Book Name |
: Ganit Prakash |

Subject |
: Mathematics (Maths) |

Class |
: 9 (Madhyamik/WB) |

Publisher |
: Prof. Nabanita Chatterjee |

Chapter Name |
: Area Of Circle (18th Chapter) |

*Let us work out – 18*

**1. Today the cow of Aminabibi is fastened to a post with a rope of length 2.1 meter in the vacant field. Let us see by calculating how much maximum area the cow will graze.**

* Solution:* Radius of rope = 2.1m

\text { Area of the cow will graze } =\pi(\text { radius })^2 \\

=\frac{22}{7} \times(2.1)^2 \\

=\frac{22}{7} \times 2.1 \times 2.1 \text { sq.m } \\

= 13.86 \text { sq.m (Ans.) }

**2. Suhana will draw a circle of which perimeter will be 35,2 cm. Let us see by calculating what length of radius Suhana takes to draw a circle and what will be the area of that?**

* Solution :* Let ‘ r ‘ be the radius of the circle

Then Perimeter of a circle = 35.2 cm.

or, 2 \pi \mathrm{r}=35.2

or, 2 \times \frac{22}{7} r=\frac{352}{10}

or, r=\frac{352 \times 7}{2 \times 22 \times 10}=\frac{56}{10}

\therefore \mathrm{r}=5.6 cm.

\therefore \quad Radius of the circle = 5.6 cm (Ans.)

Area of that circle =\pi r^2

=\frac{22}{7} \times(5.6)^2 \\

=\frac{22}{7} \times 5.6 \times 5.6 \text { sq.cm } \\

=98.56 \text { sq. cm (Ans.) }

**3. Grandmother of Rekha was made a circular table cover of which area 5544 \mathrm{sq} . \mathrm{cm}. She wants to paste colouring tape surrounding this circular cover of the table, let us see by calculating how long colouring tape she will buy.**

* Solution : *Let ‘ r ‘ be the radius of the circle.

\text { Then, Area }=5544 \text { Sq.cm. } \\

\text { or, } \pi r^2=5544 \\

\text { or, } \frac{22}{7} \times(r)^2=5544 \\

\text { or, } r^2=\frac{5544 \times 7}{22} \\

\text { or, } r^2=252 \times 7 \\

\text { or, } r^2=6 \times 6 \times 7 \times 7 \\

\text { or, } r \quad=\sqrt{6 \times 6 \times 7 \times 7} \\

\text { or, } r \quad=6 \times 7 \\

\therefore r=42 \\

\therefore \quad \text { Length of colouring tape }=2 \pi \mathrm{r}=2 \times \frac{22}{7} \times 42 \mathrm{cm} \text {. } \\

\therefore \quad 264 \text { cm. (Ans.) }

**4. The cost of fencing of our village play ground with railing is Rs. 924 at the rate of Rs. 21 per meter. Let us write by calculating how much sq. meter canvas will be bought for covering the field.**

* Solution :* Let ‘ r ‘ be the radius of the circle.

The cost of fencing of our village play ground with railing is Rs. 924 at the rate of Rs. 21 per meter.

\therefore \quad \text { Circumference }=\frac{924}{21} \mathrm{~m}=44 \mathrm{~m} \\

\text { or, } 2 \pi \mathrm{r}=44 \\

\text { or, } 2 \times \frac{22}{7} \times \mathrm{r}=44

or, \frac{44 r}{7}=44

or. \frac{r}{7}=1

\therefore \quad r=7

\therefore Radius of circular field covering with Canvas

=\pi r^2 \\

=\frac{22}{7} \times(7)^2 \\

=\frac{22}{7} \times 7 \times 7 \\

=154 \text { sq. m }(Ans.)

**5. Faruk will draw a circle of which area will be 616 \mathrm{sq}. meter. Let us see by calculating what length of radius Faruk will take to draw a circle and what perimeter he will get.**

* Solution :* Let ‘ r ‘ be the radius of the circle

Then Area of a circle =616 \mathrm{sq} . \mathrm{m}.

or, \pi r^2=616

or, \frac{22}{7} \times(r)^2=616

or, 22 r^2=616 \times 7

or, r^2=\frac{616 \times 7}{22}

or, r^2=28 \times 7

or, r^2=2 \times 2 \times 7 \times 7

or, r=\sqrt{2 \times 2 \times 7 \times 7}

or, r=2 \times 7

\therefore r=14

\therefore Length of the radius =14 \mathrm{~m}.

Perimeter of the circle =2 \pi \mathrm{r}

=2 \times \frac{22}{7} \times 14 \mathrm{~m} \\

=88 \mathrm{~m} \text { (Ans.) }

**6. Palash and Piyali have drawn two circles the ratio of which length of radius is 4 : 5. Let us write by calculating the ratio of the area of two circular fields drawn by them.**

* Solution :* Let r_1 \text{ and } r_2 be be the radius of the 1st and 2nd circle

Then, \frac{r_1}{r_2}=\frac{4}{5}

Squaring both sides,

\text { or, } \frac{r_1^2}{r_2^2}=\frac{4^2}{5^2} \\

\text { or, } \frac{\pi r_1^2}{\pi r_2^2}=\frac{16}{25} \\

\therefore \quad \frac{\text { Area of } 1 \text { st. circle }}{\text { Area of } 2 \text { nd circle }}=\frac{16}{25} \\

\therefore \quad \text { Ratio of Area of two cricles }=16: 25 \text { (Ans.) }

**7. Sumit and Reba have taken two copper wire having same length. Sumi draws the wire in the form of a rectangular shape of which length and breadth are 48 cm. and 40 cm. But Reba bent the copper wire with same length in the form of a circle. Let us see by calculating which will cover maximum place between the rectangle drawn by Sumit and circle drawn by Reba.**

* Solution :* Sumit draws Rectangular shape with length 48 cm. & breadth 40 cm.

\therefore \quad \text { Area of Rectangular shape } =48 cm \times 40 cm . \\

=1920 \text { sq.cm. (Ans.) } \\

\text { Perimeter of Rectangular shape } =2(l+\mathrm{b}) \\

=2(48+40) =2 \times 88=176 cm .

Reba draws circular shape with circumference 176 cm.

Let ‘ r ‘ be the radius of the circle

\therefore \quad Circumference = 176

or, 2 \pi \mathrm{r}=176

or, 2 \times \frac{22}{7} \times r=176

or, 44 r=176 \times 7

or, r=\frac{176 \times 7}{44}

\therefore \quad \mathrm{r}=28

\therefore \quad Area of the circle =\pi r^2

=\frac{22}{7} \times(28)^2 \\

=\frac{22}{7} \times 28 \times 28 \\

=2464 \text { sq. m (Ans.) }

**8. At the centre of rectangular field of Pioneer atheletic club there is a circular pool of which length of radius is 14 meter. The length and breadth of recatangular field are 60 meter and 42 meter respectively. Let us see by calculating how much cost it will take for planting grass of remaining palce of rectangular field except pool at the rate of Rs. 75 per square meter.**

**Solution :**

Length of rectangular field = 60m.

\therefore Breadth of rectangular field = 42m.

\therefore \quad Area of rectangular field = 60 \mathrm{~m} \times 42 \mathrm{~m}=2520 \mathrm{sq} . \mathrm{m}.

Length of radius of circular pool = 14m

\therefore \quad Area of circular pool =\pi(14)^2

=\frac{22}{7} \times 14 \times 14 \\

=616 \text { sq.m. }

Area of remaining place = (2520 – 616) sq.m

= 1904sq.m

Cost of planting grass at the rate of Rs. 75

\text { Per square meter } =\text { Rs. } 1904 \times 75 \\

=\text { Rs. } 142800 (Ans.)

**9. A 7 meter wide path runs outside a circular park of Etalgacha Friends association club along perimeter. Let us write by calculating the area of path, if the perimeter of circular park is 352 meter, let us write by calculating how much cost for concreting the path at the rate of Rs. 20 per square meter.**

**Solution :**

Let ‘ r ‘ be the radius of the circle Then Perimeter of a circle = 352m.

or, 2 \pi r=352

or, 2 \times \frac{22}{7} r=352

or, 44 r=352 \times 7

\text { or, } r=\frac{352 \times 7}{44} \\

\therefore r=56 \mathrm{~m} .

\therefore \quad Width of the path = 7 \mathrm{~m}.

\therefore \quad \text { Area of path } =\pi\{(56+7)^2-(56)^2\} \\

=\pi\{(63)^2-(56)^2\} \\

=\frac{22}{7} \times(63+56)(63-56) \\

=\frac{22}{7} \times 119 \times 7 \\

=2168 \text { Sq.m. (Ans.) }

\text { Total Cost at the rate of Rs. } 20 \text { per sq.m } =\text { Rs. } 2618 \times 20 \\

=\text { Rs. } 52360 \text { (Ans.) }

**10. Anwaribibi has spent Rs. 2664 at the rate of Rs. 18.50 per meter for fencing of her semicircular land. Let us write by calculating how much cost it will take, if she makes the semicircular land plough at the rate of Rs. 32 per sq.m.**

* Solution:* Anwaribibi has spent Rs, 2664 at the rate of Rs. 18.50 per\m.

Let, ‘ r ‘ be the radius of semicircular land.

Circumference of the semi circular land

=\frac{2664}{18.50} \mathrm{~m} \text {. } \\

=\frac{2664 \times 100}{1850} \mathrm{~m} .=144 \mathrm{~m} .

According to the condition of problem,

\pi r+2 r=144

or, r(\pi+2)=144

or, r(\frac{22}{7}+2)=144

or, r(\frac{22+14}{7})=144

or, r. \frac{36}{7}=144

or, 36 r=144 \times 7

or, r=\frac{144 \times 7}{36}

\therefore \quad \mathrm{r}=28

\therefore Area of Semicircular land.

=\frac{\pi r^2}{2} \\

=\frac{22}{7} \times \frac{1}{2} \times(28)^2 \\

=\frac{22}{7} \times \frac{1}{2} \times 28 \times 28 \text { sq.m. } \\

=1232 \text { Sq.m. }

Total cost = Rs. 1232 × 32 = Rs. 39,424 (Ans.)

**11. The time which took today my friend Rajat running with uniform speed to round once of a circular field of school is 30 seconds less while he ran diametrically with same speed. Let us write by calculating the area of field of school if his speed is 90 meters/second.**

* Solution :* Let the radius of the circular field be r

\therefore \quad The distance along the diameter = 2r

We have, speed = 90 \mathrm{~m} / \mathrm{sec}.

Distance covered in 30 \mathrm{sec}. =90 \times 30 \mathrm{~m}=2700 \mathrm{~m}.

According to the condition of the problem,

2 \pi \mathrm{r}-2 \mathrm{r}=2700 \\

\text { or, } 2 r(\pi-1)=2700 \\

\text { or, } 2 r(\frac{22}{7}-1)=2700 \\

\text { or, } 2 r(\frac{22-7}{7})=2700 \\

\text { or, } 2 r \times \frac{15}{7}=2700 \\

\text { or, } \frac{30 r}{7}=2700 \\

\text { or, } \frac{r}{7}=90 \\

\text { or, } r=630 \\

\therefore \quad \text { Area of field }=\pi r^2 \\

=\frac{22}{7} \times(630)^2 \text { sq.m } \\

=\frac{22}{7} \times 630 \times 630 \\

=1247400 \text { Sq.m. (Ans.) } \\

**12. An equally wide path runs out-side the circular field of Bakultala. The length of outer circumference exceeds the inner circumference by 132 meter. If the area of path is 14190 sq. meter, let us write by calculating the area of circular path.**

* Solution:* Let ‘ r ‘ be the radius of inner circle and ‘R’ be the radius of outer circle.

According to the condition of the problem,

2 \pi(\mathrm{R}-\mathrm{r})=132

or, 2 \times \frac{22}{7}(R-r)=132

or, \frac{R-r}{7}=3

\therefore \quad R – r = 21 ……… (i)

Again,

\pi(R^2-r^2)=14190 \\

\text { or, } \frac{22}{7}(R+r)(R-r)=14190 \\

\text { or, } \frac{22}{7}(R+r) \times 21=14190 \\

\text { or, } 66(R+r)=14190 \\

\text { or, }(R+r)=\frac{14190}{66} \\

\text { or, } R + r = 215 ……. (ii)

Adding (i) and (ii)

2R = 236

R = 118

Subtracting (i) and (ii)

2r = 194

r = 97

\therefore \quad Area of circular path =\pi r^2

=\frac{22}{7} \times(97)^2 \text { sq.m } \\

=\frac{22}{7} \times 97 \times 97 \\

=\frac{206998}{7} \\

=29571 \frac{1}{7} \text { Sq.m (Ans.) }

**13. Let us write by calculating the area of shaded region pictures below**

**(i) **

**ABCD is a square. The length of radius of circle is 7cm.**

**(ii) **

**The length of radius of each circle is 3.5 cm. The centres of four circles are A, B, C, D respectively.**

* Solution :* (i) Length of radius of circle = 7 cm.

\therefore \quad \text { diameter } =2 \times 7 cm=14 cm

\therefore \text { Area of circle } =\pi(7)^2 \\

=\frac{22}{7} \times 7 \times 7 \\

=154 \text { sq.m. }

Let ‘ a ‘ be the side of a square,

\therefore \quad \text { Length of diagonal }=a \sqrt{2} \\

\therefore \quad a \sqrt{2}=14 \\

\text { or, } \quad a \sqrt{2}=7 \times 2 \\

\text { or, } \quad a \sqrt{2}=7 \sqrt{2} \times \sqrt{2} \\

\therefore \quad a=7 \sqrt{2}

\text { Area of square } \mathrm{ABCD}=(7 \sqrt{2})^2 \\

=49 \times 2 \text { sq.cm. } \\

=98 \text { sq.cm. (Ans.) } \\

Area of shaded region = (154 – 98) = 56 sq.cm.

(ii) \text { Area of one circle } =\pi(3.5)^2 \\

=\frac{22}{7} \times 3.5 \times 3.5 \\

=38.5 \text { sq. cm }

Area of three circle i.e, area of shaded region

=3 \times 38.5 \text { sq. } \mathrm{cm} . \\

=115.5 \text { sq.cm. } (Ans)

**14. Dinesh has made a pie-chart of the students of their class who want to play which game. He has taken length of radius of circle 3.5 cm, Let us write by calculating the perimeter and area of each sector of circles.**

* Solution:* Radius of the circle = 3.5 cm.

\text { Perimieter of the circle }=2 \times \pi \times 3.5

= 2 \times \frac{22}{7} 3.5 \\

=\times 22 cm \quad \text { (Ans.) } \\

\text { Area of the circle } =\pi(3.5)^2 \\

=\frac{22}{7} \times 3.5 \times 3.5 \\

=38.5 \text { sq.cm. (Ans.) }

**15. Nitu has drawn a square ABCD of which length of each side as 12 cm. My sister has drawn four circular ares with length of radius 6 cm. centering A, B, C, D like picture besides and she has designed some portion. Let us write by calculating the area of shaded region.**

**Solution:**

Area of square ABCD = (12)^2 = 144 Sq.cm.

Four circular forms a circle

\text { Area of one circle } =\pi(6)^2 \\

=\frac{22}{7} \times 6 \times 6 \\

=\frac{792}{7} \text { Sq.cm. }

The area of shaded region

=(144-\frac{792}{7}) \text { Sq.cm } \\

=(\frac{1008-792}{7}) \text { Sq.cm } \\

=\frac{216}{7} \text { Sq.cm. } \\

=30 \frac{6}{7} \text { Sq.cm. (Ans.) }

**16. The area of circular field is 154 sq.cm. Let us write by calculating the perimeter and area of circumfering circular field with square.**

**Solution :**

Let ‘ r ‘ be the radius of the circle.

Then,

Area of circular field =154 sq. cm.

or, \pi(r)^2=154

or, \quad \frac{22}{7} \times r^2=154

or, r^2=7 \times 7

\text { or, }=r=\sqrt{7 \times 7} \\

\therefore r=7 cm

\text { Diameter of the circle } =2 \mathrm{r} \\

=2 \times 7 cm=14 cm .

Length of side of the circumscribed square = 14 cm.

\therefore \text { Perimeter of square } =4 \times \text { side } \\

=4 \times 14 cm. \\

= 56 cm.

\text { Area of square } =(14)^2 \text { sq.cm } \\

=196 \text {cm (Ans.) }

**17. Let us write perimeter and area of circular shaded sector below.**

**(i) **

**(ii) **

**Solution :**

(i) \text { Length of } \operatorname{arc} \mathrm{AB} =\frac{90}{360} \times 2 \pi \mathrm{r} \\

=\frac{1}{4} \times 2 \times \frac{22}{7} \times 12 \\

=\frac{132}{7} cm . \\

=18.86 cm .

\text { Length of } \mathrm{AB}=\sqrt{(12)^2+(12)^2}=\sqrt{2 \times(12)^2} \\

=12 \sqrt{ } 2 cm \text {. } \\

=12 \times 1.41 cm \quad[\because \sqrt{ } 2 \cong 1.41] \\

=16.968 cm \text {. } \\

\text { Perimeter of shaded region } =(18.86+16.968) \mathrm{cm} . \\

=35.83 cm \text { (Approx) }

Area of sector of circle AOB =\frac{90}{360} \times \pi \times(12)^2 \\

=\frac{1}{4} \times \frac{22}{7} \times 12 \times 12 \\

=\frac{729}{7} \mathrm{Sq} \cdot \mathrm{cm} .

\text { Area of } \triangle \mathrm{AOB} =\frac{1}{2} \times 12 \times 12 \text { sq.cm. } \\

=72 \text { sq.cm. }

The area of shaded region

=(\frac{792}{7}-72) \text { Sq.cm. } \\

=(\frac{792-504}{7}) \text { Sq.cm. } \\

=\frac{288}{7} \text { Sq. cm. } \\

=41 \frac{1}{7} \text { Sq. cm. (Ans.) }

(ii) \text { Length of } \operatorname{arc} A C =\frac{60}{360} \times 2 \pi r \\

=\frac{1}{6} \times 2 \times \frac{22}{7} \times 42 \\

= 44cm

\therefore \quad \mathrm{AB}=\mathrm{BC}=\mathrm{AC}=42 cm \text {. }

Perimeter of shaded region = Length of arc AC + AC

= (44+42) \mathrm{cm} .=86 cm \text { (Ans.) }

\text { Area of sector of circle } \mathrm{ABC} =\frac{60}{360} \times \pi \times(42)^2 \\

=\frac{1}{6} \times \frac{22}{7} \times 42 \times 42 \\

=924 \mathrm{sq} . \mathrm{cm} .

\text { Area of } \triangle \mathrm{ABC} =\frac{\sqrt{ } 3}{4} \times(\text { Side })^2 \\

=\frac{\sqrt{ } 3}{4} \times(42)^2 \\

=\frac{\sqrt{ } 3}{4} \times 42 \times 42 \\

=763.812 \mathrm{sq} . \mathrm{cm} .

Area of shaded region = (924 – 763.812) sq.cm.

=160.188 \text { Sq.cm. (Ans.) }

**18. Buying a bangle from fair Nila wears in her hand. The bangle contains 269.5 sq.cm. metal. If the length of oute diameter of bangle is 28 cm, let us write by calculating the length of inner diameter.**

* Solution :* Outer diameter = 28 cm.

\text { Outer radius }=\frac{28}{2} cm .=14 cm

Let ‘ r ‘ be the inner radius.

Area of bangle = 269.5 Sq.cm.

or, \pi\{(14)^2-(r)^2\}=269.5

or, \frac{22}{7}\{196-r^2\}=\frac{2695}{10}

or, \quad 196-r^2=\frac{2695 \times 7}{22 \times 10}

or, \quad 196-\mathrm{r}^2=\frac{343}{4}

or. \quad r^2=196-85.75

or, \quad r_x^2=110.25

or, r=\sqrt{110.25}

\therefore \quad r = 10.5

Inner diameter of bangle = 2r

=2 \times 10.5=21 cm \text {. (Ans.) }

**19. Protul has drawn an equilateral triangle ABC picture beside of which length of each side is 10 cm. Sumita has drawn three circular areas centering A, B, C with the length of radius 5 cm. and has coloured some portion at the middle. Let us write by calculatin the area of coloured portion.**

**Solution** :

Area of equilateral triangle ABC

=\frac{\sqrt{3}}{4} \times(10)^2

=\frac{\sqrt{3}}{4} \times 100 \\

=25 \sqrt{ } 3 Sq.cm . \\

=25 \times 1.732 Sq.cm . \\

=43.3 Sq.cm . \\

\therefore \quad<\mathrm{ABC} =<\mathrm{ACB}=60^{\circ}

\text { Aiea of two sectors } =\frac{60}{360} \times 2 \times \pi \times(5)^2 . \\

=\frac{1}{6} \times 2 \times \frac{22}{7} \times 25 \\

=\frac{550}{21} Sq.cm . \\

=26.19 Sq.cm . \\

\text { Area of coloured part } =(43.3-26.19) \text { Sq.cm. } \\

=17.10 Sq.cm \text { (Ans.) }

**20. Rabeya drew an equilateral triangle with sides 21 cm. on a big paper. Drawing a circle incribing that triangle coloured the circular region. I write by calculating the area of coloured region.**

**Solution :**

Let ‘ r ‘ be the radius of the circle.

Area of equilateral triangle ABC

=\frac{\sqrt{ } 3}{4} \times(21)^2 Sq.cm

or, \triangle \mathrm{AOB}+\triangle \mathrm{BOC}+\triangle \mathrm{AOC}=\frac{\sqrt{ } 3}{4} \times 441

or, \frac{1}{2}(\mathrm{AB} \times \mathrm{OE}+\mathrm{BC} \times \mathrm{OF}+\mathrm{AC} \times \mathrm{OG})=\frac{\sqrt{3}}{4} \times 441

or, \frac{1}{2} \times 21(r+r+r)=\frac{\sqrt{3}}{4} \times 441

or, \frac{1}{2} \times 21 \times 3 r=\frac{\sqrt{3}}{4} \times 441

or, r=\frac{\sqrt{3}}{4} \times \frac{441 \times 2}{21 \times 3}

=\frac{7 \sqrt{3}}{2} cm . \\

\text { Area of the circle }=\frac{22}{7} \times(\frac{7 \sqrt{ } 3}{2})^2 \text { Sq.cm. } \\

=\frac{22}{7} \times \frac{49 \times 3}{4} \text { Sq.cm. } \\

=\frac{231}{2} \text { Sq.cm. } \\

=115.5 \text { sq.cm. } \quad \text { (Ans.) }

**21. The area of circumscribing an equilateral triangle is 462 sq.cm. Let us write by calculating length of each side of this triangle.**

**Solution :**

Let ‘ r ‘ be the radius of the circle.

Then, Area = 462 sq.cm.

or, \pi(r)^2=462

or, \frac{22}{7} \times r^2=462

or, r^2=21 \times 7

or, r=\sqrt{7 \times 7 \times 3}

\therefore r=7 \sqrt{3} cm

\therefore \quad B G=7 \sqrt{3} cm

\therefore \quad \mathrm{BG}=\frac{2}{3} \times \mathrm{BD}

\therefore \quad \mathrm{BD}=\frac{3}{2} \times \mathrm{BG}

\mathrm{BD}=\frac{3}{2} \times 7 \sqrt{3}=\frac{21 \sqrt{3}}{2} cm

Let ‘ a ‘ be the side of an equilateral triangle.

Then,

\mathrm{BD}=\frac{\sqrt{ } 3}{2} \times \mathrm{a} \\

\therefore \quad \frac{\sqrt{ } 3}{2} \times \mathrm{a}=\frac{21 \sqrt{3}}{2} cm . \\

\therefore \quad \mathrm{a}=21 \\

\therefore \quad \text { Length of each side }=21 cm .(Ans.)

**22. Perimeter of a triangle is 32 cm. and area of incribing circle is 38.5 Sq.cm. Let us write by calculating the area of this triangle.**

**Solution :**

Let ‘ r ‘ be the radius of the circle.

Area of the circle = 38.5 Sq.cm.

\therefore \pi \mathrm{r}^2=38.5 Sq.cm.

or, \frac{22}{7} r^2=\frac{385}{10}

or, r^2=\frac{385 \times 7}{22 \times 10}

or, r^2=\frac{49}{4}

or, r=\sqrt{\frac{49}{4}}

\therefore \quad r=\frac{7}{2}

Numerical value of the area of \triangle A B C = Numerical value of sum of the areas of \triangle A O B, \triangle B O C, \triangle A O C

=\frac{1}{2} \mathrm{AB} \times \mathrm{r}+\frac{1}{2} \mathrm{BC} \times \mathrm{r}+\frac{1}{2} \mathrm{AC} \times \mathrm{r}[\because \mathrm{OF}=\mathrm{OD}=\mathrm{OE}=\mathrm{r}] \\

=\frac{1}{2} \mathrm{r}(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})

=\frac{1}{2} \times \frac{7}{2} \times 32 \text { Sq.cm. } \\

=56 \text { Sq.cm. }

\therefore \quad \text{ Area of }\triangle A B C \ is \ 56 Sq.cm. (Ans.)

**23. Let us write by calculating the length of radius of circle and circumcircle of a triangle of which sides are 20 cm, 15 cm and 25 cm. Let us calculate the area of incircle and circum- circle.**

**Solution :**

Let a = 20 cm.

S =\frac{a+b+c}{2} \\

=\frac{20+15+25}{2} \\

=\frac{60}{2}=30 cm

\therefore \quad \text { Area of triangle } =\sqrt{\mathrm{S}(\mathrm{S}-\mathrm{a})(\mathrm{S}-\mathrm{b})(\mathrm{S}-\mathrm{C})} \\

=\sqrt{30(30-20)(30-15)(30-25)} \\

=\sqrt{30 \times 10 \times 15 \times 5} \\

=\sqrt{10 \times 3 \times 10 \times 3 \times 5 \times 5} \\

=3 \times 5 \times 10 \text { Sq.cm }=150 \text { Sq.cm. }

Let ‘ r ‘ be the radius of incircle.

or, \text{ Area of } \triangle A O B+ \text{ Area of } \triangle B O C+ \text{ Area of } \triangle A O C =150 \mathrm{Sq.}\mathrm{cm}.

or, \frac{1}{2} A B \times r+\frac{1}{2} B C \times r+\frac{1}{2} A C \times r=150

[\because \mathrm{OD}=\mathrm{OE}=\mathrm{OF}=\mathrm{r}]

or, \frac{1}{2}r(A B+B C+A C)=150

or, \quad \frac{1}{2} \mathrm{r}(20+15+25)=150

or, \quad \frac{1}{2} \times r \times 60=150

or, 3r = 15

or, r = 5

\therefore \quad Radius of the incircle = 5 cm (Ans.)

Area of the incircle =\pi(5)^2

=\frac{22}{7} \times 25=\frac{550}{7} \text { Sq. cm } \\

=78 \frac{4}{7} \text { Sq. } \mathrm{cm} \quad \text { (Ans.) }

We see that,

(20)^2+(15)^2 \\

= 400 + 225

= 625 = (25)^2

Hence ABC is a right-angle triangle.

\therefore Radius of circumcircle = mid point of the hypotenuse

=\frac{25}{2} cm . =12.5 cm \text { (Ans.) } \\

\therefore \text { Area of circumcircle } =\pi(12.5)^2 \\

=\frac{22}{7} \times 12.5 \times 12.5 \\

=\frac{3437.5}{7} \text { Sq.cm. } \\

=491.07 \text { Sq.cm. }(Ans.)

**24. Jaya drew an incircle of a square. That circle is also circumscribe an equilateral triangle of which each length of side is 4 \sqrt{3} \mathrm{cm}. Let us write by calculating the length of diagonal of square.**

**Solution :**

Side of equilateral triangle \triangle \mathrm{ABC}=4 \sqrt{3} cm

\therefore \quad \text { Radius of circumcircle } =\frac{4 \sqrt{3}}{\sqrt{3}} cm . \\

= 4 cm.

\text { Diameter of incircle of square } =2 \times 4 cm .=8 cm . \\

\text { Length of diagonal of square } =\text { Side } \sqrt{2} cm \\

=8 \sqrt{2} cm \quad \text { (Ans.) }

**25. Sumit cut a wire into two equal parts. One part he bent in the form of square and other part bent in the form of circle. If the area of circle exceeds that of the square by 33sq.cm. Let us write by calculating the original length of the wire.**

* Solution :* Let the length of the wire be x cm.

wire is cut into two equal part in order to make a circle and a square

\therefore \text { Perimeter of square }=\frac{\mathrm{x}}{2} cm \\

\therefore \text { Length of side of square }=\frac{\mathrm{x} / 2}{4} cm=\frac{\mathrm{x}}{8} cm . \\

\therefore \quad \text { Area of square }=(\frac{\mathrm{x}}{8}) \text { Sq.cm } \\

=\frac{\mathrm{x}^2}{64} cm.

\therefore \text{ Again, Circumference of circle } =\frac{\mathrm{x}}{2} cm.

or, 2 \pi \mathrm{r}=\frac{\mathrm{x}}{2} cm.

or, r=\frac{x}{2} \times \frac{1}{2} \times \frac{7}{22}=\frac{7 x}{88} cm.

\therefore Radius of circle =\frac{7 x}{88} cm.

\therefore \quad Area of circle =\pi r^2

=\frac{22}{7} \times \frac{7 \mathrm{x}}{88} \times \frac{7 \mathrm{x}}{88} \text { Sq.cm }

= \frac{7x^2}{352} Sq.cm.

According to the condition of problem

\frac{7x^2}{352} - \frac{x^2}{64} = 33

or, \frac{14x^2-11x^2}{704}=33

or, \frac{3x^2}{704}=33

or, x = \frac{33 \times 704}{3}

\text { or, } x =11 \times 704

\text { or, } x =11 \times 11 \times 64 \text { or } 1 \times 17=\sqrt{11 \times 1 \times 11 \times 64}

\therefore \quad x=11 \times 8=88 Ans. Length of the wire = 88 cm.

## 26. M.C.Q :

**(i) If area of circular field is x sq. unit, perimeter is y unit Igrand length of diameter is zanit that the value of x/y is**

**(a) 1/2**

**(b) 1/4**

**(c) 1**

**(d) 1/8**

* Solution :* Let ‘ r ‘ be the radius and ‘ d ‘ be diameter of a circular field.

Then,

\frac{x}{y z}

\frac{\pi r^2}{2 \pi r \times 2 r}=\frac{1}{4}

(b) is correct option (Ans.)

**(ii) The ratio of area of two square circumscribe and inscribe by a circle is**

**(a) 4 : 1**

**(b) 1 : 4**

**(c) 2 : 1**

**(d) 1 : 2**

**Solution:**

Let ‘a’ be the side of a square,

Then ‘ a ‘ is the diameter of

in-circle and length of diagonal = a \sqrt{2} is the diameter of circumcircle,

Then ratio of two square circumcribe and inscribe by a circle is

=\pi(\frac{a \sqrt{2}}{2})^2: \pi(\frac{a}{2})^2 \\

= 2 : 1

(c) is correct option (Ans.)

**(iii) The numerical value of perimeter and area of circular field is equal. The length of diagonal of square circumscribe by a circle is**

**(a) 4 unit**

**(b) 2 unit**

**(c) 4 \sqrt{2} unit**

**(d) 2 \sqrt{2} unit**

* Solution :* Let ‘ r ‘ be the radius of a circle,

Then, 2 \pi r=\pi r^2

or, r = 2 unit.

\therefore \quad diameter of the circle = diagonal of a square =2 \times 2 units. = 4 units.

\therefore \quad (a) is correct option (Ans.)

**(iv) The ratio of the area of an equilateral triangle of circumscribed and inscribed circle is**

**(a) 4 : 1**

**(b) 1 : 4**

**(c) 2 : 1**

**(d) 1 : 2**

**Solution :**

Ratio of two circles

=\frac{2}{3} \times \mathrm{AD}: \frac{1}{3} \times \mathrm{AD} \\

= 2 : 1

Ratio of their areas

=(2)^2:(1)^2 \\

= 4 : 1

\therefore (a) is correct option (Ans.)

**(v) The inner diameter and external diameter of an Iron ring are 20 cm and 22 cm. The quality of iron plate in the ring is **

**(a)22 sq.cm**

**(b) 44 sq.cm**

**(c) 66 sq.cm**

**(d) 88 sq.cm**

* Solution :* Area of iron plate

=\{(\frac{22}{2})^2-(\frac{20}{2})^2\} \\

=\frac{22}{7}\{(11)^2-(10)^2\} \\

=\frac{22}{7}(121-100) \text { Sq.cm. } \\

=\frac{22}{7} \times 21 \text { Sq.cm. }=66 \text { sq.cm.(Ans.) }

\therefore (c) is correct option (Ans.)

## 27. Short answer type :

**(i) If the length of radius of a circular field was increased by 10%, let us write by calculating what percent it increase the area of circular field.**

* Solution :* Let ‘r’ be the radius of a circular field.

Then, Area =\pi r^2

If the length of radius of a circular field was increased by 10%

\text { Then, length of radius }=r+10 \% \text { of } r \\

=r+\frac{10}{100} \times r \\

=r+\frac{r}{10}=\frac{11 r}{10} \\

\text { Area }=\pi(\frac{11 \mathrm{r}}{2})^2 \\

=\pi \times \frac{121 r^2}{100} . \\

\text { Increase in area }=\pi(\frac{121 r^2}{100}-r^2) \\

\text { Percentage of increase of area }=\frac{\frac{121 \times r^2}{100}}{\pi r^2} \times 100 \%\\

\frac{21}{100} \times 100 \%

= 21% (Ans.)

**(ii) If the perimeter of a circular field was decreased by 50%, let us write by calculating what percent it decreases the area of circular field**

* Solution:* Let ‘ r ‘ be the radius of a circular field.

Then, Area of circular field = πr^{2}

Circumference = 2πr

If the perimeter of a circular field was decreased by 50%

Then, Perimeter of the circular field

=2 \pi \mathrm{r}-50 \% \text { of } 2 \pi \mathrm{r}

=2 \pi r-\frac{50}{100} \times 2 \pi r \\

=2 \pi r-\pi r \\

= πr

= 2 \pi(\frac{r}{20})

\therefore \text{ Area of the circular field } =\pi(\frac{\mathrm{r}}{2})^2

=\frac{\pi r^2}{4}

Decrease in area = \pi r^2 - \frac{2 \pi r^2}{4}

=\frac{3 \pi r^2}{4}

Percentage of decrease in area = \frac{\frac{3\pi r^2}{4}}{\pi r^2} \times 100\%

= \frac{3}{4} \times 100 \%

= 75% (Ans.)

**(iii) The length of radius of a circular field is r meter If the area of other circle is times of first circle, Let us see by calculating how length is of radius of other circle.**

* Solution :* Let ‘ r ‘ be the radius of other circle.

Then, \pi R^2=x \times \pi r^2

\therefore \quad R^2=\pi r^2 \\

\therefore \quad R=\sqrt{x} \cdot r

\therefore \quad Length of radius of other cfrcle = r\sqrt{x} meter (Ans.)

**(iv) Let us calculate the area of a circle circumscribe a triangle of which sides are 3 cm, 4 cm and 5 cm.**

* Solution:* We see that,

(3)^2+(4)^2 \\

= 9 + 16

= 25

= (5)^2

\therefore We see that the sides 3 cm, 4 cm and 5 cm is the sides of a right-angled trianglé.

\therefore \quad Length of radius of circumscribe

=\text { Mid-point of hypotenuse }=5 / 2=2.5 cm \text {. }

Area of the circle circumscribe =\pi(2.5)^2

=\frac{22}{7} \times 2.5 \times 2.5 \\

=\frac{22}{7} \times \frac{25}{10} \times \frac{25}{10}

=\frac{266}{14} \text { Sq.cm. } \\

=19 \frac{9}{14} \text { Sq.cm. (Ans.) }

**(v) Three circular plate were cut off from a tin plate with equal width if the ratio of length of diameter of three is 3 : 5 : 7, let us see by calculating the ratio of their weight.**

* Solution :* Ratio of length of diameter of three circles

= 3 : 5 : 7

=\frac{3}{2}: \frac{5}{2}: \frac{7}{2}\\

= 3 : 5 : 7

\text { The ratio of their weight } =\pi(3)^2: \pi(5)^2: \pi(7)^2 \\

= 9 : 25 : 49 (Ans.)