# CBSE Solutions for Class 10 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

What is the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm? What is the area of a circle whose circumference is 8π? What is the perimeter of a semicircular protractor whose diameter is 14 cm? What is the radius of a circle whose perimeter and area are numerically equal? Circumference of a circle is 39.6 cm. find its area.

Circumference of circle = 2 π r = 39.6 cm Circumference of a circle is 22 cm. Find the area of its quadrant. A circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, Find the Diameter If the area of a circle is numerically equal to twice its circumference, then find the diameter of the circle. Find the perimeter of a square which circumscribes a circle of radius a cm. Side of the square = 2  radius of circle = 2a cm

Then, Perimeter of the square = (4  2a) = 8a cm

What is the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre? All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is 1256 cm2. [Use π = 3.14] (2015OD)

In rhombus, AB = BC = CD = AD
AC = BD = 2r In Figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region. [Use π = 22/7] (2016D)  Area of shaded region = 2(Area of square) – 4(Area of semicircle) = 2 × 196 – 308 = 392 – 308 = 84 cm2

In Figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle ABC of side 14 cm as centres to intersect the sides BC, CA and AB at BZ their respective mid-points D, E and F. Find the area of the shaded region. [Use π = 22/7  and  √3 = 1.73] (2011D) ABC = BAC = ACB = 60° Match the radius with the resp. perimeters of circle

 1 12 A 36π 2 14 B 24π 3 16 C 28π 4 18 D 32π

1-B, 2-C, 3-D, 4-A

Match the radius with the resp. perimeters of semi-circle

 1 2 A 6(π + 2) 2 4 B 2(π + 2) 3 6 C 8(π + 2) 4 8 D 4(π + 2)

1-B, 2-D, 3-A, 4-C

Match the radius with the resp. perimeters of circle

 1 2 A 8π 2 4 B 12π 3 6 C 4π 4 8 D 16π

1-C, 2-A, 3-B, 4-D