NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.1

In NCERT Solutions Chapter 9 of Class 8 Maths, Mensuration Exercise 9.1, students will learn how to calculate the surface area of different solid figures like cubes and cuboids. This NCERT Solutions Class 8 Maths Chapter 9 exercise helps in understanding how much space the outer surface of an object covers, which is useful in real-life situations like painting walls or wrapping gifts.

Our NCERT Solutions for Exercise 9.1 provide easy-to-follow answers along with accurate solutions based on the latest NCERT textbook. These step-by-step solutions are specially kept to let the student know exactly how to solve a particular question. 

1.0Download NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.1: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.1 PDF with simple, step-by-step answers to help you understand and solve questions easily.

2.0Key Concepts in Exercise 9.1 of Class 8 Maths Chapter 9

Exercise 9.1 of Chapter 9 – Mensuration introduces the concept of area of trapezium, general quadrilateral, and polygon, laying the foundation for understanding areas of irregular and complex shapes.

• Area of a Trapezium
  A trapezium has one pair of parallel sides. The formula for finding its area is:
  $Area=\frac{2}{1}\times \text{(Sum of parallel sides)}\times Height$
• Area of a General Quadrilateral
  For a quadrilateral divided into two triangles using a diagonal, the total area is the sum of areas of both triangles. Students use Heron’s formula or triangle area formulas to find this.
• Area of a Polygon
  Complex polygons are broken down into simpler shapes (like triangles or trapeziums), and their areas are added to get the total area.
• Use of Diagrams
  Students are encouraged to draw and label figures accurately to identify the required dimensions such as base, height, or sides.
• Application-Based Word Problems
  The exercise includes practical scenarios such as calculating land area, fields, and plots, helping students apply these formulas in real-life contexts.

3.0NCERT Class 8 Maths Chapter 9: Other Exercises

4.0NCERT Class 8 Maths Chapter 9 Exercise 9.1: Detailed Solutions

1. The shape of the top surface of a table is trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m .
   
   Sol. Area of top surface of a table = Area of the trapezium $=\frac{1}{2}\times ($ sum of parallel sides $)\times ($ distance between them) $=\left[\frac{1}{2}\times (1+1.2)\times 0.8\right]{\mathrm{m}}^2$ $=\left(\frac{1}{2}\times 2.2\times 0.8\right){\mathrm{m}}^2$ $=(1.1\times 0.8){\mathrm{m}}^2=0.88{\mathrm{m}}^2$
2. The area of a trapezium is $34{\mathrm{cm}}^2$ and the length of one of the parallel sides is 10 cm and its height is 4 cm . Find the length of other parallel side. Sol. Let the required side be xcm . Then, area of the trapezium $=\left[\frac{1}{2}\times (10+x)\times 4\right]{\mathrm{cm}}^2=2(10+\mathrm{x}){\mathrm{cm}}^2$ But, the area of the trapezium $=34{\mathrm{cm}}^2$ (given) $\therefore 2(10+\mathrm{x})=34$ $\Rightarrow 10+\mathrm{x}=17$ $\Rightarrow \mathrm{x}=17-10=7$ Hence, the other side $=7\mathrm{cm}$
3. Length of the fence of a trapezium shaped field $ABCD$ is 120 m . If $\mathrm{BC}=48\mathrm{m},\mathrm{CD}=17$ m and $\mathrm{AD}=40\mathrm{m}$, find the area of this field. Side $AB$ is perpendicular to the parallel sides AD and BC .
   
   Sol. Let ABCD be the given trapezium in which $\mathrm{BC}=48\mathrm{m},\mathrm{CD}=17\mathrm{m}$ and $\mathrm{AD}=40\mathrm{m}$
   
   Through D, draw DL $\perp$ BC. Now, BL = AD = 40 m and $\mathrm{LC}=\mathrm{BC}-\mathrm{BL}=(48-40)\mathrm{m}=8\mathrm{m}$ Applying Pythagoras theorem in right $△$ DLC, we have ${\mathrm{DL}}^2={\mathrm{DC}}^2-{\mathrm{LC}}^2=17^2-8^2=289-64=225$ $\Rightarrow \mathrm{DL}=\sqrt{225}=15\mathrm{m}$ Now, area of the trapezium ABCD $=\frac{1}{2}\times (BC+AD)\times DL$ $=\frac{1}{2}\times (48+40)\times 15{\mathrm{m}}^2$ $=(44\times 15){\mathrm{m}}^2=660{\mathrm{m}}^2$.
4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m . Find the area of the field.
   
   Sol. Let ABCD be the given quadrilateral in which $\mathrm{BE}\perp \mathrm{AC}$ and $\mathrm{DF}\perp \mathrm{AC}$. It is given that $\mathrm{AC}=24\mathrm{m},\mathrm{BE}=8\mathrm{m}$ and $\mathrm{DF}=13\mathrm{m}$.
   
   Now, area of quad. ABCD $=$ area of $△ABC+$ area of $△ACD$ $=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{DF}$ $=\left(\frac{1}{2}\times 24\times 8+\frac{1}{2}\times 24\times 13\right){\mathrm{m}}^2$ $=(12\times 8+12\times 13){\mathrm{m}}^2$ $=(96+156){\mathrm{m}}^2=252{\mathrm{m}}^2$
5. The diagonals of a rhombus are 7.5 cm and 12 cm . Find its area. Sol. Area of a rhombus $=\frac{1}{2}\times$ (product of diagonals) $=\left(\frac{1}{2}\times 7.5\times 12\right){\mathrm{cm}}^2=45{\mathrm{cm}}^2$
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm . If one of its diagonals is 8 cm long, find the length of the other diagonal. Sol. Let ABCD be a rhombus of side 5 cm and whose altitude $\mathrm{DE}=4.8\mathrm{cm}$. Also, one of its diagonals, $\mathrm{BD}=8\mathrm{cm}$. Area of the rhombus ABCD $=2\times$ Area $(△ABD)=2\times \frac{1}{2}\times AB\times DE$ $=(5\times 4.8){\mathrm{cm}}^2=24{\mathrm{cm}}^2$ Also, area of the rhombus ABCD $=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}$ $\Rightarrow \frac{1}{2}\times \mathrm{AC}\times 8=24$ $\Rightarrow \mathrm{AC}=6\mathrm{cm}$
   
   Hence, the other diagonal is 6 cm .
7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per ${\mathrm{m}}^2$ is ₹ 4 . Sol. Area of the floor $=3000\times$ Area of one tile $=3000\times \frac{1}{2}\times 45\times 30{\mathrm{cm}}^2$ $=1500\times 45\times 30{\mathrm{cm}}^2$ $=\frac{1500\times 45\times 30}{100\times 100}{\mathrm{m}}^2=202.5{\mathrm{m}}^2$ Cost of polishing the floor ₹ 4 per m ${}^2$ $=₹(4\times 202.5)=₹810$
8. Mohan wants to buy a trapezium shaped field. It's side along the river is parallel to and twice the side along the road. If the area of this field is $10500{\mathrm{m}}^2$ and the perpendicular distance between the two parallel sides is 100 m , find the length of the side along the river.
   
   Sol. Let the parallel sides of the trapezium shaped field be x m and 2 x m . Then, its area. $=\frac{1}{2}(\mathrm{x}+2\mathrm{x})\times 100{\mathrm{m}}^2$ $=50\times 3\mathrm{x}{\mathrm{m}}^2=150\mathrm{x}{\mathrm{m}}^2$ But, it is given that the area of the field is $10500{\mathrm{m}}^2$. $\therefore 150\mathrm{x}=10500$ $\Rightarrow \mathrm{x}=\frac{10500}{150}=70$ $\therefore$ The length of the side along the river is $2\times 70$, i.e., 140 metres.
9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
   
   Sol. Area of the octagonal surface ABCDEFGH = Area (trap. ABCH) + Area (rect. HCDG) + Area (trap. GDEF)
   
   $=\left[\frac{1}{2}(5+11)\times 4+11\times 5+\frac{1}{2}(11+5)\times 4\right]{\mathrm{m}}^2$ $=(32+55+32){\mathrm{m}}^2=119{\mathrm{m}}^2$
10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area? Sol. Taking Jyoti's diagram : Area of the pentagonal shaped park
    
    $=2\times$ Area of trapezium ABEF $=2\times \frac{1}{2}\times (15+30)\times \frac{15}{2}{\mathrm{m}}^2$ $=\left(45\times \frac{15}{2}\right){\mathrm{m}}^2=337.5{\mathrm{m}}^2$ Taking Kavita's diagram : Area of the pentagonal shaped park $=$ Area (square ACDF) + Area ( $△\mathrm{DEF}$ ) $=15\times 15+\frac{1}{2}\times 15\times 15$ $=225+\frac{225}{2}=225+112.5=337.5{\mathrm{m}}^2$
    

• Diagram of the adjacent picture frame has outer dimensions $24\mathrm{cm}\times 28\mathrm{cm}$ and inner dimensions $16\mathrm{cm}\times 20\mathrm{cm}$. Find the area of each section of the frame, if the width of each section is same.
  
  
  Sol. Width of each section $=4\mathrm{cm}$. Area of trapezium ABQP = Area of trapezium RCDS $=\frac{1}{2}\times 4\times (16+24)\mathrm{cm}=80{\mathrm{cm}}^2$ Area of trapezium $BQRC=$ Area of trapezium APSD $=\frac{1}{2}\times 4\times (28+20){\mathrm{cm}}^2$ $=96{\mathrm{cm}}^2$

5.0Key Features and Benefits of Class 8 Maths Chapter 9 Exercise 9.1

• Surface Area of Solids: Exercise 9.1 helps students learn how to calculate the surface area of 3D shapes like cubes and cuboids.
• Useful in Real Life: These concepts are helpful in everyday situations like painting walls, wrapping boxes, or covering surfaces.
• Simple Formulas and Steps: Students get to practice easy-to-remember formulas and apply them step by step to solve problems.
• Strengthens Visual Understanding: By working with 3D figures, students improve their ability to imagine and work with solid shapes.
• Based on NCERT Textbook: The questions follow the latest NCERT syllabus, making them great for school exams, homework, and quick revision.