Class 10th, Area Related Circle Exe-12.3 (Solution2)

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Maths NCERT Class 10th Chapter 12 – Areas Related to Circles

 Class 10th Maths Chapter 12 Exercise: 12.3 (Q5 to Q8)

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5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

Solution:

Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)2 = 42 = 16 cm2

Area of the quadrant = (π R2)/4 cm2 = (22/7 × 12)/4 = 11/14 cm2

∴ Total area of the 4 quadrants = 4 × (11/14) cm2 = 22/7 cm2

Area of the circle = π R2 cm2 = (22/7 × 12) = 22/7 cm2

Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)

= 16 cm2 – (22/7 + 22/7) cm2

= 68/7 cm2

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

Solution:

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By Pythagoras theorem,

AB2 = AD2 + BD2

⇒ AB2 = 482 + (AB/2)2

⇒ AB2 = 2304 + AB2/4

⇒ 3/4 (AB2) = 2304

⇒ AB2 = 3072

⇒ AB = 32√3 cm

Area of ΔADB = √3/4 × (32√3)2 cm2 = 768√3 cm2

Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2

Area of the design = Area of circle – Area of ΔADB

= (22528/7 – 768√3) cm2

7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:

Side of square = 14 cm

Four quadrants are included in the four sides of the square.

∴ Radius of the circles = 14/2 cm = 7 cm

Area of the square ABCD = 142 = 196 cm2

Area of the quadrant = (π R2)/4 cm2 = (22/7 × 72)/4 cm2

= 77/2 cm2

Total area of the quadrant = 4 × 77/2 cm2 = 154 cm2

Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm2 – 154 cm2

= 42 cm2

8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If

the track is 10 m wide, find :

(i) the distance around the track along its inner edge

(ii) the area of the track.

Solution:

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O’C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B

= 30 + 10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)

= 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m

= 212 + 1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB × CD) + (EF × GH) + 2 × (πr2/2) – 2 × (πR2/2) m2

= (106 × 10) + (106 × 10) + 2 × π/2 (r2 -R2) m2

= 2120 + 22/7 × 70 × 10 m2

= 4320 m2

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