第10类RD Sharma解决方案–第15章与圆相关的区域–练习15.1 |套装2

问题11：两个圆的半径分别是19厘米和9厘米。求出具有等于两个圆的周长之和的圆的半径和面积。

解决方案：

Radius of circle 1 = r₁ = 19 cm

Radius of circle 2 = r₂ = 9 cm

So, C₁ = 2πr_1, C₂ = 2πr₂

C = C₁ + C₂

2πr = 2πr₁ + 2πr₂

r = r₁ + r₂ 

r = 19 + 9

r = 28 cm

Therefore, the radius of the circle = 28 cm

Therefore, area of required circle = πr² 

                                                     = (22/7) × 28 × 28 

                                                     = 2464 cm²

问题12.圆形操场的面积为22176 m ² 。找出以每米₹50的速度护栏的成本。

解决方案：

Area of the circular playground = 22176 m²

Area = πr²

πr² = 22176

r² = 22176(7/22) 

    = 7056

r = 84 m

Circumference of the ground = 2πr 

                                               = 2(22/7)84 

                                               = 528 m

Cost of fencing 528 m = ₹50 x 528 

                                   = ₹26400

Therefore, the cost of fencing the ground = ₹26400.

问题13.正方形的边长为10厘米。找到外切圆和内切圆的区域。

解决方案：

Diagonal of the square = AC = √2 x side

                                     = 10√2 cm

Radius of circumscribed circle = Diagonal/2

R = 5√2cm

R = 7.07cm

Area= πR² 

        = (22/7) × 7.07 × 7.07 

        = 157.09 cm² 

Therefore, the Area of the Circumscribed circle = 157.09 cm² 

For inscribed circle diameter of circle = side of square = AB

Radius = side of square/2

           = 10/2 

     r     = 5 m

Area = πr² 

        = (22/7) × 5 × 5 

        = 78.5 cm² 

Therefore, the area of the circumscribed circle = 157.09 cm² and the area of the inscribed circle = 78.5 cm².

问题14.如果正方形被刻在圆上，求出该圆与正方形的面积之比。

解决方案：

Let side of square AB be x cm which is inscribed in a circle.

Radius of circle (r) = 1/2 (diagonal of square)

                              = 1/2(a√2)

                              r = a/√2

Area of the square = a²

Area of the circle = πr²

                           = π(a²/2)

Ratio of areas = Area of circle:Area of square

                        = π(a²/2) : a²

                                 ⁼ π : 2

Therefore, the ratio of areas of the circle and the square = π : 2

问题15.等边三角形内切圆的面积为154 cm ² 。找到三角形的周长。

解决方案：

Area of a Circle = πr²

(22/7) × r² = 154

r² = (154 x 7)/22 

   = 7 × 7 

   = 49

r = 7 cm

OP is perpendicular bisector of BC (as BP is tangent and it is a equilateral triangle)

BP = ½ x BC

Consider the side of the equilateral triangle be a cm.

In right-angled triangle OPB

OB² = OP² + BP² (By Pythagoras theorem)

OB² = r² + (a/2)²    (BP is half of a)

OB² = 49 + a²/4

OB = √(49 + a²/2)    ….. (1)

AP = (√3/2)a (height of an equilateral triangle) 

OA = (√3/2)a – r

Similarly 

OB = (√3/2)a – r    …. (2)

From (1) and (2)

Squaring both sides

49 + a²/4 = (3/4)a² + r² – √3ar

r = 7

49 + a²/4 = (3/4)a² + 49 – 7√3a

a²/4- (3/4)a² = -7√3a

Taking 4 as LCM

(a² – 3a²) / 4 = -7√3a

-2a²/4 = -7√3a

a = 14√3 cm

Perimeter of equilateral triangle = 3a

                                                    = 42√3

Therefore, the perimeter of the triangle = 42(1.73) 

                                                             = 72.7 cm