NCERT Solutions Class 7 Maths Chapter 9 - Perimeter and Area

NCERT Solutions Class 7 Maths Chapter 9 Perimeter and Area consists of all the questions in the NCERT Textbook, along with their step-by-step solutions. Practising these questions to improve your problem-solving skills and mathematical reasoning would be best. Perimeter and Area is an important chapter of class 7 Maths. 

In this chapter, students will learn two basic concepts of geometry that will help them understand the dimensions of different shapes. You need to master these concepts not only to clear the exam with good marks but also for practical application in daily life like landscaping, design, etc., and that can be possible only by practising NCERT questions. 

You can easily access the downloadable PDF of the NCERT Solution of Class 7th Maths Chapter Perimeter And Area. Along with PDF links, other details like number questions, subtopics, and how NCERT solutions help you prepare for the class 7 maths test. 

1.0Download Class 7 Maths Chapter 9 NCERT Solutions PDF Online

Here, we are providing the NCERT solution class 7 maths PDF free. These solutions are prepared based on the latest syllabus and cover concepts like calculating the area and perimeter of rectangles, triangles, squares, and circles. By downloading this PDF, you can access in-depth explanations and practice questions that reinforce your learning and build confidence in your mathematical skills.

2.0NCERT Solutions for Class 7 Chapter Perimeter and Area: Overview

In the chapter NCERT solutions class 7 maths chapter 9, students will learn about the concepts of perimeter and area for various shapes such as rectangles, triangles, squares, circles, and parallelograms.

The NCERT solutions contain a range of questions, including multiple-choice, short-answer, and long-answer questions, to improve students' comprehension of the fundamentals of perimeter and area and their applications. All the solutions provided are to the point with step-by-step explanations so you can easily understand them and improve your problem-solving skills within a certain amount of time. So let's explore the NCERT solutions class 7 maths chapter 9 and its more details. 

3.0NCERT Solutions Class 7 Maths Perimeter and Area Subtopics

The NCERT Solutions for Class 7 Maths Chapter on Perimeter and Area covers key subtopics that enhance students' understanding of geometry. This chapter emphasizes the concepts of perimeter and area, offering clear explanations and practical applications for different geometric shapes. Explore the various subtopics included in Chapter 9 of Class 7 Maths to strengthen your grasp of these essential concepts.

• Area of a Parallelogram
• Area of a  Triangle
• Circles

4.0Maths Class 7 Chapter Perimeter and Area: Exercises of NCERT Solutions 

The NCERT solutions for Class 7 Maths Chapter 9 include numerous exercises that enhance students' geometric knowledge. Each exercise is designed to challenge students' understanding and allow them to apply the formulas they've learned to real-life situations, enabling them to calculate perimeter and area in various contexts. Here are the details about the exercises and the number of questions in each exercise. 

5.0How to Prepare with NCERT Solutions for Class 7 Chapter Perimeter and Area

• Preparing with the NCERT Solutions for Class 7 Chapter on Perimeter and Area can significantly enhance your understanding and performance in mathematics. 
• It familiarises you with the key concepts, such as the definitions of perimeter and area, and the formulas for calculating them for various shapes, including squares, rectangles, and triangles.
• As you work through the exercises, focus on understanding the step-by-step solutions provided. This will help you grasp the underlying principles and improve your problem-solving skills. 
• To solidify your knowledge, practice the questions regularly by attempting different types of problems, including those involving composite shapes.
• Use the examples of solutions to see how these concepts apply in the real world. Completing the exercises helps you prepare for the tests and lays a solid foundation for geometry's more difficult topics.

6.0NCERT Questions with Solutions for Class 7 Maths Chapter 9 - Detailed Solutions

Exercise : 9.1

• Find the area of each of the following parallelograms :

• (a)

• (b)

• (c)

• (d)

• (e) Sol. We know that the area of parallelogram $=$ base $\times$ height (a) Here base $=7\mathrm{cm}$ and height $=4\mathrm{cm}$ $\therefore$ Area of parallelogram $=7\times 4$ $=28{\mathrm{cm}}^2$ (b) Here base $=5\mathrm{cm}$ and height $=3\mathrm{cm}$ $\therefore$ Area of parallelogram $=5\times 3$ $=15{\mathrm{cm}}^2$ (c) Here base $=2.5\mathrm{cm}$ and height $=3.5\mathrm{cm}$ $\therefore$ Area of parallelogram $=2.5\times 3.5$ $=8.75{\mathrm{cm}}^2$ (d) Here base $=5\mathrm{cm}$ and height $=4.8\mathrm{cm}$ $\therefore$ Area of parallelogram $=5\times 4.8=24{\mathrm{cm}}^2$ (e) Here base $=2\mathrm{cm}$ and height $=4.4\mathrm{cm}$ $\therefore$ Area of parallelogram $=2\times 4.4=8.8{\mathrm{cm}}^2$
• Find the area of each of the following triangles:

• (a)
  
  (b)
  
  (c)
  
  (d) Sol. We know that the area of triangle $=\frac{1}{2}\times$ base $\times$ height (a) Here, base $=4\mathrm{cm}$ and height $=3\mathrm{cm}$ $\therefore$ Area of triangle $=\frac{1}{2}\times 4\times 3=6{\mathrm{cm}}^2$ (b) Here, base $=5\mathrm{cm}$ and height $=3.2\mathrm{cm}$ $\therefore$ Area of triangle $=\frac{1}{2}\times 5\times 3.2=8$ ${\mathrm{cm}}^2$ (c) Here, base $=3\mathrm{cm}$ and height $=4\mathrm{cm}$ $\therefore$ Area of triangle $=\frac{1}{2}\times 3\times 4=6{\mathrm{cm}}^2$ (d) Here, base $=3\mathrm{cm}$ and height $=2\mathrm{cm}$ $\therefore$ Area of triangle $=\frac{1}{2}\times 3\times 2=3{\mathrm{cm}}^2$
• Find the missing values:

Sol. We know that the area of parallelogram $=$ base $\times$ height (a) Here, base $=20\mathrm{cm}$ and area $=246{\mathrm{cm}}^2$ $\therefore$ Area of parallelogram $=$ base $\times$ height $\Rightarrow 246=20\times$ height $\Rightarrow$ height $=\frac{246}{20}=12.3\mathrm{cm}$ (b) Here, height $=15\mathrm{cm}$ and area $=154.5{\mathrm{cm}}^2$ $\therefore$ Area of parallelogram $=$ base $\times$ height $\Rightarrow 154.5=$ base $\times 15$ $\Rightarrow$ base $=\frac{154.5}{15}=10.3\mathrm{cm}$ (c) Here, height $=8.4\mathrm{cm}$ and area $=48.72{\mathrm{cm}}^2$ $\therefore$ Area of parallelogram $=$ base $\times$ height $\Rightarrow 48.72=$ base $\times 8.4$ $\Rightarrow$ base $=\frac{48.72}{8.4}=5.8\mathrm{cm}$ (d) Here, base $=15.6\mathrm{cm}$ and area $=16.38{\mathrm{cm}}^2$ $\therefore$ Area of parallelogram $=$ base $\times$ height $\Rightarrow 16.38=15.6\times$ height $\Rightarrow$ height $=\frac{16.38}{15.6}=1.05\mathrm{cm}$ Thus, the missing values are :

• Find the missing values :

Sol. We know that the area of triangle $=\frac{1}{2}\times$ base $\times$ height In first row, base $=15\mathrm{cm}$ and area $=87{\mathrm{cm}}^2$ $\therefore 87=\frac{1}{2}\times$ height $\times 15$ $\Rightarrow$ height $=\frac{87\times 2}{15}=11.6\mathrm{cm}$ In second row, height $=31.4\mathrm{mm}$ and area $=1256{\mathrm{mm}}^2$ $\therefore 1256=\frac{1}{2}\times 31.4$ base $\Rightarrow$ height $=\frac{1256\times 2}{31.4}=80\mathrm{mm}$ In third row, base $=22\mathrm{cm}$ and area $=170.5{\mathrm{cm}}^2$ $\therefore 170.5=\frac{1}{2}\times 22\times$ height $\Rightarrow$ height $=\frac{170.5\times 2}{22}=15.5\mathrm{cm}$ Thus, the missing values are:

• PQRS is a parallelogram QM is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$. If $SR=12\mathrm{cm}$ and $QM=7.6\mathrm{cm}$. Find: (a) The area of the parallelogram PQRS (b) $QN$, if $PS=8\mathrm{cm}$
  
  Sol. $\mathrm{SR}=12\mathrm{cm},\mathrm{QM}=7.6\mathrm{cm},\mathrm{PS}=8\mathrm{cm}$. (a) Area of parallelogram $=$ base $\times$ height $=12\times 7.6=91.2{\mathrm{cm}}^2$ (b) Area of a parallelogram $$\begin{gathered} =\text{ base }\times \text{ height } \\ \Rightarrow 91.2=8\times \mathrm{QN} \\ \Rightarrow \mathrm{QN}=\frac{91.2}{8}=11.4\mathrm{cm} \end{gathered}$$
• $DL$ and $BM$ are the heights on sides $AB$ and AD respectively of parallelogram ABCD . If the area of the parallelogram is $1470{\mathrm{cm}}^2,\mathrm{AB}=35\mathrm{cm}$ and $\mathrm{AD}=49\mathrm{cm}$, find the length of BM and DL .
  
  Sol. Area of parallelogram $=1470{\mathrm{cm}}^2$ Base (AB) $=35\mathrm{cm}$ and base (AD) $=49\mathrm{cm}$ Since Area of parallelogram $=$ base $\times$ height $\Rightarrow 1470=35\times$ DL $\Rightarrow \mathrm{DL}=\frac{1470}{35}$ $\Rightarrow$ DL $=42\mathrm{cm}$ Again, Area of parallelogram $=$ base $\times$ height $\Rightarrow 1470=49\times \mathrm{BM}$ $\Rightarrow \mathrm{BM}=\frac{1470}{49}$ $\Rightarrow \mathrm{BM}=30\mathrm{cm}$ Thus, the length of DL and BM are 42 cm and 30 cm respectively.
• $△\mathrm{ABC}$ is right angled at $\mathrm{A}.\mathrm{AD}$ is perpendicular to BC . If $\mathrm{AB}=5\mathrm{cm},\mathrm{BC}=13$ cm and $AC=12\mathrm{cm}$, Find the area of $△ABC$. Also find the length of AD.
  
  Sol. In right angle triangle $\mathrm{BAC},\mathrm{AB}=5\mathrm{cm}$ and $\mathrm{AC}=12\mathrm{cm}$ Area of triangle $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times \mathrm{AB}\times \mathrm{AC}$ $=\frac{1}{2}\times 5\times 12=30{\mathrm{cm}}^2$ Now, In $△\mathrm{ABC}$, Area of triangle $\mathrm{ABC}=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}$ $\Rightarrow 30=\frac{1}{2}\times 13\times \mathrm{AD}$ $\Rightarrow \mathrm{AD}=\frac{30\times 2}{13}=\frac{60}{13}\mathrm{cm}$
• $△\mathrm{ABC}$ is isosceles with $\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}$ and $BC=9\mathrm{cm}$. The height $AD$ from $A$ to $BC$, is 6 cm . Find the area of $△ABC$. What will be the height from C to AB i.e., CE ?
  
  Sol. In $△\mathrm{ABC},\mathrm{AD}=6\mathrm{cm}$ and $\mathrm{BC}=9\mathrm{cm}$ Area of triangle $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}=\frac{1}{2}\times 9\times 6=27{\mathrm{cm}}^2$ Again, Area of triangle $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times \mathrm{AB}\times \mathrm{CE}$ $\Rightarrow 27=\frac{1}{2}\times 7.5\times \mathrm{CE}$ $\Rightarrow \mathrm{CE}=\frac{27\times 2}{7.5}$ $\Rightarrow \mathrm{CE}=7.2\mathrm{cm}$ Thus, height from C to AB i.e., CE is 7.2 cm .

Exercise : 9.2

• Find the circumference of the circle with the following radius : (Take $\pi =\frac{22}{7}$ ) (a) 14 cm (b) 28 mm (c) 21 cm Sol. (a) Circumference of the circle $=2\pi r$ $=2\times \frac{22}{7}\times 14=88\mathrm{cm}$ (b) Circumference of the circle $=2\pi \mathrm{r}$ $=2\times \frac{22}{7}\times 28=176\mathrm{mm}$ (c) Circumference of the circle $=2\pi \mathrm{r}$ $=2\times \frac{22}{7}\times 21=132\mathrm{cm}$
• Find the area of the following circles, given that : (a) radius $=14\mathrm{mm}($ Take $\pi =\frac{22}{7})$ (b) diameter $=49\mathrm{m}$ (c) radius $=5\mathrm{cm}$ Sol. (a) Area of circle $$\begin{gathered} =\pi r^2=\frac{22}{7}\times 14\times 14 \\ =22\times 2\times 14=616{\mathrm{mm}}^2 \end{gathered}$$ (b) Diameter $=49\mathrm{m}$ $\therefore$ radius $=\frac{49}{2}=24.5\mathrm{m}$ $\therefore$ Area of circle $=\pi r^2$ $=\frac{22}{7}\times 24.5\times 24.5$ $=22\times 3.5\times 24.5$ $=1886.5{\mathrm{m}}^2$ (c) Area of circle $=\pi r^2$ $=\frac{22}{7}\times 5\times 5=\frac{550}{7}{\mathrm{cm}}^2$
• If the circumference of a circular sheet is 154 m , find its radius. Also find the area of the sheet. (Take $\pi =\frac{22}{7}$ ) Sol. Circumference of the circular sheet $=154$ m $\Rightarrow 2\pi r=154\Rightarrow r=\frac{154}{2\pi }$ $\Rightarrow \mathrm{r}=\frac{154\times 7}{2\times 22}=24.5\mathrm{m}$ Now Area of circular sheet $=\pi r^2=\frac{22}{7}\times 24.5\times 24.5$ $=22\times 3.5\times 24.5=1886.5{\mathrm{m}}^2$ Thus, the radius and area of circular sheet are 24.5 m and $1886.5{\mathrm{m}}^2$ respectively.
• A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also the cost of the rope, if it costs Rs 4 per meter. (Take $\pi =\frac{22}{7}$ )
  
  Sol. Diameter of the circular garden $=21\mathrm{m}$ $\therefore$ Radius of the circular garden $=\frac{21}{2}\mathrm{m}$ Now Circumference of circular garden $=2\pi \mathrm{r}=2\times \frac{22}{7}\times \frac{21}{2}$ $=22\times 3=66\mathrm{m}$ The gardener makes 2 rounds of fence so that the total length of the rope of fencing $=2\times 2\pi r=2\times 66=132\mathrm{m}$ Since the cost of 1 meter rope $=₹4$ Therefore, cost of 132 meters rope $=4\times 132=$ Rs. 528
• From a circular sheet of radius 4 cm , a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take $\pi =3.14$ ) Sol. Radius of circular sheet (R) $=4\mathrm{cm}$ and radius of removed circle ( r ) $=3\mathrm{cm}$ Area of remaining sheet = Area of circular sheet - Area of removed circle $=\pi {\mathrm{R}}^2-\pi {\mathrm{r}}^2=\pi \left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$ $=\pi \left(4^2-3^2\right)=\pi (16-9)$ $=3.14\times 7=21.98{\mathrm{cm}}^2$ Thus, the area of remaining sheet is $21.98{\mathrm{cm}}^2$.
• Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m . Find the length of the lace required and also find its cost if one meter of the lace costs Rs 15. (Take $\pi =3.14$ ) Sol. Diameter of the circular table cover $=1.5\mathrm{m}$ $\therefore$ Radius of the circular table cover $=\frac{1.5}{2}\mathrm{m}$ Circumference of circular table cover $=2\pi \mathrm{r}$ $=2\times 3.14\times \frac{1.5}{2}=4.71\mathrm{m}$ Therefore, the length of required lace is 4.71 m . Now the cost of 1 m lace $=₹15$ Then the cost of 4.71 m lace $=15\times 4.71$ = ₹ 70.65 Hence, the cost of 4.71 m lace is ₹ 70.65 .
• Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
  
  Sol. Diameter $=10\mathrm{cm}$ $\therefore$ Radius $=\frac{10}{2}=5\mathrm{cm}$ According to question, Perimeter of figure $=$ Circumference of semi-circle + diameter $=\pi r+D$ $=\frac{22}{7}\times 5+10=\frac{110}{7}+10$ $=\frac{110+70}{7}=\frac{180}{7}=25.71\mathrm{cm}$ Thus, the perimeter of the given figure is 25.71 cm .
• Find the cost of polishing a circular tabletop of diameter 1.6 m , if the rate of polishing is Rs $15/{\mathrm{m}}^2$. (Take $\pi =3.14$ ) Sol. Diameter of the circular table top $=1.6\mathrm{m}$ $\therefore$ Radius of the circular table top $=\frac{1.6}{2}=0.8\mathrm{m}$ Area of circular table top $=\pi r^2$ $=3.14\times 0.8\times 0.8=2.0096{\mathrm{m}}^2$ Now cost of $1{\mathrm{m}}^2$ polishing $=$ Rs. 15 Then cost of $2.0096{\mathrm{m}}^2$ polishing $=15\times 2.0096=$ Rs. 30.14 (approx.) Thus, the cost polishing a circular table top is Rs. 30.14 (approx.)
• Shazli took a wire of length 44 cm and bent it into the shape of circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its side? Which figure encloses more area, the circle or the square? (Take $\pi =22/7$ ) Sol. Total length of the wire $=44\mathrm{cm}$ $\therefore$ The circumference of the circle $=2\pi \mathrm{r}$ $=44\mathrm{cm}$ $\Rightarrow 2\times \frac{22}{7}\times \mathrm{r}=44\Rightarrow \mathrm{r}=\frac{44\times 7}{2\times 22}=7\mathrm{cm}$ Now, Area of the circle $=\pi {\mathrm{r}}^2=\frac{22}{7}\times 7\times 7=154{\mathrm{cm}}^2$ Now the wire is converted int square. Then perimeter of square $=44\mathrm{cm}$ $\Rightarrow 4\times$ side $=44$ $\Rightarrow$ side $=\frac{44}{4}=11\mathrm{cm}$ Now area of square $=$ side $=11\times 11$ $=121{\mathrm{cm}}^2$ Therefore, on comparing, the area of circle is greater than that of square, so the circle enclosed more area.
• From a circular card sheet of radius 14 cm , two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi =\frac{22}{7}$ )
  
  Sol. Radius of circular sheet $(\mathrm{R})=14\mathrm{cm}$ and Radius of smaller circle ( r ) $=3.5\mathrm{cm}$ Length of rectangle ( $\ell$ ) $=3\mathrm{cm}$ and breadth of rectangle (b) $=1\mathrm{cm}$ According to question, Area of remaining sheet = Area of circular sheet - (Area of two small circles + Area of rectangle) $=\pi R^2-\left[2\left(\pi r^2\right)+(\ell \times b)\right]$ $=\frac{22}{7}\times 14\times 14-\left[\left(2\times \frac{22}{7}\times 3.5\times 3.5\right)+(3\times 1)\right]$ $=22\times 14\times 2-[44\times 0.5\times 3.5+3]$ $=616-80$ $=536{\mathrm{cm}}^2$ Therefore, the area of remaining sheet is $536{\mathrm{cm}}^2$.
• A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm . What is the area of the left-over aluminium sheet. (Take $\pi =3.14$ ) Sol. Radius of circle $=2\mathrm{cm}$ and side of aluminium square sheet $=6\mathrm{cm}$ According of aluminium sheet left = Total area of aluminium sheet - Area of circle $=$ side $\times$ side $-\pi r^2=6\times 6-3.14\times 2\times 2$ $=36-12.56=23.44{\mathrm{cm}}^2$ Therefore, the area of aluminium sheet left is $23.44{\mathrm{cm}}^2$.
• The circumference of a circle is 31.4 cm . Find the radius and the area of the circle? (Take $\pi =3.14$ ) Sol. The circumference of the circle $=31.4\mathrm{cm}$ $\Rightarrow 2\pi r=31.4\Rightarrow 2\times 3.14\times r=31.4$ $\Rightarrow \mathrm{r}=\frac{31.4}{2\times 3.14}=5\mathrm{cm}$ Then area of the circle $=\pi r^2=3.14\times 5^2$ $=78.5{\mathrm{cm}}^2$ Therefore, the radius and the area of the circle are 5 cm and $78.5{\mathrm{cm}}^2$ respectively.
• 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? ( $\pi =3.14$ )
  
  Sol. Diameter of the circular flower bed $=66\mathrm{m}$ $\therefore$ Radius of circular flower bed $(\mathrm{r})=\frac{66}{2}=33\mathrm{m}$ $\therefore$ Radius of circular flower bed with $4m$ wide path $(R)=33+4=37$ According to the question. Area of path $=$ Area of bigger circle - Area of smaller circle $=\pi {\mathrm{R}}^2-\pi {\mathrm{r}}^2=\pi \left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$
  
  $=\pi \left[(37{)}^2-(33{)}^2\right]$ $=3.14[(37+33)(37-33)]$ $\left[\because a^2-b^2=(a+b)(a-b)\right]$ $=3.14\times 70\times 4=879.20{\mathrm{m}}^2$ Therefore, the area of the path is 879.20 ${\mathrm{m}}^2$.
• A circular flower garden has an area of $314{\mathrm{m}}^2$. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m . Will the sprinkler water the entire garden? (Take $\pi =3.14$ ) Sol. Circular area by the sprinkler $=\pi r^2=3.14\times 12\times 12$ $=3.14\times 144$ $=452.16{\mathrm{m}}^2$ Area of the circular flower garden $=314{\mathrm{m}}^2$ Since Area of circular flower garden is smaller than area by sprinkler. Therefore, the sprinkler will water the entire garden.
• Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take $\pi =3.14$ )
  
  Sol. Radius of outer circle ( r ) $=19\mathrm{m}$ $\therefore$ Circumference of outer circle $=2\pi r=2\times 3.14\times 19$ $=119.32\mathrm{m}$ Now radius of inner circle (r') = 19 - 10 = 9 $\therefore$ Circumference of inner circle $=2\pi r^{\prime }=2\times 3.14\times 9=56.52\mathrm{m}$ Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 $m$ respectively.
• How many times a wheel of radius 28 cm must rotate to go 352 m ? (Take $\pi =\frac{22}{7}$ ) Sol. Let wheel must be rotate n times of its circumference. Radius of wheel $=28\mathrm{cm}$ and Total distance $=352\mathrm{m}=35200\mathrm{cm}$ $\therefore$ Distance covered by wheel $=\mathrm{n}\times$ circumference of wheel $\Rightarrow 35200=n\times 2\pi r$ $\Rightarrow 35200=\mathrm{n}\times 2\times \frac{22}{7}\times 28$ $\Rightarrow \mathrm{n}=\frac{35200\times 7}{2\times 22\times 28}$ $\Rightarrow \mathrm{n}=200$ revolutions Thus, wheel must rotate 200 times to go 352 m.
• 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 $\pi =3.14$ ) Sol. In 1 hour, minute hand completes one round means it makes a circle. Radius of the circle ( r ) $=15\mathrm{cm}$ Circumference of circular clock $=2\pi \mathrm{r}$ $=2\times 3.14\times 15=94.2\mathrm{cm}$ Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.