7th Class Mental Ability Related to Competitive Exam Mensuration

Mensuration

Category : 7th Class

Mensuration

Learning Objectives

Mensuration
Introduction to Solid Shapes
Important Facts

Mensuration

Mensuration is the branch of mathematics which deals with the study of geometrical shapes, their area, volume and related parameters. Here, we will discuss the areas and perimeter of plane figures.

Some important mensuration formula are listed in the table given below.

Name	Figure	Perimeter	Area
Rectangle		\[2\left( a + b \right)\]	\[\operatorname{a}~\,\,\times \,\,b\]
Square		4a	\[{{a}^{2}}\]
Right Triangle		\[\operatorname{a} + b + h\]	\[\frac{1}{2}\,\,\times \,\,a\,\,\times \,\,h\]
Scalene Triangle		\[\operatorname{a} + b + c\]	\[\sqrt{s(s-a)(s-b)(s-c)}\] Where\[s=\frac{a+b+c}{2}\]
Isosceles Triangle		\[2a+b\]	\[\frac{b}{4}\sqrt{4{{a}^{2}}-{{b}^{2}}}\]
Equilateral Triangle		3a	\[\frac{\sqrt{3}}{4}\times {{a}^{2}}\]
Parallelogram		\[2\left( a +b \right)\]	\[\left( \operatorname{a}~\,\,\times \,\,h \right)\]
Rhombus		4a	\[\frac{1}{2}\,\,\times \,\,{{d}^{1}}\,\,\times \,\,{{d}^{2}}\]
Trapezium		Sum of all sides	\[\frac{1}{2}\,\,\times \,\,(a+b)\,\,\times \,\,h\]
Circle		\[2\pi r\]	\[\pi {{r}^{2}}\]

Example

A girl walking at the rate of 9 km per hour crosses a square field diagonally in 12 seconds. The area of the field is: The area of the field is:

(a) 460 sq. m (b) 600 sq. m

(e) None of these

Ans. (d)

Explanation: Distance covered by girl = \[\frac{9\times 1000}{3600}\,\,\times \,\,12\]= 30 m

So, the diagonal of square field = 30 m

Thus, the area of square field = \[\frac{1}{2}{{d}^{2}}\] (where d is diagonal)

\[=\,\,\frac{{{30}^{2}}}{2}\,\,=\,\,\frac{900}{2}\,\,\text{= }450\text{ }sq.\text{ }m.\]

The inner circumference of a circular race track which is 18 m wide is 880 m. Find the radius of the outer circle.

(a) 140 m (b) 150 m

(e) None of these

Ans. (a)

Explanation: Let the inner radius be r metre.

Then \[2\pi r\,\,=\,\,640\,\,\Rightarrow \,\,2\,\,\times \,\,\frac{22}{7}\,\,\times \,\,r\,\,=\,\,880\]

\[\Rightarrow \,\,\frac{44}{7}\,\,\,r\,\,=\,\,880\,\,\,\,\Rightarrow \,\,r\,\,=\,\,880\,\,\times \,\,\frac{7}{44}\,\,=\,\,140\,m.\]

The sides of a quadrilateral taken in order are 20 cm, 27 cm, 7 cm and 24 cm. The angles between the last two sides is a right angle, find the area of quadrilateral.

(a) \[875.96\text{ }c{{m}^{2}}\] (b) \[975.50\text{ }c{{m}^{2}}\]

(e) None of these

Ans. (c)

Let ABCD be a quadrilateral of sides AB = 20 cm, BC = 27 cm, CD = 7 cm,

DA = 24 cm and \[\angle CDA=\,90{}^\circ \]

Join AC, since \[\Delta \]ADC is a right angled triangle.

By using Pythagoras theorem, we have

\[{{\operatorname{AC}}^{2}}= {{7}^{2}}+ 2{{4}^{2}}= 49 + 576 = 625~\,\,\Rightarrow \,\,\,AC = 25 cm\]

Calculating area of \[\Delta \]ABC:

\[S=\frac{26+27+25}{2}=39\,cm\]

\[\therefore \,\,Area\text{ }of\,\,\Delta \,\,\,=\,\sqrt{\,39(39-26)(39-27)(39-25)}\]

= \[\sqrt{39\,\,\times \,\,13\,\,\times \,\,12\,\,\times \,\,14}\,\,=\,\,\sqrt{85,176}=291.85\,\,c{{m}^{2}}\]

Calculating area of \[\Delta \]ADC:

Area of \[\Delta \text{ }=\text{ }\frac{1}{2}\,\,\times \text{ }7\text{ }\times \text{ }24\text{ }=\text{ }84\text{ }c{{m}^{2}}\]

Area of quadrilateral ABCD = Area of \[\Delta \]ABC + Area of \[\Delta \]ADC

\[= 291.85 + 84 = 375.85 c{{m}^{2}}\].

The area of a parallelogram is \[\mathbf{70}\text{ }\mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}\]and its altitude is 7 cm. Find the perimeter of a rectangle having equal area on the same base.

(a) 36 cm (b) 99 cm

(e) None of these

Ans. (d)

Explanation: \[\operatorname{Area} of the parallelogram = base~\,\,\times \,\,\,height\]

\[\therefore \text{ }70\text{ }=\text{ }a~\,\,\times \,\,7\], where a is the base.

\[\Rightarrow \]a = 10 cm

If rectangle has one side as b cm, the base as 10 cm, the area is 70 cm2,

\[\therefore \,\,\,\,10b=70\,\,\Rightarrow \,\,b\,\,=\,\,7\,cm\]

Hence, perimeter of rectangle = 2 (10 + 7) = 34 cm.

Introduction to Solid Shapes

Cube, cuboid, sphere, cylinder, cone and pyramid are the examples of solid shape. These are three-dimensional shapes. One can view these shapes from three sides (left, right and top).
The corners of a solid shape are called its vertices, the line segments are called its edges and its flat surfaces are its faces.
The skeleton-outline that can be folded to make the solid is called a net. The same solid can have several types of nets.
Two types of sketches of a solid can be drawn:

Oblique sketch: It is a type of sketch of the solid which does not have proportional lengths but still it conveys all important aspects of the appearance of the solid

Isometric sketch: It is a type of sketch of the solid in which the measurements are kept proportional.

Example

How many edges does a cube has?

(a) 4 (b) 6

(e) None of these

Ans. (d)

Explanation: A cube has 12 edges.

The net shown below belongs to which solid shape?

(a) Cube (b) Cuboid

(e) None of these

Ans. (b)

Explanation: The two faces (x mark) of the net are of equal sizes while the other four (3 mark) are equal in size but different from the first two. Thus, the net belongs to cuboid.

Important Facts

The distance around a closed figure is called perimeter whereas area is the part of plane occupied by the closed figure.
One should keep in mind that the values of length and breadth should be in the same units while placing their values in a formula for calculating area or perimeter. Based on the conversion of units for lengths, the units of areas can also be converted:
\[1\text{ }c{{m}^{2}}=\text{ }100\text{ }m{{m}^{2}},\text{ }1\text{ }{{m}^{2}}=\text{ }10000\text{ }c{{m}^{2}},\text{ }1\text{ }hectare\text{ }=\text{ }10000\text{ }{{m}^{2}}\].
Plane figures are of two dimensions (2 - D).
Solid shapes can be drawn on a flat surface (like paper) realistically. It is called 2?D representation of a 3-D solid.

7th Class Mental Ability Related to Competitive Exam Mensuration

Other Topics

Notes - Mensuration