Area Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 7

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Class 8 Maths Ganita Prakash Part 2 Chapter 7 Solutions

Class 8 Maths Area Solutions

Class 8 Ganita Prakash Part 2 Chapter 7 Solutions Area

NCERT Page 148-150

Question 1.
Try to think of different creative ways to divide a square into 4 parts of equal area.
Solution:
Do yourself

Question 2.
Why do we count the number of unit squares to assign measures for area? Couldn’t we have just used the perimeter of a region i.e. the length of its boundary as a measure of its area?
Solution:
Perimeter is not a suitable measure for area because different shapes can have the same perimeter but vastly different areas and vice-verse.
Since, area measures the amount of two-dimensional space within a boundary, while perimeter measures the one-dimensional length of the boundary itself.
So, counting unit squares provides a consistent, standardised way to qualify the surface coverage, as each unit square represents the same amount of area.

Question 3.
(i) Find two rectangles that are examples of such regions like perimeter of region 1> perimeter of region 2 but area of region 1< area of region 2.
If needed, use a grid paper (given at the end of the book) for this.
(ii) Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area. This property should be visually clear in your example.
Solution:
(i) Region 1
Let the sides of rectangle ABCD be 1 cm and 10 cm.
We know that the perimeter of rectangle
= 2 (Length + Width)

So, perimeter of rectangle ABCD
= 2(1 + 10)
= 2 × 11
= 22 cm

We know that the area of rectangle = Length × Width
= 1 × 10
= 10 cm²

Region 2
Let the sides of rectangle PQRS be 4 cm and 4 cm.
So, perimeter of rectangle PQRS = 2(4 + 4)
= 2 × 8
= 16 cm

and the area of rectangle PQRS = 4 × 4 = 16 cm²
So, from Region 1 and Region 2.
Perimeter of Region 1 > Perimeter of Region 2 but area of Region 1 < Area of Region 2.

(ii) Region 1
Consider a square ABCD with side 4 cm.

Now, perimeter of square ABCD = 4 × Side
= 4 × AB
= 4 × 4
= 16 cm
and area of square ABCD = Side × Side
= 4 × 4
= 16 cm²

Region 2
Consider a rectangle PQRS.
So, length of rectangle = 8 cm
and width of rectangle = 1cm
Now, perimeter of rectangle PQRS
= 2 [length + width]
= 2 [8 + 1]
= 2 × 9
= 18 cm
and area of rectangle PQRS = length × width
= 8 × 1
= 8 cm²
Here, rectangle PQRS has larger perimeter but a smaller area than square ABCD.
Hence, a region with a larger perimeter need not have a larger area.

Figure it Out (page 150-152)

Question 1.
Identify the missing sldelengths.

Solution:
We know that area of rectangle = Length x Width
Now, in rectangle GCDE, we have
Area = 21 in²

and width (GE) =7 in
So, length (CG) = \(\frac{21}{2}\) = 3 in
Now, BG = BC + CG = 4 + 3 = 7 in [∵ BC = 4 in]

Now, in rectangle ABGM,
Area of ABGM
AB = \(=\frac{\text { Area of } A B G M}{B G}=\frac{28}{7}\) = 4 in
[∵BG = BC + CG and area of ABGM = 28 in²]
AB = MG

So, MG = 4 in
HG = HM + MG
= 3 + 4
= 7 1n [∵ HM = 3 in]

Now, in rectangle HIJG,
HI = \(\frac{\text { Area of } H I J G}{H G}\) [∵area of HIJG = 35 in²]
= \(\frac{35}{7}\)
= 5 in
HI = GJ
GJ = 5 in
GK = GJ + JK
= 5 + 2
= 7 in

Now, in rectangle GFLK,
KL = \(\frac{\text { Area of } G F L K}{G K}=\frac{14}{7}\)
[∵ area of GFLK = 14 in²]

Solution:
In rectangle ABEF,
AB = \(\frac{\text { Area of } A B E F}{A F}=\frac{29}{4}\) = 7.25 m
[∵ area of ABEF = 29 m² and AF = 4m]

Since, AF = BE
So, BE = 4 m
Now, in rectangle BCDE,
BC = \(\frac{\text { Area of } B C D E}{B E}=\frac{11}{4}\) = 2.75 m
[∵ area of BCDE = 11m²]
Now, in rectangle EFGH
EH = \(=\frac{\text { Area of } E F G H}{E F}=\frac{50}{7.25}\) = 6.896 m
[∵ area of EFGH = 50 m² and AB = EF = 725 m]

Question 2.
The figure shows a path (the shaded portion) laid around a rectangular park EFGH.

(i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area.
An example of a formula,
area of a rectangle = Length × Width.
[Hint There is a relation between the areas of EFGH, the path and ABCD]
Solution:
Measurements required
To find the area of the path, we need the length and breadth of the outer rectangle ABCD and the inner rectangle (park) EFGH.
Reason
The path is the region between the two rectangles.
Area of path = Area of outer rectangle ABCD – Area of inner rectangle EFGH

Assigning possible values
Consider outer rectangle ABCD
Length = 20 m, Breadth = 14 m
and for inner rectangle EFGH
Length = 14 m, Breadth = 8 m
Area of ABCD = 20 × 14 = 280 m²
Area of EFGH =14 × 8 = 112m²
Hence, area of path = 280 – 112 = 168 m²

Formula
Area of path = (L x B) – (1 x b)
Where, L,B = Outer rectangle ABCD dimensions
l, b = Inner rectangle EFGH dimensions

(ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements.
[Hint Break the path into rectangles.]
Solution:
No, we cannot find the area with only the path width. We need atleast the length and width of the inner rectangle or the outer rectangle.
Suppose, width of path on each side = 2 m.
Inner park ETGLf has length = 14 m and breadth = 8 m.
Other measurements needed
We need the outer rectangle dimensions, which are obtained by adding twice the width of path.
Outer length =14 + 2 + 2 = 18 m
Outer breadth = 8 + 2 + 2 = 12 m
Area of outer rectangle = 18 × 12 = 216m²
Area of inner rectangle = 14 × 8 = 112 m²
Area of path = 216 – 112 = 104m²

Formula
If width of path = w,
Area of path = (l + 2w)(b + 2w) – lb

(iii) Does the area of the path change, when the outer rectangle is moved, while keeping the inner rectangular park EFGH inside it, as shown?

Solution:
No, the area does not change.
Reason
The dimensions of the inner rectangle EFGH remain the same.
The dimensions of the outer rectangle ABCD remain the same.
Only the position of the inner rectangle changes not the sizes.
Since, area depends only on dimensions and not on position, the area of the path remains unchanged.

Question 3.
The figure shows a plot with sides 14 m and 12 m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

Solution:
Given, the length and width of given rectangular plot is 14 m and 12 m.

To find the area of the cross path, we need

• Length of the vertical path
• Length of the horizontal path

Let width of each path = 2 m
and we have, length of vertical path = 12 m
[∵ AD = IJ = 12 m]
Then, area of vertical path JLKI = 12 × 2 = 24 m²
and the length of horizontal path = 14 m
[∵ CD = EF = 14 m]
Then, area of horizontal path EFFG = 14 × 2 = 28 m²

Now, area of overlapping square when side is 2 m
⇒ area = 2 × 2 = 4 m²
Hence, the area of crosspath = 24 + 28 – 4 = 48 m²

Formula
Let length of plot = L, width of plot = B
and width of path = W
Then, area of crosspath = L × W + B × W – W²

Question 4.
(i) Find the area of the spiral tube shown in the figure. The tube has the same width throughout.

[Hint: There are different ways of finding the area. Here is one method]
Answer:
The length of the spiral tube is the sum of the lengths of all its straight segments.
From the figure, the segments have lengths 20, 20, 20, 15, 15, 10, 10, 5 and 5.
So, length = 20 + 20 + 20 + 15 + 15 + 10 + 10 + 5 + 5 = 120 units
and width = 1 unit
So, the area of spiral tube = Length × Width
= 120 × 1
= 120 sq unit

(ii) What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

Solution:
The length of the straight tube = 5 + 5 = 10 unit.

Question 5.
In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Solution:
When the sidelength of the square is doubled, the area of the square becomes four times.
Since, regions 1,2 and 3 are fixed parts of the square, the area of each region also becomes four times.
Therefore, the increase in the area of each region is three times its original area.

Question 6.
Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure.

Rearrange the pieces to get a larger square, with a hole inside.
You can try this activity by constructing the square using cardboard, thick chart paper or similar materials.
Solution:
Do yourself

NCERT Page (153-154)

Question 1.
In the given figure, which triangle has a greater area: ΔXDC or ΔYDC, if both the rectangles are identical?

Solution:
Both ΔXDC and ΔYDC have equal areas.
Reason

• In both figures, DC is the same base.
• Points X and Y lie on the top side of identical rectangles.
• Since, the rectangles are identical, the perpendicular distance (height) from X to DC is equal to the perpendicular distance from Y to DC.

Using the area formula of a triangle,
Area of a triangle = \(\frac{1}{2}\) × Base × Height

For both triangles

• Base = DC (same)
• Height = Height of the rectangle (same)

∴ Area (ΔXDC) = Area (ΔYDC)

Question 2.
In the given figure, which triangle has a greater area ΔXDC or ΔYBC, if both the rectangles are identical?

In each case, by dropping the altitudes from X and Y, it becomes clear that each triangle has exactly half the area of the rectangle ABCD.
Solution:
∆XDC and ∆YBC have equal areas.
Reason
In both cases, each triangle lies inside an identical rectangle ABCD.

We know that the area of a triangle = \(\frac{1}{2}\) × Base × Height
In the first rectangle, the base of ∆XDC is the side DC and the height of the triangle is the perpendicular distance from point X to the base DC, which is equal to the side AD or BC of the rectangle.

In the second rectangle, the base of ∆YBC is side BC and the height of the triangle is the perpendicular distance from point Y to the base BC, which is equal to the side AB or DC of the rectangle.

Since, both rectangles are identical, their corresponding sides have the same lengths, so the product of their respective base and height are equal.

Question 3.
Find the area of ∆XDC.

Solution:
Given, the base of ∆XDC = 5
and the height of ∆XDC = 4
We know that the area of triangle = \(\frac{1}{2}\) × Base × Height
So, area of ∆XDC = \(\frac{1}{2}\) × 5 × 4
= 10 sq units

NCERT (Page 156)

Question 1.

Line l ∥ BC. Consider the different triangles that have BCas their base and with their third vertex lying anywhere on /.
Which of these triangles has the maximum area and which has the minimum area?
Solution:
All triangles have the same area, there is no single triangle with a maximum or minimum area.
Reason The area of a triangle is calculated using the formula.
Area = \(\frac{1}{2}\) × Base × Height
The base of all triangles is the line segment BC, which has a fixed length.
The third vertex of each triangle lies on line l, which is parallel to BC. The perpendicular distance between two parallel lines is constant. This means the height (h) of all triangles with their base on one line and the opposite vertex on the other line is the same.
Since, both the base and height are constant for all such triangles, so their areas are equal.

Figure it Out (Page 157-159)

Question 1.
Find the areas of the following triangles.

Solution:
(i) Given, the base of triangler, BC = 4 cm and the height of triangle, AE = 3 cm
We know that the area of triangle = \(\frac{1}{2}\) × Base × Height
So, area of ΔABC = \(\frac{1}{2}\) × 4 × 3= 6 cm

(ii) Given, the base of triangle, EF = 5 cm and the height of triangle, DN = 3.2 cm
So, area of ΔDEE = \(\frac{1}{2}\) × EF × DN = \(\frac{1}{2}\) × 5 × 32 = 8 cm²

(iii) Given, the base of triangle, AT =3 cm and the height of triangle, AN = 4 cm
So, area of ΔATN = \(\frac{1}{2}\) × AT × AN= \(\frac{1}{2}\) × 3 × 4= 6 cm²

Question 2.
Find the length of the altitude BY.

Solution:
In ΔABC, we have
Base, BC = 6 units
and height, AX = 4 units
Then, area of ΔABC using base BC = \(\frac{1}{2}\) × BC × AX
= \(\frac{1}{2}\) × 6 × 4
= 12sq units … (i)

Now, area of ΔABC using base AC
= \(\frac{1}{2}\) × AC × BY
= \(\frac{1}{2}\) × 8 × BY [∵ AC = 8]
= 4BY …(ii)
From Eqs. (i) and (ii), we get
12 = 4BY ⇒ BY = 3units

Question 3.
Find the area of ASUB, given that it is isosceles. SE is perpendicular to UBand the area of ASEBis 24 sq units.

Solution:
Since, ΔSUB is an isosceles triangle with altitude SE to the base UB.
The altitude drawn from the vertex angle (the angle between two equal sides) to the base of an isosceles triangle always bisects the base.
So ,UE = EB
andSE is a common side in both ΔSUE and ΔSEB.
So, ΔSUE = ΔSEB
Given, area of ΔSEB = 24 sq units
Now, area of ΔSUB = Area of ΔSUE + Area of ΔSEB
= 24 + 24
= 48 sq units

Question 4.
[Sulba-Sutras] Give a method to transform a rectangle into a triangle of equal area.
Solution:
We know that area of rectangle
= Length × Width
= Base × Height

Reason Area of ΔABC = \(\frac{1}{2}\) × BC × AD

Then, Base × Height = \(\frac{1}{2}\) × Base × Height
Base × Height = Base × Height
Hence, if the base are the same then height must be double of triangle than rectangle.

Question 5.
[Sulba-Sutras) Give a method to transform a triangle Into a rectangle of equal area.
Solution:
Method
1. Consider a ΔABC

2. Draw the altitude from vertex A to base B meeting it atD.
3. Construct a rectangle having base
= BC and height = half of AD.
4. The rectangle formed will have the same area as the triangle.
Reason Area of ABC = \(\frac{1}{2}\) × BC ×AD
Area of rectangle = BC × \(\frac{AD}{2}\)
Both areas are equal.

Question 6.
ABCD, BCEF, and BFGH are identical squares

(i) If the area of the red region is 49 sq units then what is the area of the blue region?
(ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq units then what is the area of each square?
Solution:
(i) Given, the area of ΔHDC = 49 sq units
⇒ \(\frac{1}{2}\) × DC × HC = 49

Since, squares ABCD, BCEF and BFGHare identical.
So, BC = BH and AI = IB
From Eq. (i),
\(\frac{1}{2}\) × AB × 2BC = 49 [∵ DC = AB sides of square]
⇒ \(\frac{1}{2}\) × AI × 2BC = 49
⇒ \(\frac{1}{2}\) × AI × BC = \(\frac{49}{4}\)
⇒ \(\frac{1}{2}\) × AI × AD = \(\frac{49}{4}\)
[∵ BC = AD side of same square]
⇒ Area of ΔAID= 12.25 sq units
Hence, the area of blue regioni is 12.25 sq units.

(ii) Given, area of HBCDA = 180 sq units
From figure, the region HBCDA consists of 5 identical triangle.
So, area of (ΔADI + ΔDIJ + ΔJIC + ΔCIS + ΔIBH) = 180 …(i)
⇒ 5 × Area of ΔIBH = 180 Area of ΔIBH = 36 and from figure,

Area of square ABCD = Area of (ΔDAI + ΔDJI + ΔJIC + ΔCIB)
From Eq. (i),
Area of square ABCD+ Area of A IBH = 180
⇒ Area of square AS CD + 36 = 180
⇒ Area of square ABCD = 144 sq units
Since, all squares are identical.
So, the area of each square is 144 sq units.

Question 7.
If M and N are the mid-points of XV and XZ, what fraction of the area of ΔXYZ is the area of ΔXMN? [Hint Join NY]

Solution:
We know that a median of a triangle divides it into two triangles of equal area.

Here, M is the mid-point of X Y, so ZM is a median of ΔXYZ
So, area of ΔXZM = Area of ΔYZM
= \(\frac{1}{2}\) × Area of ΔXYZ

Now, in ΔZXM, N is the mid-point of XZ, so MN is a median, so MN divides ΔXZM into two equal triangles i.e. ΔXMN and ΔZMN.
Area of ΔXMN = Area of ΔZMN
= \(\frac{1}{2}\) × Area of ΔXZM

Now, from Eqs. (i) and (ii),
Area of ΔXMN = \(\frac{1}{2}\) × (\(\frac{1}{2}\) × Area of ΔXYZ)
⇒ Area of ΔXMN = \(\frac{1}{4}\) × Area of ΔXYZ
Hence, the area of ΔXMN is \(\frac{1}{4}\) of the area of ΔXYZ.

Question 8.
Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

Solution:
The shortest path is a straight line from the house to the river then another straight line from the same point on the river to the water tank.

NCERT (Page 159)

Question 1.
How do we find the area of this pentagon?

Solution:
Method
Divide the pentagon into triangles.
1. Choose any one vertex of the pentagon.
2. Join this vertex to all the non-adjacent vertices using straight lines.
3. The pentagon gets divided into 3 triangles.
4. Find the area of each triangle using the formula.
Area of a triangle = \(\frac{1}{2}\) × Base × Height
5. Add the areas of the three triangles.

Figure it Out (NCERT Page 160)

Question 1.
Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN= 3 cm, BM is perpendicular to AC and DN is perpendicular to AC.

Solution:
Since, the diagonal AC divides the rectangle ABCD into two triangles ADC and ABC.

So, area of rectangle ABCD = Area of ΔABC + Area of ΔADC
= \(\frac{1}{2}\) × AC × BM + \(\frac{1}{2}\) × AC × DN
[∵ area of triangle = \(\frac{1}{2}\) × base × height]
= \(\frac{1}{2}\) × 22 × 3 + \(\frac{1}{2}\) × 22 × 3
= 33 + 33
= 66 cm²

Question 2.
Find the area of the shaded region given that ABCD is a rectangle.

Solution:
We know that area of rectangle = Length × Width
So, area of rectangle ABCD = 18 × 10 = 180 cm2
We know that area of triangle = \(\frac{1}{2}\) × Base × Height
So, area of ΔAEF= \(\frac{1}{2}\) × 10 × 6
[∵ base = 10 cm, height = 6cm]
= 30 cm²
and area of AEBC = \(\frac{1}{2}\) × 8 × 10 = 40 cm²
[∵ base = 8 cm and height =10 cm]

Now, area of shaded region EFDC
= Area or rectangle ABCD – Area of ΔAEF – Area of ΔEBC
= 180 – 30 – 40
= 110 cm²

Question 3.
What measurements would you need to find the area of a regular hexagon?
Solution:
To find the area of a regular hexagon, we need the length of one side of the hexagon.
Reason A regular hexagon can be divided into 6 congruent equilateral triangles.
If the sidelength is known the area of one equilateral triangle can be found and then multiplied by 6.

Question 4.
What fraction of the total area of the rectangle is the area of the blue region?

Solution:

Area of ADEC = \(\frac{1}{2}\) × DC × PE 2
[∵ area of triangle = \(\frac{1}{2}\) × base × height]
= \(\frac{1}{2}\) × w × h₁
and area of ΔABE = \(\frac{1}{2}\) × AB × QE = \(\frac{1}{2}\) × w × h₂ … (ii)

Now, area of blue region = Area of ΔDEC + Area of ΔABE
= \(\frac{1}{2}\) × w × h₁ + \(\frac{1}{2}\) × w × h₂
= \(\frac{1}{2}\) × w [h₁ + h₂] [∵ h₁ + h₂ = h]
= \(\frac{1}{2}\) × w × h
and area of rectangle ABCD (total area) = w × h
\(\frac{\text { Area of blue region }}{\text { Area of rectangle }}=\frac{\frac{1}{2} \times w \times h}{w \times h}=\frac{1}{2}\)
Hence, the fraction of the total area that is blue is \(\frac{1}{2}\).

Question 5.
Give a method to obtain a quadrilateral, whose area is half that of a given quadrilateral.
Solution:
The quadrilateral formed by joining the mid-points of the sides of any quadrilateral has an area equal to half the area of the given quadrilateral. This quadrilateral is always a parallelogram.

Here, the area of quadrilateral PQRS is the half of the area of quadrilateral ABCD.

NCERT (Page 161-162)

Question 1.
Can the parallelogram be cut along CZ and rearranged to form a rectangle?
Solution:
Yes, cut the parallelogram along the line CZ (as shown in the figure).

The triangular portion can be shifted to the opposite side. This rearrangement forms a rectangle.
The rectangle has length = Base of the parallelogram and breadth = Height of the parallelogram
Since, rearrangement does not change area.
Area of parallelogram = Area of rectangle

Figure it Out (Page 162-164)

Question 1.
Observe the parallelograms in the figure below.
(i) What can we say about the areas of all these parallelograms?
(ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter

Solution:
(i) The areas of all these parallelograms are the same.
Each parallelogram lies on the same grid and has the same base length and the same corresponding height (the perpendicular distance between the pair of parallel sides).
Since, area of a parallelogram = Base x Height
and both base and height are the same for all, their areas are equal, even though their shapes are different (they are slanted differently).

(ii) The perimeters are not equal.
Even if areas are the same perimeters can be different because the lengths of the slant sides change.
Observation from the figures
The parallelogram that looks most slanted has the longest sidelengths, so it has the maximum perimeter.
The parallelogram that is closest to a rectangle (least slanted) has the shortest side lengths, so it has the minimum perimeter.
Figure (a), appears to have the minimum perimeter (shortest slanted sides) and figure (g), appears to have the maximum perimeter (longest slanted sides).

Question 2.
Find the areas of the following parallelograms:

Solution:
We know that the area of parallelogram = Base × Height
(i) Given, base = 7 cm and height = 4 cm
Then, the area of parallelogram = 7 × 4 = 28 cm²

(ii) Given, base = 5 cm and height = 3 cm
Then, the area of parallelogram = 5 × 3 = 15 cm²

(iii) Given, base = 5 cm and height = 4.8 cm
Then, the area of parallelogram = 5 × 4.8 = 24 cm²

(iv) Given, base = 2 cm and height = 4.4 cm
Then, the area of parallelogram = 2 × 4.4 = 8.8 cm²

Question 3.
Find QN.

Solution:
From figure, in parallelogram PQRS, we have
Base, SR = 12 cm and height, QM = 6 cm
Now, area of parallelogram PQRS using base SR
= SR × QM
= 12 × 6 = 72 cm² …(i)
Now, area of parallelogram PQRS using base PS = PS × QN
= 7.6 × QN …(ii) [∵ PS = 7.6 cm]
From Eqs. (i) and (ii),
72 = 7.6 × QN
QN = \(\frac{72}{7.6}\)
⇒ QN = 9.47 cm
Hence, the length of QN is 9.47 cm.

Question 4.
Consider a rectangle and a parallelogram of the same sidelengths 5 cm and 4 cm. Which has the greater area?
[Hint Imagine constructing them on the same base.]

Solution:
We know that area of rectangle = Length × Width
Given, the length of rectangle = 5 cm
and width of rectangle = 4 cm
Area of rectangle = 4 × 5 = 20 cm²
and we know that the area of parallelogram = Base × Height
Here, the height of the parallelogram must be less than the length of its slanted side 4 cm.
So, its area will be less than 20 cm²

Hence, rectangle has the greater area.

Question 5.
(i) Give a method to obtain a rectangle, whose area is twice that of a given triangle.
Solution:
1.Take a ΔABC.

2. Draw the altitude fromAto base BC.
3. Construct a rectangle with base = BC and height = altitude of the triangle.
4. Since and area of triangle = \(\frac{1}{2}\) × BC × h
and area of rectangle = BC × h

Hence, the rectangle has twice the area of the triangle,

(ii) What are the different methods that you can think of?
Solution:
Do yourself.

Question 6.
[Sulba-Sutras] Give a method to obtain a rectangle of the same area as a given triangle.
Solution:
A rectangle with the same area as a given triangle can be constructed by taking half the length of the base of the triangle as one side of the rectangle and the height of the triangle as the other side.

Question 7.
[Sulba-Sutras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

[Hint Show that ∆ADB and ∆ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cutouts if necessary.]
Solution:
Method
1. Let ∆ABC be an isosceles triangle with AB = AC.
2. Draw the altitude AD from the vertex A to the base BC.
In an isosceles triangle, this altitude also bisects the base.
3. The triangle is now divided into two congruent right-angled triangles ADB and ADC.
4. Each of these triangles has half the area of the original triangle.
5. Place the two congruent triangles along their slant height.
6. They fit exactly to form a rectangle.

Question 8.
[Sulba-Sutras] Give a method to convert a rectangle into an isosceles triangle by dissection.
Solution:
Method
1. Take a rectangle ABCD.
2. Draw one diagonal, say BD.
3. The diagonal divides the rectangle into two congruent right triangles.
4. Join these two right triangles along their heights.
5. The figure obtained is an isosceles triangle.

Question 9.
Which has greater area an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area two identical equilateral s triangles together or a square of the same sidelength as the triangle? Give reasons.
Solution:
For the same sidelength, a square covers more area than an equilateral triangle.
Reason
The square spreads the same sidelength in two directions.
The equilateral triangle is narrower.
So, the square has a greater area.

NCERT (Page 164-166)

Question 1.
Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.
Solution:
Area of rhombus ABCD
= Area of ∆ADB + Area of ∆BCD
= \(\frac{1}{2}\) × AO × BD + \(\frac{1}{2}\) × CO × BD
= \(\frac{1}{2}\) × BD × (AO + CO) … (i)

From figure, AO + OC = AC
From Eq. (i),
Area of rhombus ABCD = \(\frac{1}{2}\) × BD × AC
Hence, the area of rhombus is half the product of its diagonals.

Figure it Out (Page 169-170)

Question 1.
Find the area of a rhombus, whose diagonals are 20 cm and 15 cm.
Solution:
Given, diagonals of the rhombus are 20 cm and 15 cm.
We know that the area of rhombus
= \(\frac{1}{2}\) × Product of the diagonals
= \(\frac{1}{2}\) × 20 × 15
= 150 cm²

Question 2.
Give a method to convert a rectangle into a rhombus of equal area using dissection.
Solution:
To convert a rectangle into a rhombus of equal area using dissection.
1. Draw diagonals AC and BD of a rectangle ABCD. They intersect at point O.

2. The diagonals divide the rectangle into four triangles ∆AOB, ∆BOC, ∆COD and ∆DOA. All four triangles are equal in area.
3. Cut the rectangle along diagonals AC andBD. We you now have four congruent triangles.
4. Rearrange the four triangles so that their equal sides meet consecutively.
5. Arrange the triangles so that their hypotenuse sides form the boundary.
Hence, a rectangle can be converted into a rhombus of equal area by cutting it along its diagonal.

Question 3.
Find the areas of the following figures.

Solution:
We know that the area of trapezium
= \(\frac{1}{2}\) × Height × Sum of the parallel sides
(i) Given, the height of trapezium = 16 ft
and the length of parallel sides = 10 ft and 7 ft
∴ So, the area of trapezium = \(\frac{1}{2}\) × 16 × (10 + 7)
= \(\frac{1}{2}\) × 16 × 17
= 136 ft²

(ii) Given, the height of trapezium = 14 m
and the length of parallel sides = 24 m and 36 m
So, the area of trapezium = \(\frac{1}{2}\) × 14 × (24 + 36)
= 7 × 60
= 420 m²

(iii) Given, the height of trapezium = 10 in
and the length of parallel sides = 14 in and 6 in
So, the area of trapezium = \(\frac{1}{2}\) × 10 × (14 + 6)
= 5 × 20
= 100 in²

(iv) Given, the height of trapezium = 8 ft .
and the length of parallel sides = 18 ft and 12 ft
So, the area of trapezium = \(\frac{1}{2}\) × 8 × (18 + 12)
= 4 × 30
= 120 ft²

Question 4.
[Sulba-Sutras] Give a method to convert an isosceles trapezium to a rectangle using dissection.
Solution:
Construction/Method (Dissection)
1. Drop perpendiculars (heights)
From vertices A and B, draw perpendiculars to the longer parallel side CD.
Let the feet of the perpendiculars beF andF, respectively.

2. Identify the figures formed
The central region AEFB is a rectangle (since, opposite sides are parallel and all angles are right angles).
On the two sides of the rectangle, we get two right angled triangles AED and BFC.

3. Use the isosceles property
Since, the trapezium is isosceles,
AD = BC ⇒ ΔAED ≅ ΔBFC.
(They have equal hypotenuse and equal height and one right angle.)

4. Dissection and rearrangement
Cut one of the triangular pieces (say AAED) and place it exactly over the other ABFC on the opposite side of the rectangle.

5. Resulting figure
After rearrangement, the rectangle AEFB together with the moved triangle forms a larger rectangle.

Question 5.
Here is one of the ways to convert trapezium ABCD into a rectangle EFGHof equal area.

Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH?
[Hint If ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ then the trapezium and rectangle have equal areas.]
Solution:
(i) Extend the side AB from both sides.
(ii) From points A and B, draw perpendicular to these extensions to get points H and E.
(iii) Cut ΔAHIfrom the left side and place it at ΔDGI.
(iv) Cut ΔBEJ from the right side and place it at ΔCFJ.
(v) Since, ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, only rearrangement is done, so the area remains the same.
(vi) The new figure formed is rectangle EFGH.
Hence, rectangle EFGH has the same area as trapezium ABCD because congruent triangles are rearranged without changing their size.

Question 6.
Using the idea of converting a trapezium into a rectangle of equal area and vice versa, construct a trapezium of area 144 cm².
Solution:
Method
(i) First, construct a rectangle, whose area is 144 cm2.
e.g. Take length = 16 cm and breadth = 9 cm.
(ii) Now, convert this rectangle into a trapezium by cutting a triangular part from one side.
(iii) Rearrange this triangular part on the other side.
(iv) The figure obtained is a trapezium.
(v) Since, only cutting and rearranging is done, the area remains 144 cm2.

Question 7.
A regular hexagon is divided into a trapezium, an equilateral triangle and a rhombus, as shown. Find the ratio of their areas.

Solution:
Method
(i) A regular hexagon can be divided into 6 equal equilateral triangles.

(ii) From the figure
The equilateral triangle consists of 1 small triangle. The rhombus consists of 2 small triangles.
The trapezium consists of 3 small triangles.

(iii) Therefore, the ratio of their areas is Trapezium: Equilateral triangle: Rhombus
= 3 : 1 : 2

Question 8.
ZYXW is a trapezium with ZY ∥ WX. A is the mid-point of XV. Show that the area of the trapezium ZYXW is equal to the area of ΔZWB.
Solution:
Given, ZYXW is a trapezium with ZY ∥ WX A is the mid-point of XY.
Line ZA is produced to meet the extension of WX at B.
To prove
Area of trapezium ZYXW = Area of ΔZWB
Proof
Since, ZY ∥ WX, triangles and trapeziums formed between the same parallel lines can be compared using the concept of equal heights. Point Ais the mid-point ofXY.
⇒ XA = AY
In ΔZAY and ΔBAX we have
YA = XA [given]
∠YAZ = ∠XAB
[vertically opposite angles]
∠YZA = ∠XBA
[alternate interior angles]
∴ ΔZAY = ΔBAX [by A AS congruency]
So, area of AZAY = araofABAX …(i)
Now, area of trapezium ∠YXW
= Area of quadrilateral ZAXW + Area of ΔZAY = Area of quadrilateral ZAXW + Area of ΔBAX [v from Eq. (i)]
= Area of ΔZWB
So, area of trapezium ZYXW = Area of ΔZWB
Hence proved.

NCERT (Page 170 and 171)

Question 1.
What do you think is the area of an A4 sheet? Its sidelengths are 21 cm and 29.7 cm. Now, find its area.
Solution:
Given, length = 29.7 cm
and breadth = 21cm
Now, area of A4 sheet = Length × Breadth
= 29.7 × 21
= 623.7 cm²

Question 2.
What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise, how many A4 sheets can fit on your table?
The dimensions of furniture like tables and chairs are sometimes measured in inches (in) and feet (ft).
1 in = 2.54 cm
1 ft = 12 in
Solution:
Do yourself

Question 3.
Express the following lengths in centimetres.
(i) 5 in
(ii) 7.4 in
Solution:
We know that 1 in = 254 cm
(i) We have, 5 in = 5 × 254 = 12.7 cm
(ii) We have, 7.4 in = 7.4 × 254 = 18.796 cm

Question 4.
Express the following lengths in inches.
(i) 5.08 cm
(ii) 11.43 cm
Solution:
We know that
1 cm = \(\frac{1}{2.54}\) in 254
(i) We have, 5.08 cm = \(\frac{5.08}{2.54}\) = 2 in
(ii) We have, 11.43 cm = \(\frac{11.43}{2.54}\) = 45 in

Question 5.
Convert 161.29 cm² to in².
Every 6.4516 cm² givens an in².
Hence, 161.29 cm² = \(\frac{161.29}{6.4516}\)in².
Evaluate the quotient.
Solution:
We have, \(\frac{161.29}{6.4516}\) = 25

Question 6.
What do you think is the area of your classroom? Areas of classroom, house etc. are generally measured in ft2 orm2.
Solution:
Do yourself

Question 7.
How many in² is 1 ft²?
Solution:
We know that
1 ft = 12 in
So, 1 ft² = (12)² in²
= (12 × 12) in²
= 144 in²

Question 8.
What do you think is the area of your school? Make an estimate and compare it with the actual data.
Solution:
Do yourself

Question 9.
Find out the local unit of area measurement in your region.
Solution:
Do yourself

Question 10
What do you think is the area of your village/town /city? Make an estimate and compare it with the actual data.
Solution:
Do yourself

Question 11.
How many m² is a km²?
Solution:
We know that 1 km = 1000 m
So, 1 km² =(1000 × 1000) m² = 1000000 m²

Question 12.
How many times is your village/town/city bigger than your school?
Solution:
Do yourself

Question 13.
Find the city with the largest area in
(i) India and
(ii) the world.
Solution:
(i) In India → Delhi has the largest area.
(ii) In world → New York city (USA) has the largest area.

Question 14.
Find the city with the smallest area in
(i) India and
(ii) the world.
Solution:In India → Mahe district (Puducherry) has the smallest area.
In world → Vatican city has the smallest area.