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Area Exercise 7.1 Class 8 Maths – Chapter 7 NCERT Solutions

Class 8 Maths NCERT detailed solutions for Term 2 Chapter 7, Area Exercise 7.1, are given below and are specially prepared by the ALLEN expert team. These solutions provide step-by-step explanations of the questions for practice and revision.Exercise 7.1. The dimensions and areas of the inner and outer rectangles are given, and you have to find the missing side lengths of the rectangles and the area of the rectangular paths. Also it contains paths of constant width, outside rectangle movement, and crosspaths with overlapping parts.

1.0Download Class 8 Maths Chapter 7 Ex 7.1 NCERT Solutions PDF

Get the Exercise 7.1 NCERT Solutions PDF and refer to the worked-out answers while solving questions from the Area chapter.

Chapter 7 : Area Exercise 7.1

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.1 Class 8 Maths (Ganita Prakash Term 2)

1. Identify the missing side lengths. (i)

(ii)

Sol.

(i) After naming the figure in rectangle ABCD, Area of rectangle

=

Length × Breadth

7 \times BC = 21

\Rightarrow BC = 3 in

∴ AF = AD + DF = 3 in + 4 in = 7 in

In the rectangle EFAG,

EF \times AF = 28 in^{2}

\Rightarrow EF \times 7 in = 28 in^{2}

\Rightarrow EF = 4 in

∴ HA = HG + GA = 3 in + 4 in = 7 in

In rectangle HIJA, Area $= HA \times AJ$ $\Rightarrow 35 in^{2} = 7 in \times AJ$ $\Rightarrow AJ = 5 in$ $∴ AK = AJ + JK = 5 in + 2 in = 7 in$ In rectangle KLMA, Area $=$ KL × LM $\Rightarrow x in \times 7 in = 14 in^{2}$ $\Rightarrow x = 2$ in Thus, the missing side length $= 2$ in (ii) After naming the figure

Area $= 50 m^{2}$ In rectangle ABGH , Area $= AB \times AH$ $AB \times 4 m = 29 m^{2}$ $AB = \frac{29}{4} m$ or 7.25 m Area of rectangle HGDC $=$ Area of rectangle ABDC - Area of rectangle ABGH $= 50 m^{2} - 29 m^{2} = 21 m^{2}$

In rectangle $HGDC, CD \times GD = 21 m^{2}$ $\Rightarrow \frac{29}{4} \times GD = 21 m^{2}$ $\Rightarrow GD = \frac{84}{29} m$ or 2.9 m In rectangle $BEFG, BG \times BE = 11 m^{2}$ $\Rightarrow 4 m \times M = 11 m^{2}$ $\Rightarrow BE = \frac{11}{4} m$ or 2.75 m Thus, $AB = \frac{29}{4} m$ ; $BE = \frac{11}{4} m$ and $GD = \frac{84}{29} m$

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.] (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.] (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?

Sol. (i) Measurements needed: Length of outer rectangle $ABCD = A$ Width of outer rectangle $ABCD = B$ Length of inner rectangle $EFGH = a$

Width of inner rectangle $EFGH = b$ Let $A = 10, B = 8, a = 6, b = 4$ .

Calculation:

Area of path $=$ Area of ABCD - Area of EFGH $= 10 \times 8 - 6 \times 4 = 80 - 24 = 56 m^{2}$

Formula: Area of path $= (A \times B) - (a \times b)$ (ii) Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle. If the width of the path $d = 2 m$ (Uniform on all sides)

Length of inner path EFGH $= ℓ = 16 m$ Width of inner path EFGH $= w = 11 m$ Length of outer rectangle $= 1 + 2 d$ $= 16 + 2 (2) = 20 m$ Width of outer rectangle $= w + 2 d$ $= 11 + 2 (2) = 15 m$ Breaking the path into rectangles Now there are 4 rectangles Left rectangle $= w \times d = 11 \times 2 = 22 m^{2}$ Right rectangle $= w \times d = 11 \times 2 = 22 m^{2}$ Top rectangle $= (ℓ + 2 d) d = 20 \times 2 = 40 m^{2}$ Bottom rectangle $= (ℓ + 2 d) d = 20 \times 2$ $= 40 m^{2}$ Total area of path $= 22 + 22 + 40 + 40$ $= 124 m^{2}$ Formula: Area of path $= 2 d (ℓ + w) + 4 d^{2}$ where d - width of path $ℓ$ - length of inner path w - width of inner path (iii) No, the area of the path does not change. Reason: The area of the path depends only on: Area of outer rectangle $A BC D$ . Area of inner rectangle EFGH.

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.

Sol. Measurements needed: Length of plot $= 14 m$ Width of plot $= 12 m$ Width of horizontal path $= x 1$ Width of vertical path $= x 2$ Assign values: Let $x 1 = 2 m$ $x_{2} = 2 m$ Now area of horizontal path $= 14 \times 2 = 28 m$ Area of vertical path $= 12 \times 2 = 24 m$ Area of overlapping square $= 2 \times 2 = 4 m^{2}$ ∴ Area of cross path $= 28 + 24 - 4 = 48 m^{2}$

Formula:

Area of cross path $= (L \times w_{1}) + (W \times w_{2}) - (w_{1} \times w_{2})$ Here, $L =$ length of plot W = Width of plot $w_{1} =$ Width of horizontal path $w_{2} =$ Width of vertical path

4. 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.]

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

Sol. After naming the figure

The area of the spiral tube = Area of the rectangle, $ABEC +$ Area of the rectangle, DEGF + Area of the rectangle, GHIJ + Area of the rectangle, JKML + Area of the rectangle, NOPL + Area of the rectangle, PQRS + Area of the rectangle, STUV + Area of the rectangle, VWYX + Area of the rectangle, $XZA_{1} B_{1}$ $= AC \times AB + EG \times DE + IH \times JI + LJ \times LM + NO \times NL + PQ \times PS + UT \times ST + VX \times VW + ZA_{1} \times A_{1} B_{1} = 20 \times 1 + 18 \times 1 + 20 \times 1 + 13 \times 1 + 15 \times 1 + 8 \times 1 + 10 \times 1 + 3 \times 1 + 5 \times 1 = 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5 = 112$ sq. units

Thus, the area of the spiral tube is 112 sq . units.

Let the length of the straight tube be x . The area of the bent tube on the left = Area of rectangle, $BACD +$ Area of rectangle, $BGFE = AC \times CD + BG \times BE = 5 \times 1 + 4 \times 1 = 9$ sq. units

Area of straight tube $= x \times 1$ Area of bent tube $= 9$ sq. units According to the question, both are the same. So, $x \times 1 = 9$ $\Rightarrow x = 9$ units.

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

Sol. Let the side of the square be $a$ . Area of square $= a^{2}$ When side is doubled ⇒ new side = 2a New area $= (2 a)^{2} = 4 a^{2}$ So, total area becomes 4 times.

The square is divided into 3 regions $(1, 2, 3)$ by lines.

These lines divide the square into fixed proportions.

When the square is enlarged, all lengths double, so each region's area also becomes 4 times.

Let original areas be:

Region $1 = A_{1}$
Region $2 = A_{2}$
Region $3 = A_{3}$

After doubling:

New area of Region $1 = 4$ $A_{1} \Rightarrow$ Increase in area $= 4 A_{1} - A_{1} = 3 A_{1}$
New area of Region $2 = 4$ $A_{2} \Rightarrow$ Increase in area $= 3 A_{2}$
New area of Region $3 = 4$ $A_{3} \Rightarrow$ Increase in area $= 3 A_{3}$ The area of each region ( 1,2 and 3 ) becomes 4 times, so the increase in the areas of each region is 3 times its original area.

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.

Sol. 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.

Let us take a square of cardboard (8 $cm \times 8 cm$ ).
Draw two perpendicular lines inside the square (not through the center), dividing it into 4 rectangular pieces.
Cut along these lines to get 4 pieces.
Rearrange these 4 pieces.

Place them at the four corners of a larger imaginary square. The pieces should be arranged so that they form a square with a hole in the middle.

3.0Key Concepts of Chapter 7 Area Exercise 7.1

Finding unknown side lengths from the given area of a rectangle.
Finding the area of a rectangular path from the outer and inner rectangle areas.
Finding outer rectangle dimensions when the inner rectangle and path width are given.
Dividing a uniform-width rectangular path into smaller rectangles.
Using 2d(l + w) + 4d² to calculate the area of a uniform-width path.
Examining whether moving the outer rectangle changes the area of the path.
Finding the area of a crosspath by subtracting the overlapping portion.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.1

Finds unknown rectangle dimensions from the given area and one side.
Calculates the area of a rectangular path using the outer and inner rectangles.
Determines outer rectangle dimensions from the path width and inner dimensions.
Calculates a uniform-width path by dividing it into four rectangular parts.
Applies the formula 2d(l + w) + 4d² to rectangular paths.
Compares the path area before and after moving the outer rectangle.
Accounts for the common overlapping portion while calculating crosspath area.

NCERT Class 8 Maths Ch. 7 Area Other Exercises:-

Exercise 7.1

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 : Fractions in Disguise

Chapter 2 : The Baudhayana - The Pythagoras Theorem

Chapter 3 : Proportional Reasoning - 2

Chapter 4 : Exploring Some Geometric Themes

Chapter 5 : Tales by Dots and Lines

Chapter 6 : Algebra Play

Chapter 7 : Area

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Area Exercise 7.1 Class 8 Maths – Chapter 7 NCERT Solutions

Class 8 Maths NCERT detailed solutions for Term 2 Chapter 7, Area Exercise 7.1, are given below and are specially prepared by the ALLEN expert team. These solutions provide step-by-step explanations of the questions for practice and revision.Exercise 7.1. The dimensions and areas of the inner and outer rectangles are given, and you have to find the missing side lengths of the rectangles and the area of the rectangular paths. Also it contains paths of constant width, outside rectangle movement, and crosspaths with overlapping parts.

1.0Download Class 8 Maths Chapter 7 Ex 7.1 NCERT Solutions PDF

Get the Exercise 7.1 NCERT Solutions PDF and refer to the worked-out answers while solving questions from the Area chapter.

Chapter 7 : Area Exercise 7.1

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.1 Class 8 Maths (Ganita Prakash Term 2)

1. Identify the missing side lengths. (i)

(ii)

Sol.

(i) After naming the figure in rectangle ABCD, Area of rectangle

=

Length × Breadth

7 \times BC = 21

\Rightarrow BC = 3 in

∴ AF = AD + DF = 3 in + 4 in = 7 in

In the rectangle EFAG,

EF \times AF = 28 in^{2}

\Rightarrow EF \times 7 in = 28 in^{2}

\Rightarrow EF = 4 in

∴ HA = HG + GA = 3 in + 4 in = 7 in

In rectangle HIJA, Area $= HA \times AJ$ $\Rightarrow 35 in^{2} = 7 in \times AJ$ $\Rightarrow AJ = 5 in$ $∴ AK = AJ + JK = 5 in + 2 in = 7 in$ In rectangle KLMA, Area $=$ KL × LM $\Rightarrow x in \times 7 in = 14 in^{2}$ $\Rightarrow x = 2$ in Thus, the missing side length $= 2$ in (ii) After naming the figure

Area $= 50 m^{2}$ In rectangle ABGH , Area $= AB \times AH$ $AB \times 4 m = 29 m^{2}$ $AB = \frac{29}{4} m$ or 7.25 m Area of rectangle HGDC $=$ Area of rectangle ABDC - Area of rectangle ABGH $= 50 m^{2} - 29 m^{2} = 21 m^{2}$

In rectangle $HGDC, CD \times GD = 21 m^{2}$ $\Rightarrow \frac{29}{4} \times GD = 21 m^{2}$ $\Rightarrow GD = \frac{84}{29} m$ or 2.9 m In rectangle $BEFG, BG \times BE = 11 m^{2}$ $\Rightarrow 4 m \times M = 11 m^{2}$ $\Rightarrow BE = \frac{11}{4} m$ or 2.75 m Thus, $AB = \frac{29}{4} m$ ; $BE = \frac{11}{4} m$ and $GD = \frac{84}{29} m$

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.] (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.] (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?

Sol. (i) Measurements needed: Length of outer rectangle $ABCD = A$ Width of outer rectangle $ABCD = B$ Length of inner rectangle $EFGH = a$

Width of inner rectangle $EFGH = b$ Let $A = 10, B = 8, a = 6, b = 4$ .

Calculation:

Area of path $=$ Area of ABCD - Area of EFGH $= 10 \times 8 - 6 \times 4 = 80 - 24 = 56 m^{2}$

Formula: Area of path $= (A \times B) - (a \times b)$ (ii) Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle. If the width of the path $d = 2 m$ (Uniform on all sides)

Length of inner path EFGH $= ℓ = 16 m$ Width of inner path EFGH $= w = 11 m$ Length of outer rectangle $= 1 + 2 d$ $= 16 + 2 (2) = 20 m$ Width of outer rectangle $= w + 2 d$ $= 11 + 2 (2) = 15 m$ Breaking the path into rectangles Now there are 4 rectangles Left rectangle $= w \times d = 11 \times 2 = 22 m^{2}$ Right rectangle $= w \times d = 11 \times 2 = 22 m^{2}$ Top rectangle $= (ℓ + 2 d) d = 20 \times 2 = 40 m^{2}$ Bottom rectangle $= (ℓ + 2 d) d = 20 \times 2$ $= 40 m^{2}$ Total area of path $= 22 + 22 + 40 + 40$ $= 124 m^{2}$ Formula: Area of path $= 2 d (ℓ + w) + 4 d^{2}$ where d - width of path $ℓ$ - length of inner path w - width of inner path (iii) No, the area of the path does not change. Reason: The area of the path depends only on: Area of outer rectangle $A BC D$ . Area of inner rectangle EFGH.

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.

Sol. Measurements needed: Length of plot $= 14 m$ Width of plot $= 12 m$ Width of horizontal path $= x 1$ Width of vertical path $= x 2$ Assign values: Let $x 1 = 2 m$ $x_{2} = 2 m$ Now area of horizontal path $= 14 \times 2 = 28 m$ Area of vertical path $= 12 \times 2 = 24 m$ Area of overlapping square $= 2 \times 2 = 4 m^{2}$ ∴ Area of cross path $= 28 + 24 - 4 = 48 m^{2}$

Formula:

Area of cross path $= (L \times w_{1}) + (W \times w_{2}) - (w_{1} \times w_{2})$ Here, $L =$ length of plot W = Width of plot $w_{1} =$ Width of horizontal path $w_{2} =$ Width of vertical path

4. 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.]

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

Sol. After naming the figure

The area of the spiral tube = Area of the rectangle, $ABEC +$ Area of the rectangle, DEGF + Area of the rectangle, GHIJ + Area of the rectangle, JKML + Area of the rectangle, NOPL + Area of the rectangle, PQRS + Area of the rectangle, STUV + Area of the rectangle, VWYX + Area of the rectangle, $XZA_{1} B_{1}$ $= AC \times AB + EG \times DE + IH \times JI + LJ \times LM + NO \times NL + PQ \times PS + UT \times ST + VX \times VW + ZA_{1} \times A_{1} B_{1} = 20 \times 1 + 18 \times 1 + 20 \times 1 + 13 \times 1 + 15 \times 1 + 8 \times 1 + 10 \times 1 + 3 \times 1 + 5 \times 1 = 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5 = 112$ sq. units

Thus, the area of the spiral tube is 112 sq . units.

Let the length of the straight tube be x . The area of the bent tube on the left = Area of rectangle, $BACD +$ Area of rectangle, $BGFE = AC \times CD + BG \times BE = 5 \times 1 + 4 \times 1 = 9$ sq. units

Area of straight tube $= x \times 1$ Area of bent tube $= 9$ sq. units According to the question, both are the same. So, $x \times 1 = 9$ $\Rightarrow x = 9$ units.

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

Sol. Let the side of the square be $a$ . Area of square $= a^{2}$ When side is doubled ⇒ new side = 2a New area $= (2 a)^{2} = 4 a^{2}$ So, total area becomes 4 times.

The square is divided into 3 regions $(1, 2, 3)$ by lines.

These lines divide the square into fixed proportions.

When the square is enlarged, all lengths double, so each region's area also becomes 4 times.

Let original areas be:

Region $1 = A_{1}$
Region $2 = A_{2}$
Region $3 = A_{3}$

After doubling:

New area of Region $1 = 4$ $A_{1} \Rightarrow$ Increase in area $= 4 A_{1} - A_{1} = 3 A_{1}$
New area of Region $2 = 4$ $A_{2} \Rightarrow$ Increase in area $= 3 A_{2}$
New area of Region $3 = 4$ $A_{3} \Rightarrow$ Increase in area $= 3 A_{3}$ The area of each region ( 1,2 and 3 ) becomes 4 times, so the increase in the areas of each region is 3 times its original area.

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.

Sol. 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.

Let us take a square of cardboard (8 $cm \times 8 cm$ ).
Draw two perpendicular lines inside the square (not through the center), dividing it into 4 rectangular pieces.
Cut along these lines to get 4 pieces.
Rearrange these 4 pieces.

Place them at the four corners of a larger imaginary square. The pieces should be arranged so that they form a square with a hole in the middle.

3.0Key Concepts of Chapter 7 Area Exercise 7.1

Finding unknown side lengths from the given area of a rectangle.
Finding the area of a rectangular path from the outer and inner rectangle areas.
Finding outer rectangle dimensions when the inner rectangle and path width are given.
Dividing a uniform-width rectangular path into smaller rectangles.
Using 2d(l + w) + 4d² to calculate the area of a uniform-width path.
Examining whether moving the outer rectangle changes the area of the path.
Finding the area of a crosspath by subtracting the overlapping portion.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.1

Finds unknown rectangle dimensions from the given area and one side.
Calculates the area of a rectangular path using the outer and inner rectangles.
Determines outer rectangle dimensions from the path width and inner dimensions.
Calculates a uniform-width path by dividing it into four rectangular parts.
Applies the formula 2d(l + w) + 4d² to rectangular paths.
Compares the path area before and after moving the outer rectangle.
Accounts for the common overlapping portion while calculating crosspath area.

Frequently Asked Questions

What is covered in Exercise 7.1 of Class 8 Maths Chapter 7?

How are missing side lengths found in Exercise 7.1?

How is the area of a rectangular path calculated?

How is the area of a path with uniform width calculated?

Does moving the outer rectangle change the area of the path?

How is the area of a crosspath calculated?

Where can students find NCERT Solutions for Exercise 7.1?

Join ALLEN!

Area Exercise 7.1 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.1 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.1 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.1

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.1

Frequently Asked Questions

What is covered in Exercise 7.1 of Class 8 Maths Chapter 7?

How are missing side lengths found in Exercise 7.1?

How is the area of a rectangular path calculated?

How is the area of a path with uniform width calculated?

Does moving the outer rectangle change the area of the path?

How is the area of a crosspath calculated?

Where can students find NCERT Solutions for Exercise 7.1?

Join ALLEN!

Area Exercise 7.1 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.1 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.1 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.1

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.1