NCERT Solutions Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area on Padhaiwise

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.1#

Unless stated otherwise, use $\pi \approx \dfrac{22}{7}$.

$$\text{Circumference} = 2\pi r = \pi d$$

$$\text{Arc length} = \dfrac{\theta}{360°} \times 2\pi r$$
Perimeter is the length of the whole boundary — add every curved and straight part, but leave out lines that are inside the shape.

1. The perimeter of a circle is 44 cm. What is its radius? #

Solution

The perimeter (circumference) is $2\pi r$, so:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.2#

$$\text{Area of a triangle} = \dfrac{1}{2} \times \text{base} \times \text{height}$$
Heron’s formula: for sides $a$, $b$, $c$ and semi-perimeter $s = \dfrac{a + b + c}{2}$,
$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$
Key idea: triangles with the same base and equal heights (their third vertices on a line parallel to the base) have equal areas.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.3#

Area of a circle: $\pi r^2$

Area of a sector with angle $\theta$ at the centre: $\dfrac{\theta}{360°} \times \pi r^2$

Area of a segment = area of the sector − area of the triangle formed by the chord and the two radii.

Use $\pi = \dfrac{22}{7}$ unless the question says otherwise.

1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°. #

Solution

$$\text{Area} = \dfrac{60}{360} \times \dfrac{22}{7} \times 7^2$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — End-of-Chapter Exercises#

Unless stated otherwise, use $\pi \approx \dfrac{22}{7}$.

$$\text{Triangle} = \dfrac{1}{2}bh, \qquad \text{Heron: } \sqrt{s(s - a)(s - b)(s - c)}$$

$$\text{Trapezium} = \dfrac{1}{2}(a + b)h, \qquad \text{Rhombus or kite} = \dfrac{1}{2}d_1d_2$$

$$\text{Circle} = \pi r^2, \qquad \text{Sector} = \dfrac{\theta}{360°}\pi r^2$$

1. Identities in algebra can sometimes be shown as area relationships. For example, Fig. 6.41 corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. #

How Fig. 6.41 shows (a + b)²

The big square has side $a + b$, so its area is $(a + b)^2$. It is cut into a square $a^2$, a square $b^2$ and two rectangles $ab$. Adding the pieces gives $a^2 + 2ab + b^2$, so $(a + b)^2 = a^2 + 2ab + b^2$.