How many times a wheel of radius 28 cm must rotate to go 352 m? (Take \(\pi\) = \(\frac{22}{7}\))
Solution and Explanation
\(r = 28\;cm\)
\(\text{Circumference}\) = \(2\pi r\)
= \(2\times \frac{22}{7}\times 28\)
= \(176 \;cm\)
\(\text{Number of rotations}=\frac{\text{Total distance to be covered}}{\text{Circumference of wheel}}\)
= \(\frac{352\;m}{176\;cm}\)
= \(\frac{35200}{176}\)
= \(200\)
Top Questions on Circles
What is the diameter of the circle in the figure ?
Consider the above figure and read the following statements.
Statement 1: The length of the tangent drawn from the point P to the circle is 24 centimetres. If OP is 25 centimetres, then the radius of the circle is 7 centimetres.
Statement 2: A tangent to a circle is perpendicular to the radius through the point of contact.
Now choose the correct answer from those given below.
- Write the polynomial x² - 12x + 32 as the product of two first degree polynomials.
- A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
OR
A solid is in the shape of a cone standing on a right circular cylinder with both their radii being equal to 7.5 cm and the height of the cone is equal to its radius. If the total height of the solid is 22.5 cm, find the volume of the solid. (Use $\pi = 3.14$, $\sqrt{3} = 1.732$)
- The two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2 \angle OPQ$.
Questions Asked in CBSE Class VII exam
- Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192 (ii) 270 (iii) 729 × 64 (iv) 768- CBSE Class VII
- Exponents and Powers
- The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.
- CBSE Class VII
- Need For Rational Numbers
- Multiply and reduce to lowest form (if possible) :
(i) \(\frac{2}{3}\) × 2 \(\frac{2}{3}\)
(ii) \(\frac{2}{7}\) × \(\frac{7}{9}\)
(iii) \(\frac{3}{8}\) × \(\frac{6}{4}\)
(iv) \(\frac{9}{4}\) × \(\frac{3}{5}\)
(v) \(\frac{1}{3}\)× \(\frac{15}{8}\)
(vi) \(\frac{11}{2}\) × \(\frac{3}{10}\)
(vii) \(\frac{4}{5}\) × \(\frac{12}{7}\)- CBSE Class VII
- Multiplication Of Fractions
- Find the product:
(i)\(\frac{9}{2}\)×(-\(\frac{7}{4}\))
(ii) \(\frac{3}{10}\)×(-9)
(iii) -\(\frac{6}{5}\)×\(\frac{9}{11}\)
(iv) \(\frac{3}{7}\)×(-\(\frac{2}{5}\))
(v) \(\frac{3}{11}\)×\(\frac{2}{5}\)
(vi) \(\frac{3}{-5}\) ×\(\frac{-5}{3}\)- CBSE Class VII
- Operations On Rational Numbers
- Draw, wherever possible, a rough sketch of
- a triangle with both line and rotational symmetries of order more than 1.
- a triangle with only line symmetry and no rotational symmetry of order more than 1.
- a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
- a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
- CBSE Class VII
- Line Symmetry And Rotational Symmetry
Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
Also Check: