# NCERT Solutions Class 7 Maths Chapter 11

## NCERT Solutions Class 7 Maths Chapter 11: Perimeter and Area

Class 7 introduces students to several new Mathematics topics and concepts, which will be further built upon in higher grades. This makes it very important that Class 7 students focus on building a strong foundation in this subject at this stage.

Students can use the NCERT Solutions for Class 7 Mathematics Chapter 11 by Extramarks for their preparations. They can get detailed and accurate solutions to questions given in the NCERT textbook, which will help them understand how to solve the different kinds of problems in a step-by-step manner. If they ever get stuck on a question, they can always refer to the solutions prepared by subject matter experts.

### Access NCERT Solutions for Class 7 Maths Chapter 11 – Perimeter and Area

 Chapter 11 – Perimeter and Area Exercise Exercise 11.1 Questions & Solutions Exercise 11.2 Questions & Solutions Exercise 11.3 Questions & Solutions Exercise 11.4 Questions & Solutions

### NCERT Solutions for Class 7 Maths Chapter 11 –

NCERT Class 7 Mathematics Chapter 11 teaches students how to calculate the perimeters and areas of different types of basic shapes, including squares, circles, triangles and parallelograms. Students are introduced to a lot of different formulas that they must remember and retain to perform well in the exam. They  are also taught about how to handle the conversion of measurement units.

Some of the core concepts and formulas included in the chapter are as follows:

• Perimeter is the total length of the boundary of a closed rectilinear figure.
• The surface circled by a two-dimensional figure is its area.
• Area of a Parallelogram = Base × Altitude
• Area of a Triangle = ½ × Base × Altitude corresponding to the base
• Circumference of a circle = 2πr.
• Area of a Circle = πr²
• 1 cm² = 100 mm²
• 1m² = 10000 cm²
• 1 hectare = 10000 m²
• The perimeter of a regular polygon = Length of a side × Number of the sides

Triangle: A triangle is a plane figure surrounded by three lines segments.

Rectilinear Figure: A plane figure encompassed by line segments is called a rectilinear figure. None of the two sides of it intersects at a point other than the vertex.

Quadrilateral: It is a plane figure surrounded by four long segments.

Parallelogram: Parallelogram is a plane figure encircled by four long segments. The opposite sides of a parallelogram are equal and parallel in length.

Area of a Triangle: The area of a triangle is ½ × Base × Height (Altitude corresponding to the base).

Example: base = 6 cm, height = 12cm. What is the area of a triangle?

As the formula goes; ½ × Base × Height

= (½ × 6 × 12) cm² = 36 cm²

Area of a Parallelogram: Base × Altitude (height)

Example: base = 10 cm and height = 5 cm

The formula of parallelogram: base × height

= (10 × 5) cm²

= 50 cm²

Circumference of a Circle: Perimeter of a circle is similar to the circumference of a circle. The students would draw four circles of different radii and find their circumference with the ratio of the circumferences to the diameter. The ratio will be more than three times its diameter.

Area of a Circle: The area a circle occupies on a flat plane which has its length and breadth, is known as the area of a surface. The formula to find the area is πr².

### Conversion of Units

 Units of Length Units of Area 1 cm = 10 mm 1 cm² = (10 x 10) mm² = 100 mm² 1 dm = 10 cm 1 dm² = (10 x 10) cm² = 100 cm² 1 m = 10 dm 1 m² = (10 x 10) dm² = 100 dm² 1 dam = 10 m 1 dam² = (10 x 10) m² = 100 m² 1 hm = 10 dam 1 hm² = ( 10 x 10 ) dam² = 100 dam² 1 km = 10 hm 1 km² = ( 10 x 10) hm² = 100 hm²

Note:

1m = 100 cm ∴ 1 m² = 10000 cm²

1 km = 1000 m ∴ 1 km² = 10, 00, 000 m²

1 hectare (ha) = 100 m x 100m = 10000 m²

### NCERT Solutions for Class 7 Maths

Mathematics is that one subject that many students struggle with. This is because getting good at Mathematics requires one to practise solving many problems and most students don’t do that. With NCERT Solutions of Class 7 Mathematics, Extramarks tries to provide students with a very valuable resource that they can use to get better at the subject and eventually score well in their final exams. Students can use these solutions if they ever get stuck on a problem or to learn the approach to how to solve the different kinds of problems encountered in any NCERT Mathematics Class 7 Chapter 11.

The NCERT Solutions by Extramarks are prepared by subject matter experts who give special attention to providing detailed and accurate solutions in a step-by-step manner so that students can easily grasp the problem-solving approach. Students can use these resources for their preparations, last-minute revisions, and for help with their assignments.

### NCERT Solutions for Class 7

Apart from Mathematics, Extramarks provides comprehensive  solutions to textbook questions for all of the other subjects covered in Class 7. All the solutions are prepared by their respective subject experts and extra care is given to providing accurate solutions which students can rely on. Students can download all these solutions from the Extramarks website.

Q.1 Find the area of a square park whose perimeter is 320 m.

Ans

$\begin{array}{l}\text{Let the side of the square park be ‘}a\text{‘}\text{.}\\ \text{Since perimeter of a square}=\text{4}a\\ \text{So, we get}\\ 4a=3\text{20}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}a=\frac{320}{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=80\text{\hspace{0.17em}}\text{m}\\ \text{Therefore, the area of the square park}={a}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(80\text{\hspace{0.17em}}\text{m}\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6400\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}$

Q.2 Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.

Ans

$\begin{array}{l}\text{Let the breadth of the rectangular plot be}x\text{.}\\ \text{Area of the rectangular plot}=\text{length}×\text{breadth}\\ \text{SO, we get}\\ {\text{440m}}^{2}=22\text{\hspace{0.17em}}\text{m}×x\\ \text{}x=\frac{440\text{\hspace{0.17em}}{\text{m}}^{2}}{22\text{\hspace{0.17em}}\text{m}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20\text{\hspace{0.17em}}\text{m}\\ \text{Perimeter of the rectangular plot}=\text{2}\left(\text{length+breadth}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(20\text{\hspace{0.17em}}\text{m}+22\text{\hspace{0.17em}}\text{m}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(42\text{\hspace{0.17em}}\text{m}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=84\text{\hspace{0.17em}}\text{m}\end{array}$

Q.3 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans

$\begin{array}{l}\text{Let the breadth of the rectangular sheet be}x\text{.}\\ \text{Perimeter of the rectangular sheet}=\text{2}\left(\text{length+breadth}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}100\text{cm}=2\left(35\text{\hspace{0.17em}}\text{cm}+x\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\frac{100}{2}\text{cm}=\left(35\text{cm+x}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}50\text{\hspace{0.17em}}\text{cm}=35\text{cm+x}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=50\text{cm}-35\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=15\text{cm}\\ \text{Area of rectangular sheet}=\text{length}×\text{breadth}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{35 cm}×\text{15 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{525 cm}}^{2}\end{array}$

Q.4

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{square}\mathrm{park}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{of}\mathrm{a}\mathrm{rectangular}\\ \mathrm{park}.\mathrm{If}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{park}\mathrm{is}60\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{length}\\ \mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{park}\mathrm{is}90\mathrm{m},\mathrm{find}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{the}\\ \mathrm{rectangular}\mathrm{park}.\end{array}$

Ans

$\begin{array}{l}\text{Let the breadth of rectangular park be}x\text{.}\\ \text{Since, area of a square park is the same as of a rectangular}\\ \text{park}.\\ \text{Area of square}=\text{Area of rectangular park}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(\text{60m}\right)}^{2}=90m×x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3600{m}^{2}=90m×x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{3600{m}^{2}}{90m}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=40m\end{array}$

Q.5

$\begin{array}{l}\mathrm{A}\mathrm{wire}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{rectangle}.\mathrm{Its}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{breadth}\mathrm{is}22\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{rebent}\mathrm{in}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{each}\mathrm{side}.\\ \mathrm{Also}\mathrm{find}\mathrm{which}\mathrm{shape}\mathrm{encloses}\mathrm{more}\mathrm{area}?\end{array}$

Ans

$\begin{array}{l}\text{Perimeter of a rectangle}=\text{Perimeter of square}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\left(\text{length+breadth}\right)=4×\text{side}\\ \text{}\text{}\text{\hspace{0.17em}}\text{2}\left(40\text{\hspace{0.17em}}\text{m}+22\text{\hspace{0.17em}}\text{m}\right)=4×\text{side}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{side}=\frac{124}{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=31\text{\hspace{0.17em}}\text{cm}\\ So,\\ \text{Area of rectangle}=\text{length}×\text{breadth}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{40cm}×\text{22cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{880cm}}^{2}\\ \text{Area of square}={\left(\text{side}\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(31\text{\hspace{0.17em}}\text{cm}\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=961\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{Therefore, the square-shaped wire encloses more area}\text{.}\end{array}$

Q.6

$\begin{array}{l}\mathrm{The}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{is}130\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{breadth}\\ \mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{is}30\mathrm{cm},\mathrm{find}\mathrm{its}\mathrm{length}.\mathrm{Also}\mathrm{find}\mathrm{the}\\ \mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}.\end{array}$

Ans

$\begin{array}{l}\text{Let the length of the rectangle be}x\text{cm}\text{.}\\ \text{Perimeter of rectangle}=\text{2 (length}+\text{breadth)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}130\text{\hspace{0.17em}}\text{cm}=\text{2(30 cm}+x\text{)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{130}{2}\text{cm}=\text{30 cm}+x\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\text{65 cm}-\text{30 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{35 cm}\\ \text{Now,}\\ \text{area of rectangle}=\text{length}×\text{breadth}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{30 cm}×\text{35 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{1050 cm}}^{2}\end{array}$

Q.7 Find the area of each of the following parallelogram : Ans

$\begin{array}{l}\text{(a)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{7 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{7}×\text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{28cm}}^{2}\\ \text{(b)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{3 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}×\text{3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{15 cm}}^{2}\\ \text{(c)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{3}\text{.5 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{2}\text{.5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\text{.5}×\text{3}\text{.5}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}{\text{.75 cm}}^{2}\end{array}$ $\begin{array}{l}\text{(d)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4}\text{.8 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}×\text{4}\text{.8}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{24 cm}}^{2}\\ \text{(e)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4}\text{.4 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{2 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}×\text{4}\text{.4}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}{\text{.8 cm}}^{2}\end{array}$

Q.8 Find the area of each of the following triangles : Ans

$\begin{array}{l}\text{(a) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×4×3\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×12=6{\text{cm}}^{2}\\ \text{(b) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×5×3.2\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×16=8{\text{cm}}^{2}\\ \text{(c) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×3×4\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×12=6{\text{cm}}^{2}\\ \text{(d) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×3×2\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×6=3{\text{cm}}^{2}\end{array}$

Q.9 Find the missing values :

 S. NO. Base Height Area of the Parallelogram a. 20 cm 246 cm2 b. 15 cm 154.5 cm2 c. 8.4 cm 78.72 cm2 d. 15.6 cm 16.38 cm2

Ans

$\begin{array}{l}\text{Let the height be}h\text{and base be}b.\\ \text{(a) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ {\text{246 cm}}^{\text{2}}=\text{20 cm}×\text{h}\\ \text{h=}\frac{246\text{\hspace{0.17em}}{\text{cm}}^{2}}{20\text{\hspace{0.17em}}\text{cm}}\\ =12.3\text{\hspace{0.17em}}\text{cm}\\ \text{(b) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{154}{\text{.5 cm}}^{\text{2}}=\text{b}×\text{15 cm}\\ \text{b=}\frac{154.5\text{\hspace{0.17em}}{\text{cm}}^{2}}{15\text{\hspace{0.17em}}\text{cm}}\\ =10.3\text{\hspace{0.17em}}\text{cm}\end{array}$ $\begin{array}{l}\text{(c) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{78}{\text{.72 cm}}^{\text{2}}=\text{b}×\text{8}\text{.4 cm}\\ \text{b=}\frac{48.72\text{\hspace{0.17em}}{\text{cm}}^{2}}{8.4\text{\hspace{0.17em}}\text{cm}}\\ =5.8\text{\hspace{0.17em}}\text{cm}\\ \text{(d) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{16}{\text{.38 cm}}^{\text{2}}=\text{15}\text{.6 cm}×\text{h}\\ \text{h=}\frac{16.38\text{\hspace{0.17em}}{\text{cm}}^{2}}{15.6\text{\hspace{0.17em}}\text{cm}}\\ =1.05\text{\hspace{0.17em}}\text{cm}\\ \text{So, we get}\\ \begin{array}{cccc}\text{S}\text{.No}\text{.}& \text{Base}& \text{Height}& \text{Area of the Parallelogram}\\ \text{a}\text{.}& 20\text{cm}& 12.3\text{\hspace{0.17em}}\text{cm}& 246{\text{cm}}^{2}\\ \text{b}\text{.}& 10.3\text{\hspace{0.17em}}\text{cm}& 15\text{cm}& 154.5{\text{cm}}^{2}\\ \text{c}\text{.}& 5.8\text{\hspace{0.17em}}\text{cm}& 8.4\text{cm}& 78.72{\text{cm}}^{2}\\ \text{d}\text{.}& 15.6\text{cm}& 1.05\text{\hspace{0.17em}}\text{cm}& 16.38{\text{cm}}^{2}\end{array}\end{array}$

Q.10 Find the missing values :

 Base Height Area of Triangle 15 cm …………….. 84 cm2 …………….. 31.4 mm 1256 mm2 22 cm …………….. 170.5 cm2

Ans

$\begin{array}{l}\text{(a) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}{\text{87cm}}^{2}=\frac{1}{2}×15\text{\hspace{0.17em}}\text{cm}×\text{h}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{h}=\frac{87{\text{cm}}^{2}×2}{15\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11.6\text{cm}\\ \text{(b) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}{\text{1256 mm}}^{2}=\frac{1}{2}×b×\text{31}\text{.4 mm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=\frac{1256{\text{mm}}^{2}×2}{31.4\text{\hspace{0.17em}}\text{mm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=80\text{mm}\end{array}$ $\begin{array}{l}\text{(c) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}\text{170}{\text{.5 cm}}^{2}=\frac{1}{2}×22\text{\hspace{0.17em}}\text{cm}×\text{h}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{h}=\frac{170.5{\text{cm}}^{2}×2}{22\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=15.5\text{cm}\\ \text{So we get}\\ \begin{array}{ccc}\text{Base}& \text{Height}& \text{Area of Triangle}\\ 15\text{cm}& 11.6\text{cm}& 84{\text{cm}}^{2}\\ 80\text{\hspace{0.17em}}\text{mm}& 31.4\text{mm}& 1256\text{\hspace{0.17em}}{\text{mm}}^{2}\\ 22\text{cm}& 15.5\text{\hspace{0.17em}}\text{cm}& 170.5{\text{cm}}^{2}\end{array}\end{array}$

Q.11

$\begin{array}{l}\mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{parallelogram}. \mathrm{QM}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{SR}\\ \mathrm{and}\mathrm{QN}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{PS}.\mathrm{If}\mathrm{SR}=12\mathrm{cm}\mathrm{and}\\ \mathrm{QM}= 7.6\mathrm{cm}.\mathrm{Find}:\\ \left(\mathrm{a}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallegram}\mathrm{PQRS}\\ \left(\mathrm{b}\right)\mathrm{QN},\mathrm{if}\mathrm{PS}= 8\mathrm{cm}\end{array}$ Ans

$\begin{array}{l}\text{(a)}\\ \text{Area of a parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{SR}×\text{QM}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7.6×12\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=91.2\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{(b)}\\ \text{Area of a parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{PS}×\text{QN}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=91.2\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{QN}×8=91.2\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{QN}=\frac{91.2}{8}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11.4\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.12

$\begin{array}{l}\mathrm{DL}\mathrm{and}\mathrm{BM}\mathrm{are}\mathrm{the}\mathrm{heights}\mathrm{on}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{AD}\mathrm{respectively}\\ \mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}.\mathrm{If}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{is}\\ 1470{\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}= 35\mathrm{cm}\mathrm{and}\mathrm{AD}= 49\mathrm{cm},\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\\ \mathrm{BMand}\mathrm{DL}.\end{array}$ Ans

$\begin{array}{l}\text{Area of parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{AB}×\text{DL}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1470}=\text{35}×\text{DL}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{DL}=\frac{1470}{35}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=42\text{\hspace{0.17em}}\text{cm}\\ \text{Also,}\\ 1470=\text{AD}×\text{BM}\\ \text{1470}=\text{49}×\text{BM}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{BM}=\frac{1470}{49}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=30\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.13

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{AD}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}.\\ \mathrm{If}\mathrm{AB}=5\mathrm{cm}, \mathrm{BC}=13\mathrm{cm}\mathrm{and}\mathrm{AC}=12\mathrm{cm},\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\\ △\mathrm{ABC}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{AD}.\end{array}$ Ans

$\begin{array}{l}\text{Since, Area}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×\text{5}×\text{12}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{30 cm}}^{2}\\ \text{Also, area of triangle}=\frac{1}{2}×\text{AD}×\text{BC}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{30}=\frac{1}{2}×\text{AD}×\text{13}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{AD}=\frac{30×2}{12}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.6\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.14

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{with}\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}\mathrm{and}\mathrm{BC}=9\mathrm{cm}.\\ \mathrm{The}\mathrm{height}\mathrm{AD}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{is}6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}△\mathrm{ABC}.\\ \mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{C}\mathrm{to}\mathrm{AB}\mathrm{i}.\mathrm{e}.,\mathrm{CE}?\end{array}$ Ans

$\begin{array}{l}\text{Area of}\Delta \text{ABC}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}=\frac{1}{2}\text{BC}×\text{AD}=\frac{1}{2}×\text{9}×\text{6}={\text{27cm}}^{\text{2}}\\ \text{Also, area of}\Delta \text{ABC}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{AB}×\text{CE}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}27=\frac{1}{2}\text{7}\text{.5}×\text{CE}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{CE}=\frac{27×2}{7.5}=\text{7}\text{.2 cm}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{with}\mathrm{the}\mathrm{following}\\ \mathrm{radius}:\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{a}\right)14\mathrm{cm}\left(\mathrm{b}\right)28\mathrm{mm}\left(\mathrm{c}\right)21\mathrm{cm}\end{array}$

Ans

$\begin{array}{l}\text{(a)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\frac{22}{7}×14\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×22×2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=88\text{\hspace{0.17em}}\text{cm}\\ \text{(b)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\frac{22}{7}×28\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×22×4\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=176\text{\hspace{0.17em}}\text{mm}\end{array}$ $\begin{array}{l}\text{(c)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\frac{22}{7}×21\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×22×3\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=132\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{circles},\mathrm{given}\mathrm{that}:\\ \left(\mathrm{a}\right)\mathrm{radius}= 14\mathrm{mm}\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{b}\right)\mathrm{diameter}= 49\mathrm{m}\left(\mathrm{c}\right)\mathrm{radius}= 5\mathrm{cm}\end{array}$

Ans

$\begin{array}{l}\text{(a)}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(14\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=616\text{\hspace{0.17em}}{\text{mm}}^{2}\end{array}$ $\begin{array}{l}\text{(b)}\\ \text{Radius}=\frac{\text{Diameter}}{2}=\frac{49}{2}=24.5\text{\hspace{0.17em}}\text{mm}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(24.5\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1886.5\text{\hspace{0.17em}}{\text{m}}^{2}\\ \text{(c)}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(5\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{550}{7}\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$

Q.17 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

$\frac{22}{7}$

)

Ans

$\begin{array}{l}\text{Since, circumference}=\text{2}\pi \text{r}\\ \text{So, we get}\end{array}$ $\begin{array}{l}\text{154}=\text{2}×\frac{22}{7}×r\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{154×7}{2×22}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{49}{2}\text{\hspace{0.17em}}\text{m}=4.5\text{\hspace{0.17em}}\text{m}\\ \text{Now,}\\ \text{Area}=\pi {\text{r}}^{2}\\ \text{}\text{\hspace{0.17em}}=\frac{22}{7}×\frac{49}{2}×\frac{49}{2}\\ \text{}\text{\hspace{0.17em}}=1886.5\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}$

Q.18 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

$\begin{array}{l}\text{Outer radius of circular sheet}=\text{4 cm}\\ \text{Inner radius of circular sheet}=\text{3 cm}\end{array}$ $\begin{array}{l}\text{Remaining area}=-\left(3.14×3×3\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=50.24-28.26\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=21.98\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$

Q.19  Find the perimeter of the adjoining figure, which is a semicircle including its diameter. Ans

$\begin{array}{l}\text{Radius}=\text{5 cm}\\ \text{Length of curved part}=\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}×5\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=15.71\text{cm}\\ \text{Total permieter}=\text{Length of the curved part+Length of diameter}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{15}\text{.71+10}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{25}\text{.71 cm}\end{array}$

Q.20

$\begin{array}{l}\mathrm{Shazli}\mathrm{took}\mathrm{a}\mathrm{wire}\mathrm{of}\mathrm{length}44\mathrm{cm}\mathrm{and}\mathrm{bent}\mathrm{it}\mathrm{into}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{circle}.\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{that}\mathrm{circle}.\mathrm{Also}\mathrm{find}\\ \mathrm{its}\mathrm{area}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{bent}\mathrm{into}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\\ \mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{its}\mathrm{sides}?\\ \mathrm{Which}\mathrm{figure}\mathrm{encloses}\mathrm{more}.\end{array}$

Ans

$\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}=\text{44 cm}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}×\frac{22}{7}×r=44\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=7\text{cm}\end{array}$ $\begin{array}{l}\text{Area}=\pi {\text{r}}^{2}\\ \text{}\text{\hspace{0.17em}}=\frac{22}{7}×7×7\\ \text{}\text{\hspace{0.17em}}=154{\text{cm}}^{2}\\ \text{If the wire is bent into a square, then the length of each}\\ \text{side would be}=\frac{44}{4}=11\text{cm}\\ \text{Area of square}={\left(11\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=121{\text{cm}}^{2}\\ \text{Therefore, circle encloses more area}\text{.}\end{array}$

Q.21

$\begin{array}{l}\mathrm{From}\mathrm{a}\mathrm{circular}\mathrm{card}\mathrm{sheet}\mathrm{of}\mathrm{radius}14\mathrm{cm},\mathrm{two}\mathrm{circles}\mathrm{of}\\ \mathrm{radius}3.5\mathrm{cm}\mathrm{and}\mathrm{a}\mathrm{rectangle}\mathrm{of}\mathrm{length}3\mathrm{cm}\mathrm{and}\mathrm{breadth}\\ 1\mathrm{cm}\mathrm{are}\mathrm{removed}. \left(\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{adjoiningfigure}\right).\\ \mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{sheet}.\left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$ Ans

$\begin{array}{l}\text{Area of bigger cirle}=\frac{22}{7}×14×14\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=616{\text{cm}}^{2}\\ \text{Area of 2 small circles}=\text{2}×\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{}\text{}=2×\frac{22}{7}×3.5×3.5\\ \text{}\text{}\text{}\text{}\text{}=77{\text{cm}}^{2}\\ \text{Area of rectangle}=\text{Length}×\text{Breadth}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3}×\text{1}={\text{3 cm}}^{2}\\ \text{Remaining area of sheet}={\text{616 cm}}^{2}-{\text{77 cm}}^{2}-{\text{3 cm}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{536 cm}}^{2}\end{array}$

Q.22

$\begin{array}{l}\mathrm{A}\mathrm{circle}\mathrm{of}\mathrm{radius}2\mathrm{cm}\mathrm{is}\mathrm{cut}\mathrm{out}\mathrm{from}\mathrm{a}\mathrm{square}\mathrm{piece}\mathrm{of}\mathrm{an}\\ \mathrm{aluminium}\mathrm{sheet}\mathrm{of}\mathrm{side}6\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{left}\\ \mathrm{over}\mathrm{aluminium}\mathrm{sheet}?\left(\mathrm{Take}\mathrm{\pi }=3.14\right)\end{array}$

Ans

$\begin{array}{l}\text{Area of square-shaped sheet}={\left(\text{side}\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(6{\text{cm}}^{2}\right)}^{2}=36\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of circle}=\text{3}\text{.14}×2×2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12.56{\text{cm}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Remaining area}={\text{36 cm}}^{\text{2}}-\text{12}{\text{.56 cm}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{23}{\text{.44 cm}}^{\text{2}}\end{array}$

Q.23

The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

$\begin{array}{l}\text{Circumference}=\text{2}\pi r=31.4\text{\hspace{0.17em}}\text{cm}\\ \text{}\text{2}×\text{3}\text{.14}×\text{r}=\text{31}\text{.4}\\ \text{}\text{}\text{}\text{r}=\frac{31.4}{2×3.14}\\ \text{}\text{}\text{}=5\text{cm}\\ \text{}\text{\hspace{0.17em}}\text{So, Area}=\text{3}\text{.14}×\text{5}×\text{5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}=\text{78}{\text{.50 cm}}^{2}\end{array}$

Q.24 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14) Ans

$\begin{array}{l}\text{Radius of the flower bed}=\frac{66}{2}=33\text{\hspace{0.17em}}\text{m}\\ \text{So, radius of flower bed and path together}=\text{33+4}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{37m}\\ \text{Area of flower baed and path together}=\text{3}\text{.14}×37×37\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4298.66{\text{m}}^{2}\\ \text{Area of flower bed}=\text{3}\text{.14}×\text{33}×\text{33}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3419}{\text{.46 m}}^{2}\\ \text{Area of path}=\text{4298}{\text{.66 m}}^{\text{2}}-\text{3419}{\text{.46m}}^{\text{2}}\\ \text{}\text{}\text{}=\text{879}{\text{.20m}}^{2}\end{array}$

Q.25

$\begin{array}{l}\mathrm{A}\mathrm{circular}\mathrm{flower}\mathrm{garden}\mathrm{has}\mathrm{an}\mathrm{area}\mathrm{of}314{\mathrm{m}}^{\mathrm{2}}.\mathrm{A}\mathrm{sprinkler}\\ \mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{garden}\mathrm{can}\mathrm{cover}\mathrm{an}\mathrm{area}\mathrm{that}\mathrm{has}\mathrm{a}\\ \mathrm{radius}\mathrm{of}12\mathrm{m}.\mathrm{Will}\mathrm{the}\mathrm{sprinkler}\mathrm{water}\mathrm{the}\mathrm{entire}\mathrm{garden}?\\ \left(\mathrm{Take\pi }=3.14\right)\end{array}$

Ans

$\begin{array}{l}\text{Area}=\pi {\text{r}}^{2}=314{\text{m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}3.14×{r}^{2}=314\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=10\text{m}\\ \text{Yes, the sprinkle will water the whole garden}\text{.}\end{array}$

Q.26 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14) Ans

$\begin{array}{l}\text{Radius of outer circle}=\text{19 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\text{3}\text{.14}×\text{19}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{119}\text{.32 m}\\ \text{Radius of the inner circle}=\text{19}-\text{10}=\text{9 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\text{3}\text{.14}×\text{9}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{56}\text{.52 m}\end{array}$

Q.27 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

$\frac{22}{7}$

)

Ans

$\text{Radius, r}=\text{28 cm}$ $\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\frac{22}{7}×28\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=176\text{cm}\\ \text{Number of rotatinos}=\frac{\text{Total distance to be covered}}{\text{Circumference of the wheel}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{352\text{m}}{176\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{35200}{176}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=200\\ \text{Therefore, it will rotate 200 times}\text{.}\end{array}$

Q.28 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

$\begin{array}{l}\text{Since, Distance travelled by the tip of minute hand}\\ \text{}=\text{Circumference of the clock}\\ \text{}=\text{2}\pi \text{r}\\ \text{}=\text{2}×\text{3}\text{.14}×1\text{5}\\ \text{}=\text{94}\text{.2 cm}\\ \text{Therfore, minute hand move 94}\text{.2 cm in 1 hour}\text{.}\end{array}$

Q.29 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare. Ans

$\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of garden}=\text{90 m}\\ \text{Breadth (B) of garden}=\text{75 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of garden}=\text{L}×\text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{90 m}×\text{75 m}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{6750 m}}^{2}\end{array}$ $\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the garden,when path is also included are}\\ \text{100 m and 85 m respectively}\\ \text{Area of the garden including path}=\text{100 m}×\text{80 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{8000 m}}^{2}\\ \text{Area of the path}\\ =\text{Area of the garden including path}-\text{Area of garden}\\ =8000{\text{m}}^{2}-{\text{6750 m}}^{2}\\ =1750{\text{m}}^{2}\\ 1\text{hectare}={\text{10000 m}}^{2}\\ \text{Therefore,}\\ \text{area of the garden in hectare}=\frac{6750}{10000}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.675\text{hectare}\end{array}$

Q.30 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans $\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of park}=\text{125 m}\\ \text{Breadth (B) of park}=\text{65 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of park}=\text{L}×\text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{125 m}×\text{65 m}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{8125 m}}^{2}\end{array}$ $\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the park,when path is also included are}\\ \text{131 m and 71 m respectively}\\ \text{Area of the park including path}=\text{131 m}×\text{71 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}={\text{9301 m}}^{2}\\ \text{Area of the path}\\ =\text{Area of the park including path}-\text{Area of park}\\ =9301{\text{m}}^{2}-{\text{8125 m}}^{2}\\ =1176{\text{m}}^{2}\end{array}$

Q.31 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans $\begin{array}{l}\text{\hspace{0.17em}}\text{Length (L) of cardboard}=\text{125 m}\\ \text{Breadth (B) of cardboard}=\text{65 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of cardboard including margin}=\text{L}×\text{B}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8 cm}×\text{5 cm}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=40{\text{cm}}^{2}\end{array}$ $\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the cardboard,when margin is not included}\\ \text{are 5 cm and 2 cm respectively}\\ \text{Area of the cardboard not including margin}=\text{5 cm}×\text{2 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{10 cm}}^{2}\\ \text{Area of the margin}\\ =\text{Area of the cardboard including margin}\\ -\text{Area of cardboard not including margin}\\ =40{\text{cm}}^{2}-{\text{10 cm}}^{2}\\ =30{\text{cm}}^{2}\end{array}$

Q.32

$\begin{array}{l}\mathrm{Two}\mathrm{cross}\mathrm{roads},\mathrm{each}\mathrm{of}\mathrm{width}10\mathrm{m},\mathrm{cut}\mathrm{at}\mathrm{right}\mathrm{angles}\\ \mathrm{through}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{park}\mathrm{of}\mathrm{length}700\mathrm{m}\\ \mathrm{and}\mathrm{breadth}300\mathrm{m}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{its}\mathrm{sides}.\mathrm{Find}\mathrm{the}\mathrm{area}\\ \mathrm{of}\mathrm{the}\mathrm{roads}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{park}\mathrm{excluding}\mathrm{cross}\\ \mathrm{roads}.\mathrm{Give}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{hectares}.\end{array}$

Ans $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length (L) of the park}=\text{700 m}\\ \text{Breadth (B) of the park}=\text{300 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of park}=\text{700}×\text{300}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}={\text{210000 m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road PQRS}=\text{700 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road ABCD}=\text{300 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Width of each road}=\text{10 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(700×10\right)+\left(300×10\right)-\left(10×10\right)\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7000+3000-100\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=9900{\text{m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.99\text{hectare}\end{array}$ $\begin{array}{l}\\ \text{Area of park excluding roads}=\text{210000-9900}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{200100 m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20.01\text{hectare}\end{array}$

Q.33 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14) Ans

$\begin{array}{l}\text{Perimeter of the square}=\text{4}×\text{side of the square}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{4}×\text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{16 cm}\\ \text{Perimeter of circular pip}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3}×\text{3}\text{.14}×\text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{25}\text{.12 cm}\\ \text{Therefore,}\\ \text{length of chord left with Pragya}=\text{25}\text{.12 cm}-\text{16 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}=\text{9}\text{.12 cm}\end{array}$

Q.34 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.

(iv) the circumference of the flower bed.

Ans $\begin{array}{l}\text{(i) Area of whole land}=\text{Length}×\text{Breadth}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{10}×\text{5}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{50 m}}^{2}\\ \text{(ii) Area of flower bed}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3.14×2×2\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12.56{\text{m}}^{2}\\ \text{(iii) Area of the lawn excluding the flower bed}\\ =\text{Area of whole land}-\text{Area of flower bed}\\ =\text{50-12}\text{.56}\\ =\text{37}{\text{.44 m}}^{2}\end{array}$

Q.35 In the following figures, find the area of the shaded portions: Ans

$\begin{array}{l}\text{(i)}\\ \text{Area of EFDC}=\text{area(ABCD)}-\text{area(BCE)}–\text{area(AFE)}\\ \text{}\text{}\text{}=\left(18×10\right)-\frac{1}{2}\left(10×8\right)-\frac{1}{2}\left(6×10\right)\\ \text{}\text{}\text{}=180-40-30\\ \text{}\text{}\text{}=110{\text{cm}}^{2}\\ \text{(ii)}\\ \text{area(QTU)}=\text{area(PQRS)}-\text{area(TSU)}-\text{area(RUQ)}-\text{area(PQT)}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(20×20\right)-\frac{1}{2}\left(10×10\right)-\frac{1}{2}\left(20×10\right)-\frac{1}{2}\left(20×10\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=400-50-100-100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=150{\text{cm}}^{2}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{ABCD}.\\ \mathrm{Here},\mathrm{AC}=22\mathrm{cm},\mathrm{BM}=3\mathrm{cm},\mathrm{DN}=3\mathrm{cm},\\ \mathrm{and}\mathrm{BM}\perp \mathrm{AC},\mathrm{DN}\perp \mathrm{AC}.\end{array}$ Ans

$\begin{array}{l}\text{area(ABCD)}=\text{area(ABC)+area(ADC)}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(3×22\right)\frac{1}{2}\left(3×22\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=33+33\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=66{\text{cm}}^{2}\end{array}$

Q.37

$\begin{array}{l}\mathrm{The}\mathrm{length}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{piece}\mathrm{of}\mathrm{land}\\ \mathrm{are}500\mathrm{m}\mathrm{and}300\mathrm{m}\mathrm{respectively}.\mathrm{Find}\\ \left(\mathrm{i}\right)\mathrm{its}\mathrm{area}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{land},\mathrm{if}1{\mathrm{m}}^{\mathrm{2}}\mathrm{of}\mathrm{the}\mathrm{land}\mathrm{costs}₹10,000.\end{array}$

Ans

$\begin{array}{l}\text{(i)}\\ \text{Area}=\text{Length}×\text{breadth}\\ \text{}\text{\hspace{0.17em}}=\text{500 m}×\text{3000}×\text{m}\\ \text{}\text{\hspace{0.17em}}={\text{150000m}}^{2}\\ \text{(ii)}\\ {\text{1 m}}^{2}\text{land cost}=\text{₹ 10,000}\\ {\text{So, cost of 150000 m}}^{2}\text{land}=\text{₹ 10,000}×\text{150000}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹ 1,500,000,000}\end{array}$

Q.38

$\begin{array}{l}\mathrm{A}\mathrm{door}\mathrm{of}\mathrm{length}2\mathrm{m}\mathrm{and}\mathrm{breadth}1\mathrm{m}\mathrm{is}\mathrm{fitted}\mathrm{in}\mathrm{a}\mathrm{wall}.\\ \mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wall}\mathrm{is}4.5\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{breadth}\mathrm{is}3.6\mathrm{m}.\\ \mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\\ \mathrm{white}\mathrm{washing}\mathrm{the}\mathrm{wall}\mathrm{is}₹20\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$ Ans

$\begin{array}{l}\text{\hspace{0.17em}}\text{Area of the wall}=\text{4}\text{.5 cm}×\text{3}\text{.6 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{16}{\text{.2 m}}^{\text{2}}\\ \text{Area of the door}=\text{2}×\text{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{2m}}^{\text{2}}\\ \text{Area to be white-washed}=\text{16}{\text{.2 m}}^{2}-{\text{2 m}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{14}{\text{.2 m}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost of white-washing 1m}}^{\text{2}}\text{area}=\text{₹20}\\ \text{So, cost of white-washing 14}\text{.2m2 area}=\text{14}\text{.2}×\text{20}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹ 284}\end{array}$

Q.39

$\begin{array}{l}\mathrm{A}\mathrm{gardener}\mathrm{wants}\mathrm{to}\mathrm{fence}\mathrm{a}\mathrm{circular}\mathrm{garden}\mathrm{of}\mathrm{diameter}\\ 21\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rope}\mathrm{he}\mathrm{needs}\mathrm{to}\mathrm{purchase},\mathrm{if}\mathrm{he}\\ \mathrm{makes}2\mathrm{rounds}\mathrm{of}\mathrm{fence}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{costs}\mathrm{of}\mathrm{the}\mathrm{rope},\mathrm{if}\mathrm{it}\\ \mathrm{cost}₹ 4\mathrm{per}\mathrm{meter}.\left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$ Ans

$\begin{array}{l}\text{Given,}d=\text{21 m}\\ \text{So,}r=\frac{21}{2}\text{\hspace{0.17em}}\text{m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\frac{22}{7}×\frac{21}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=66\text{\hspace{0.17em}}\text{m}\\ \text{Length of the rope required for fencing}=\text{2}×\text{66 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{132 m}\\ \text{}\text{\hspace{0.17em}}\text{Cost of 1 m rope}=\text{₹4}\\ \text{So, cost of 132 m rope}=\text{₹ 4}×\text{132}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{₹528}\end{array}$

Q.40

$\begin{array}{l}\mathrm{Saima}\mathrm{wants}\mathrm{to}\mathrm{put}\mathrm{a}\mathrm{lace}\mathrm{on}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{a}\mathrm{circular}\mathrm{table}\\ \mathrm{cover}\mathrm{of}\mathrm{diameter}1.5\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{required}\mathrm{and}\mathrm{also}\mathrm{find}\mathrm{its}\mathrm{cost}\mathrm{if}\mathrm{one}\mathrm{meter}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{costs}₹ 15.\left(\mathrm{Take\pi } = 3.14\right)\end{array}$

Ans

$\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}×\text{3}\text{.14}×\frac{d}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×3.14\frac{1.5}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.71\text{\hspace{0.17em}}\text{m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Cost of 1 m of lace}=\text{₹ 15}\\ \text{Therefore cost of 4}\text{.71 m of lace}=\text{₹15}×\text{4}\text{.71}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹70}\text{.65}\end{array}$

Q.41

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{polishing}\mathrm{a}\mathrm{circular}\mathrm{table}–\mathrm{top}\mathrm{of}\mathrm{diameter}\\ 1.6\mathrm{m},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{polishing}\mathrm{is}\mathrm{Rs}15/{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\left(\mathrm{Take\pi }\mathrm{= 3.14}\right)\end{array}$

Ans

$\begin{array}{l}\text{Diameter}=\text{1}\text{.6 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Radius}=\frac{1.6}{2}=0.8\text{m}\\ \text{Area}=\text{3}\text{.14}×\text{0}\text{.8}×\text{0}\text{.8}\\ \text{}\text{\hspace{0.17em}}=\text{2}{\text{.0096 m}}^{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Cost for polishing 1 m2 area}=\text{₹15}\\ \text{Cost for polishing 2}\text{.0096 m2 area}=\text{₹15}×\text{2}\text{.0096}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹30}\text{.14}\\ \text{So, it will cost ₹30}\text{.14 for polishing such circular table}\text{.}\end{array}$

Q.42

$\begin{array}{l}\mathrm{A}\mathrm{verandah}\mathrm{of}\mathrm{width}2.25\mathrm{m}\mathrm{is}\mathrm{constructed}\mathrm{all}\mathrm{along}\\ \mathrm{outside}\mathrm{a}\mathrm{room}\mathrm{which}\mathrm{is}5.5\mathrm{mlong}\mathrm{and}4\mathrm{m}\mathrm{wide}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{verandah}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{cementing}\mathrm{the}\mathrm{floor}\mathrm{of}\mathrm{the}\mathrm{verandah}\mathrm{at}\mathrm{the}\\ \mathrm{rate}\mathrm{of}\mathrm{Rs}200\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\end{array}$ Ans

$\begin{array}{l}\text{(i)}\\ \text{Length (L) of room}=\text{5}\text{.5 m}\\ \text{Breadth (B) of room}=\text{4 m}\\ \text{Area of room including margin}=\text{L}×\text{B}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{5}\text{.5 m}×\text{4 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=22{\text{m}}^{2}\end{array}$ $\begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the room,when verandah is also included}\\ \text{are 10 m and 8}\text{.5 m respectively}\\ \text{Area of the verandah including margin}=\text{10 m}×\text{8}\text{.5 m}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{85 m}}^{2}\\ \text{Area of the verandah}\\ =\text{Area of the room including verandah}-\text{Area of the room}\\ =85{\text{m}}^{2}-{\text{22 m}}^{2}\\ =63{\text{m}}^{2}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{(ii)}\\ {\text{Cost of cementing 1m}}^{\text{2}}\text{area of floor of verandah}=₹200\\ {\text{Cost of cementing 63m}}^{\text{2}}\text{area of floor of verandah}=₹200×63\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=₹12600\end{array}$

Q.43

$\begin{array}{l}\mathrm{A}\mathrm{path}1\mathrm{m}\mathrm{wide}\mathrm{is}\mathrm{built}\mathrm{along}\mathrm{the}\mathrm{border}\mathrm{and}\mathrm{inside}\mathrm{a}\\ \mathrm{square}\mathrm{garden}\mathrm{of}\mathrm{side}30\mathrm{m}.\mathrm{Find}:\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{path}\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{planting}\mathrm{grass}\mathrm{in}\mathrm{the}\mathrm{remaining}\mathrm{portion}\mathrm{of}\mathrm{the}\\ \mathrm{garden}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}₹40\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

Ans $\begin{array}{l}\text{(i)}\\ \text{Side (a) of square garden}=\text{30 m}\\ \text{Area of square garden including path}={\text{a}}^{2}\text{}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(\text{30 m}\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{900 m}}^{2}\\ \text{From the figure, it can be observed that the side of the square}\\ \text{garden, when path is not included, is 28 m}\text{.}\\ \text{Area of the square garden not including the path}={\left(28\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}={\text{784 m}}^{2}\\ \text{Area of path}\\ =\text{Area of the square garden including the path}\\ -\text{Area of square garden not including the path}\\ =900{\text{m}}^{2}-784{\text{m}}^{2}\\ =116{\text{m}}^{2}\\ \text{(ii)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost of planting grass in 1 m}}^{2}\text{area of the garden}=\text{₹40}\\ {\text{Cost of planting grass in 784 m}}^{2}\text{area of the garden}=\text{₹40}×\text{784}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹31360}\end{array}$

Q.44

$\begin{array}{l}\mathrm{Through}\mathrm{a}\mathrm{rectangular}\mathrm{field}\mathrm{of}\mathrm{length}90\mathrm{m}\mathrm{and}\mathrm{breadth}\\ 60\mathrm{m},\mathrm{two}\mathrm{roads}\mathrm{are}\mathrm{constructed}\mathrm{which}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\\ \mathrm{sides}\mathrm{and}\mathrm{cut}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{right}\mathrm{angles}\mathrm{through}\mathrm{the}\mathrm{centre}\\ \mathrm{of}\mathrm{the}\mathrm{fields}.\mathrm{If}\mathrm{the}\mathrm{width}\mathrm{of}\mathrm{each}\mathrm{road}\mathrm{is}3\mathrm{m},\mathrm{find}\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{area}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{roads}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{constructing}\mathrm{the}\mathrm{roads}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\\ ₹110\mathrm{per}{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}$

Ans $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length (L) of the field}=\text{90 m}\\ \text{Breadth (B) of the field}=\text{60 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of field}=\text{90}×\text{60}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}={\text{5400 m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road PQRS}=\text{90 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Length of road ABCD}=\text{60 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Width of each road}=\text{3 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(90×3\right)+\left(60×3\right)-\left(3×3\right)\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=270+180-100\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=441{\text{m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{Cost for constructing 1 m}}^{2}\text{road}=\text{₹110}\\ {\text{Cost for constructing 4411 m}}^{2}\text{road}=\text{₹110}×441\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹48510\end{array}$

## 1. Where will I find the class 7 Mathematics chapter-wise solutions?

You will find the class 7 Mathematics chapter-wise solutions on the Extramarks website.

## 2. Why choose Extramarks NCERT solutions?

The primary reasons to choose Extramarks NCERT solutions are that they are:

•  drafted by experienced faculties
• 100% accurate
•  made as per CBSE guidelines
•  self-explanatory

## 3. What is chapter 11 of class 7 about?

Chapter 11 of Class 7 is about Area and Perimeter. Perimeter is the distance around a close figure whereas area is a metric to quantify the region occupied by a figure. Students will learn how to calculate the perimeters and areas of many different shapes including triangles, circles, squares, and parallelograms in this chapter.

## 4. What are the area and perimeter?

Perimeter is the distance that surrounds a specific area whereas, an area is the part of a plane that is encircled by the perimeter. An area is a particular shape that includes a square, rectangle, triangle, etc.

## 5. What is the formula for the area of a circle?

The formula of the area of a circle is πr2.