Question

chapter 11 class 8 example subject math's​

Answer 1

Step-by-step explanation:

Chapter: 11 Mensuration {According to NCERT)

In this Chapter:

11.1 Introduction

We have flying that for a closed figure, the perimeter is the distance around its boundary and its area is the regions covered by it. We found the area and perimeter of various plane figures such as triangle, rectangle, circles etc. We have also learnt to find the area of pathways of borders in rectangular shapes.

In this chapter, we will try to solve the problem related to parameter and area of other plane closed figures like quadrilaterals.

We will also learn about surface area and volume of solids such as cube, cuboid and cylinder.

11.2 Let us Recall

Let us take an example to review about previous knowledge.

This is the figure of a rectangular park (Fig 11.1) whose length is 30 m width is 20 m.

(I) What is the total length of the fence surrounding it? define the length of the fence we need to find the perimeter of the park, which is 100 m. [Check it]

(II) How much land is occupied by the park? To find the land occupied by the spark we need to find the area of this park which is 600 square metres (m²) (How?).

(III) There is a part of 1 m width running inside along the perimeter of the park that has to be cemented. If 1 bag of cement is required to cement 4 m² area, how many bags of cement would be required to construct the cemented path?

We can say that the number of cement bags used =

$\frac{area \: of \: the \: path}{area \: cemented \: by \: 1 \: bag}$

Area of cemented path = area of Park - area of path not cemented.

Path is 1 m wide, show the rectangular area not cemented is (30 - 2) × (20 - 2) m².

That is 28 × 18 m².

Hence number of cement bags used ________.

(IV) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as shown in the diagram (Fig 11.1) and the rest has grass on it. find the area covered by grass.

Area of rectangular beds = _________.

Area of Park left after cementing the path = _________.

Area covered by grass = ________.

We can find areas of geometrical shapes other than rectangles also if certain measurements are given to us.

11.3 Area of Trapezium

Nazma owns a plot near the main road (Fig 11.2) Unlike some other rectangular plots in her neighbourhood, the plot has only one pair of parallel opposite sides. So, it is nearly a trapezium in shape. Can you find out its area?

Let us name the vertices of this plot as shown in Fig 11.3.

By drawing EC ll AB, we can divided into two parts, one of rectangular shape and other of triangular shape, (which is right angled at C), as shown in Fig 11.3.

Area of ∆ECD =

$\frac{1}{2}h \times c = \frac{1}{2} \times 12 \times 10 = 60 \: {m}^{2}$

Area of Rectangle ABCE =

$h \times a = 12 \times 20 = \: 240 \: {m}^{2}$

Area of trapezium ABDE = area of ∆ ECD + Area of rectangle ABCE = 60 + 240 = 300 m².

We can write the area by combining the two areas and write the area of trapezium as,

Area of ABDE =

$\frac{1}{2}h \times c + h \times a = h( \frac{c}{2} + a)$

$h( \frac{c + 2a}{2}) = h( \frac{c + a + a}{2})$

$= \: h \frac{(b + a)}{2} = \: height( \frac{sum \: of \: parallel \: side}{2})$

By substituting the value of h, b and a in this expression, we find that

$h \frac{(b + a)}{2} = 300 \: {m}^{2}$

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Answer 2

Answer:

Solution:

Side of a square = 60 m (Given)

And the length of rectangular field, l = 80 m (Given)

According to question,

Perimeter of rectangular field = Perimeter of square field

2(l+b) = 4×Side (using formulas)

2(80+b) = 4×60

160+2b = 240

b = 40

Breadth of the rectangle is 40 m.

Now, Area of Square field

= (side)2

= (60)2 = 3600 m2

And Area of Rectangular field

= length×breadth = 80×40

= 3200 m2

Hence, area of square field is larger.