{White }``3118.5 text{ cm}^2`
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Q4. The wheels of a car are of diameter `text{80 cm}` each. How many complete revolutions does each wheel make in `text{10 minutes}` when the car is travelling at a speed of `text{66 km per hour}`?
`text{Sol.}`
`text{The distance covered by car in 10 minutes}`
`=10/60``times``66`
`=11` `text{ km}`
`=11000``text{ m}`
`text{no. of revolutions }`=`frac{(text{covered distance})}{(text{circumference of wheel})}`
`text{n} = frac{11000}{2pi text{r}}`
`text{n} = frac{11000}{2 times 22/7times 80/100}`
`text{n} = frac{11000 times 7 times 100}{2 times 22}`
`text{n} = 4375`
`text{Answer :}`
`text{No. of revolution made by each wheel 4375.}`
Q5. Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
A) `text{2 units}` B) `pi text{units}` C) `text{4 units}` D) `text{7 units}`
`text{Sol.}`
`text{perimeter of circle = area of circle}`
`text{2pir = pir}^2`
`text{2r = r}^2`
`text{r = 2}`
`text{Answer :}`
A) `text{2 units}`
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