Chapter 7 Area

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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 7 Area

Chapter 7 - Area, Students will have a better understanding of how to measure different types of geometrical shapes with the 3 different types of parallelogram & rhombus/trapezium formulas thereby increasing their ability to quantify the amount of space occupied by various shapes used in artwork/architecture/land measurement.

A practical understanding of area is important in everyday life, from calculating the amount of paint required for walls to measuring the size of land tracts. NCERT Solutions for Class 8 provide students with a clear explanation of the derivation of the formulas for area. They also provide a visual representation for students of how to break down complicated polygons into simpler shapes so that they can develop an overall understanding of spatial measurement.

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 7 Solutions

Download the free NCERT Class 8 Maths Ganith Prakash 2 Chapter 7 Solutions PDF to learn important concepts, practice textbook questions, and improve problem-solving skills for exams and daily revision.

Chapter 7 : Area

2.0Key Concepts covered in Class 8 Maths : Area

This chapter provides the formulas and logical methods required to calculate the surface space of two-dimensional figures.

Area of a Triangle: Reconfirming the fundamental formula: $Area = \frac{1}{2} \times base \times height$ .
Area of a Parallelogram: Learning that the area is simply the product of its base and corresponding height $(Base \times Height)$ .
Polygon Decomposition: A powerful technique where any complex polygon is evaluated by breaking it into smaller triangles and quadrilaterals.
Units of Measurement: Exploring various units from $c m^{2}$ and $m^{2}$ to larger units like hectares and $k m^{2}$ .

3.0NCERT Solutions for Class 8 Maths Chapter 7 : All Exercises

Exercise 7.1 Students explore creative area decomposition. Problems involve dividing squares and rectangles into equal parts and understanding how stretching or compressing shapes impacts their total surface area while maintaining constant spatial measurements.

Exercise 7.2 This exercise uses parallelograms and rhombuses as the bases of calculating area through the use of both base-height and diagonal formula approaches. The student will calculate the area of a number of different four-sided shapes but will make sure that they are using vertical heights rather than slanted side lengths.

Exercise 7.3 This exercise looks at calculating area in a trapezium, with problems helping them work through the formula using the sum of the two parallel sides and the vertical distance between them, with solutions to problems associated with field plots or buildings.

Exercise 7.4 In the exercise, students are taught how to calculate the area of a polygon. They complete exercises where they break complex and irregular shapes into small triangles and quadrilaterals, calculate the area of each piece, and add these areas together to find the whole area of the shape.

Exercise 7.5 Area concepts are used in real-world land measurements in this final exercise. Students will calculate the area of large irregularly-shaped parcels of land by solving 'field book' type problems and converting units of measurement such as hectares to the local form of measurement.

4.0Detailed Class 8 Maths Chapter 7 Area - NCERT Solutions

Figure it out-01

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.

Figure it out-02

1. Find the areas of the following triangles: (i)

(ii)

(iii)

Sol. (i) Area of triangle ABC

$= \frac{1}{2} \times base \times height = \frac{1}{2} \times BC \times AE = \frac{1}{2} \times 4 cm \times 3 cm = 6 cm^{2}$

Thus, the area of the triangle, $ABC = 6 cm^{2}$ (ii) Area of triangle $DEF = \frac{1}{2} \times EF \times ND$ $= \frac{1}{2} \times 5 cm \times 3.2 cm = 5 \times 1.6 cm^{2} = 8 cm^{2}$ Thus, the area of the triangle $DEF = 8 cm^{2}$ (iii) Area of triangle $= \frac{1}{2} \times$ base × height

Area of $Δ NAT = \frac{1}{2} \times AT \times NA$ $= \frac{1}{2} \times 3 cm \times 4 cm = 6 cm^{2}$ Thus, the area of the triangle, NAT $= 6 cm^{2}$

2. Find the length of the altitude BY.

Sol.

Area of $△ AXC = \frac{1}{2} \times XC \times AX = \frac{1}{2} \times (XB + 6) \times 4 = 2 \times (XB + 6) = 2 \times XB + 12$ sq. units Area of $△ AXB = \frac{1}{2} \times XB \times AX = \frac{1}{2} \times XB \times 4 = 2 \times XB$ sq. units Area of $△ ABC = \frac{1}{2} \times AC \times BY = \frac{1}{2} \times 8 \times BY = 4 BY$ ∴ Area of $△ AXC =$ Area of $△ AXB +$ Area of $△ ABC$ $2 XB + 12 = 2 XB + 4 BY$ $\Rightarrow 4 BY = 12$ $\Rightarrow BY = 3$ units Thus, the length of the altitude BY is 3 units.

3. Find the area of $△ SUB$ , given that it is isosceles, SE is perpendicular to UB and the area of $△ SEB$ is 24 sq. units.

Sol. Given, The area of $△ SEB = 24$ sq. units Given that $△ SUB$ is an isosceles triangle. SU and SB are equal sides and UB is the base. $∴ SE$ is perpendicular to UB $\Rightarrow UE = EB$ SE is the common base of $△ SUE$ and $△ SEB$ . $∴$ Area of $△ SEB$

$= 24 sq. units = Area of △ SEU$

∴ The area of triangle SUB

$= Area of △ SEU + Area of △ SEB = 24 + 24 = 48 sq. units$

Thus, the area of $△ SUB$ is 48 sq. units.

4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area. Sol. 1. Let us take a rectangle ABCD , with length $a$ and breadth $b$ .

2. Now mark the midpoint E of side CD. 3. Draw a line perpendicular bisector to CD passing through E. Mark a point M on it such that

ME = b

. 4. Draw a triangle using base

= AB

(same as the rectangle's length) and join M to A and B .

5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

Sol. 1. Take a triangle ABC with base b and height h. 2. Find the midpoint M of the height. 3. Draw a line parallel to the base through M.

This line intersects the sides of the triangle. 4. Create a rectangle using.

Length = Same as the triangle's base = b Width $=$ Half of the triangle's height $= \frac{h}{2}$

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?

Sol. Given that $ABCD, BCEF$ and BFGH are identical.

The area of the red region ( $△ HBI + IBCD$ ) $= 49$ sq. units Let the side of each square be a $∴ IB = \frac{AB}{2} = \frac{a}{2}$ units Let ' $a$ ' unit be the side of the square. (i) Area of the red region

$△ HDC = \frac{1}{2} \times DC \times HC (∴ HC = HB + BC = a + a = 2 a) \Rightarrow \frac{1}{2} \times a \times 2 a \Rightarrow \frac{1}{2} \times 2 a^{2} = 49 \Rightarrow a = 7 units$

∴ The area of the blue region, $Δ IAD = \frac{1}{2} \times AI \times AD = \frac{1}{2} \times \frac{7}{2} \times 7 = \frac{49}{4}$ sq. units $= 12.25$ sq. units

Thus, the area of the blue region is 12.25 square units. (ii) Given, the total area enclosed by the blue and red regions $=$ Area of $△ HDC +$ Area of $△ AID = 180$ sq. units Let ' $a$ ' be the side of the square. $\frac{1}{2} \times DC \times HC + \frac{1}{2} \times AI \times AD = 180 sq$ . units

$= \frac{1}{2} \times a \times 2 a + \frac{1}{2} \times \frac{a}{2} \times a = 180 \Rightarrow a^{2} + \frac{a ^{2}}{4} = 180 \Rightarrow \frac{4 a ^{2} + a ^{2}}{4} = 180$ $\Rightarrow \frac{5 a ^{2}}{4} = 180 \Rightarrow 5 a^{2} = 180 \times 4 \Rightarrow a^{2} = \frac{180 \times 4}{5} = 36 \times 4 = 144 \Rightarrow a^{2} = (12)^{2} \Rightarrow a = 12 sq. units$

∴ The area of each square $= a^{2}$

$= (12)^{2} = 144 sq. units$

Thus, the area of each square is 144 sq. units.

7. If M and N are the midpoints of XY and XZ , what fraction of the area of $△ XYZ$ is the area of $△ XMN$ ? [Hint: Join NY]

Sol. Join NY, YN is median ( ∵ N is mid point of XZ ) Median YN divides $△ XYZ$ into two equal areas so, area of $△ XYN = \frac{1}{2} \times$ area of $△ XYZ$

NM is median of $△ XYN$ ( ∵ M is mid point of XY) Median NM divides $△ XYN$ into two equal areas so, area of $△ XMN = \frac{1}{2} \times$ area of $△ XYN$

From equation

Area of $△ XMN = \frac{1}{2} \times \frac{1}{2} \times$ area of $△ XYZ$

$= \frac{1}{4} \times area of Δ XYZ$

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. Sol. Join HT, Draw perpendicular bisector of HT which join riverbank at P . Shortest path from his house to the river and then to the water tank in H to P then P to T .
(House)

Figure it out-03

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 .

Sol. Area of the quadrilateral $ABCD =$ Area of triangle $CAD +$ Area of triangle ACB Area of $△ ACB = \frac{1}{2} \times AC \times BM$ $= \frac{1}{2} \times 22 cm \times 3 cm$ $= 33 cm^{2}$ Area of $△ CAD = \frac{1}{2} \times AC \times DN$ $= \frac{1}{2} \times 22 cm \times 3 cm$ $= 33 cm^{2}$ ∴ The area of the quadrilateral $ABCD = 33 cm^{2} + 33 cm^{2} = 66 cm^{2}$

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

Sol. The area of the shaded region $=$ Area of the rectangle $ABCD -$ (Area of triangle $AEF +$ Area of triangle EBC ) Area of the rectangle ABCD $=$ Length × Breadth $= AB \times AD$ $= 18 cm \times 10 cm$ $[∵ AB = AE + EB = 10 cm + 8 cm = 18 cm;$ $AD = AF + FD = 6 cm + 4 cm = 10 cm$ ] $= 180 cm^{2}$ Area of the triangle AEF $= \frac{1}{2} \times AE \times AF$ $= \frac{1}{2} \times 10 cm \times 6 cm = 30 cm^{2}$ Area of the triangle $EBC = \frac{1}{2} \times EB \times BC$ $= \frac{1}{2} \times 8 cm \times 10 cm = 40 cm^{2}$ $∴$ The area of the shaded region $= 180$ $cm^{2} - (30 cm^{2} + 40 cm^{2})$ $= 180 cm^{2} - 70 cm^{2}$ $= 110 cm^{2}$ 3. What measurements would you need to find the area of a regular hexagon?

Sol. Side length ( $ℓ$ ) of the hexagon needed to find the area of regular hexagon.

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

Sol. Let $ℓ$ be the length and b be the breadth of the rectangle ABCD .

Total area of rectangle, $ABCD = DC \times BC = ℓ \times$ b sq. units

Area of $△ AOB = \frac{1}{2} \times AB \times OE = \frac{1}{2} \times ℓ \times xsq$ . units Area of $△ DOC = \frac{1}{2} \times DC \times OF = \frac{1}{2} \times ℓ \times y$ sq. units ∴ The area of the blue region

$= Area of △ AOB + Area of △ DOC = \frac{1}{2} \times ℓ \times x + \frac{1}{2} \times ℓ \times y = \frac{1}{2} \times ℓ \times (x + y) sq. units$ $= \frac{1}{2} \times ℓ \times b sq. units [∴ x + y = b]$

∴ Area of blue region $= \frac{1}{2} \times$ Area of rectangle Thus, the required fraction is $\frac{1}{2}$ .

5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral. Sol. Let ABCD be a given quadrilateral.

Mark mid points of $AB, BC, CD$ and DA as P, Q, R and S. Join midpoints, then PQRS is the required quadrilateral with half the area of the given quadrilateral ABCD .

Figure it out-04

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 figures appears to have the maximum perimeter and which has the minimum perimeter?

Sol. (i) (a) Area of parallelogram $=$ base × height $= 5 \times 3 = 15$ sq. units (b) Area of parallelogram $= 5 \times 3 = 15$ sq. units (c) Area of parallelogram $= 5 \times 3 = 15$ sq. units (d) Area of parallelogram $= 5 \times 3 = 15$ sq. units (e) Area of parallelogram $= 5 \times 3 = 15$ sq. units (f) Area of parallelogram $= 5 \times 3 = 15$ sq. units (g) Area of parallelogram $= 5 \times 3 = 15$ sq. units

All parallelograms have equal areas. (ii) The perimeters of these parallelograms are different even though their areas are the same.

Figure (d) has the minimum perimeter and Figure (g) has the maximum perimeter.

2. Find the area of the following parallelograms. (i)

(ii)

(iii)

(iv)

Sol. Area of the parallelogram $=$ base × height (i) Here, base $= 7 cm$ and height $= 4 cm$ Area of the parallelogram $= 7 cm \times 4 cm = 28 cm^{2}$ (ii) Here, base $= 5 cm$ and height $= 3 cm$ Area of the parallelogram $= 5 cm \times 3 cm = 15 cm^{2}$ (iii) Here, base $= 5 cm$ and height $= 4.8 cm$ Area of the parallelogram $= 5 cm \times 4.8 cm = 24 cm^{2}$ (iv) Here, base $= 2 cm$ and height $= 4.4 cm$ Area of the parallelogram $= 2 cm \times 4.4 cm = 8.8 cm^{2}$

3. Find QN.

Sol. In $△ PNQ, ∠ PNQ = 9 0^{\circ}$ $PN = 7.6 cm$ $PQ = 12 cm$ By Pythagoras theorem $PQ^{2} = PN^{2} + NQ^{2}$ $\Rightarrow (12)^{2} = (7.6)^{2} + NQ^{2}$ $\Rightarrow Q N^{2} = (12)^{2} - (7.6)^{2}$ $\Rightarrow QN^{2} = 144 - 57.76$ $\Rightarrow QN^{2} = 86.24$ $\Rightarrow QN = 9.28 cm$

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

Sol. For rectangle: $ℓ = 5, w = 4$ , all angles $= 9 0^{\circ}$ ∴ Area $= ℓ \times w = 5 \times 4 = 20 cm^{2}$ For parallelogram: Base (b) $= 5 cm$ , one slanted side $= 4 cm$ Height will be less than 4 cm because the side is slanted.

Area $= 5 \times 4 < 20 cm^{2}$ Hence, the rectangle has a greater area than the parallelogram.

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Sol. Given: Triangle with area A Required: Rectangle with area $= 2 A$

Method 1:

If the triangle has base b and height h Area of triangle $= \frac{1}{2} \times b \times h = A$

To get a rectangle with an area of 2 A . Take length $= b$ , width $= h$

Area of a rectangle $= b \times h$ $= 2 \times (\frac{1}{2} \times b \times h) = 2 A$

Steps:

Measure the base and height of the given triangle.

Construct a rectangle with these measurements as length and width.

Method 2:

Scaling method:

Take the rectangle.
Create a rectangle with base = (base of triangle) and height $=$ height of triangle
This rectangle automatically has twice the area of the triangle.

6. Give a method to obtain a rectangle of the same area as a given triangle.

Sol. Given: A triangle with base b and height h . Required: Rectangle with the same area Area of triangle $= \frac{1}{2} bh$

To get a rectangle with the same area Rectangle length $= \frac{b}{2}$ (half the triangle's base) Rectangle width $= h$ (same as the triangle's height) Area $= \frac{b}{2} \times h$ $= \frac{1}{2} \times b \times h$

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

Sol. Given: Isosceles triangle ABC , where $AB =$ AC and AD is the altitude from A to BC . Method: Since the triangle is isosceles: AD is perpendicular to $BC . D$ is the midpoint of BC (property of an isosceles triangle). AD bisects the triangle into two congruent right triangles: $△ ADB$ and $△ ADC$ .

Dissection Process:

Step 1: The altitude AD divides the isosceles triangle into two congruent right triangles $△ ADB$ and $△ ADC$ . Step 2: Each of these right triangles can be cut into pieces. Step 3: Assembly: Take triangle $△ ADB$ and $△ ADC$ , rotate one triangle $18 0^{\circ}$ arrange them.

8. Give a method to convert a rectangle into an isosceles triangle by dissection.

Sol. Step-1: Take rectangle PQRS with length $ℓ$ and width w .

Step-2: Cut it along diagonal PR.

Step-3: Rotate $△ PQR$ and join with other $Δ$ in such way that PQ coincide with RS.

Allow triangle in required isosceles $Δ$ . 9. Which has greater area, an equilateral triangle or a square of the same side length as the triangle? Which has greater area two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons. Sol. Area of equilateral triangle $= \frac{3}{4} a^{2}$ Area of square $= a^{2}$ $\Rightarrow \frac{3}{4} a^{2} < a^{2}$ So, the area of a square is greater than the area of an equilateral triangle of the same side length. Area of two identical equilateral triangles $= \frac{3}{4} a^{2} + \frac{3}{4} a^{2} = \frac{2 3}{4} a^{2} = \frac{3}{2} a^{2}$ Area of square of side length $a = a^{2}$ Clearly, $\frac{3}{2} a^{2} < a^{2}$ So, the area of a square is greater than the area of two identical equilateral triangles.

Figure it out-05

1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm

Sol. Given, first diagonal $= 20 cm$ second diagonal $= 15 cm$

The area of a rhombus $= \frac{1}{2} \times$ (Product of diagonals) $= \frac{1}{2} \times$ First diagonal × second diagonal $= \frac{1}{2} \times 20 cm \times 15 cm = 150 cm^{2}$ Thus, the area of a rhombus is $150 cm^{2}$ 2. Give a method to convert a rectangle into a rhombus of equal area using dissection.

Sol. This is the reverse of the rhombus to rectangle dissection.

Method:

Given: Rectangle PQRS with length I and width W

Required:

Rhombus with the same area $= ℓ \times W$

Dissection Process:

Step 1: The rhombus will have diagonals $d_{1}$ and $d_{2}$ such that: $\frac{1}{2} \times d_{1} \times d_{2} = ℓ \times w$ So, $d_{1} \times d_{2} = 2 lw$ . Step 2: Choose convenient diagonal lengths: Let $d_{1} = 2 ℓ$ (twice the rectangle length)

Then $d_{2} = w$ (same as rectangle width) Check: $\frac{1}{2} \times 2 ℓ \times w = ℓ w$

Let $d_{1} = 2 w$ (twice the rectangle width) Then $d_{2} = ℓ$ (same as rectangle length) Step 3: Dissection process (reverse of textbook method):

Divide the rectangle into two halves Mark the center point 0 . Cut and rotate pieces to form two isosceles triangles. Arrange these triangles to share a common diagonal. This creates a rhombus.

3. Find the area of the following figures:

(ii)

(iii)

(iv)

Sol. The area of the trapezium $= \frac{1}{2} \times$ (Sum of parallel sides) $\times$ (Distance between them) $= \frac{1}{2} \times (a + b) \times h$ (i) Here, $a = 10 ft, b = 7 ft$ and $h = 16 ft$ Area of trapezium $= \frac{1}{2} \times (10 + 7) \times 16 = 17 \times 8 = 136 ft^{2}$ (ii) Here, $a = 36 m, b$ $= 24 m$ and $h = 14 m$ Area of trapezium

$= \frac{1}{2} \times (36 + 24) \times 14 = 60 \times 7 = 420 m^{2}$

(iii) Here, $a = 14 in, b = 6 in$ and $h = 10 in$

Area of trapezium $= \frac{1}{2} \times (14 + 6) \times 10 = 20 \times 5 = 100 in^{2}$ (iv) Here, $a = 18 ft, b = 12 ft$ and $h = 8 ft$

Area of trapezium

$= \frac{1}{2} \times (18 + 12) \times 8 = 30 \times 4 = 120 ft^{2}$

Give a method to convert an isosceles trapezium to a rectangle using dissection. Sol. An isosceles trapezium has special properties that make dissection simple.

Properties of Isosceles Trapezium ABCD: $AB ∥ CD$ (parallel sides) $AD = BC$ (non-parallel sides are equal) $∠ A = ∠ B$ and $∠ D = ∠ C$ (base angles are equal)

Dissection Method:

Step-1: Take an isosceles trapezium, in which $AB ∥ CD$ & $AD = BC$ .

Step 2: Draw AP & BQ perpendicular to DC . Here $△ ADP ≅ △ BCQ$ .

Step 3: Cut $△ ADP$ and join to the other side to get rectangle.

WXYZ is required rectangle.

Resulting Rectangle:

Length = CD + AP $= CD + \frac{( AB - CD )}{2}$ $= \frac{( AB + CD )}{2}$ Width $= h$ (height of trapezium) Area $= \frac{( AB + CD )}{2} \times h = \frac{1}{2} h (AB + CD)$ This matches the trapezium area formula

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

Given the trapezium ABCD , how do we find the vertices of the rectangle EFGH ? [Hint: If $△ AHI ≅ △ DGI$ and $△ BEJ ≅ △ CFJ$ , then the trapezium and rectangle have equal areas.]

Sol. Given: Trapezium ABCD with $AB ∥ CD$ Required:

Find the positions of the vertices E, F, G and H to form a rectangle EFGH with equal area

Using the Hint:

If $Δ AHI ≅ Δ DGI$ and $Δ BEJ ≅ Δ CFJ$ , then areas are equal.

Method:

Step 1: The rectangle $EFG H$ should have: EF as one side (top side) GH is the opposite parallel side (bottom side)

EH and FG are the other pair of sides.

Step 2: Position the rectangle such that: Points I and J are strategically chosen on the trapezium.

Step 3: For congruency:

Mark I on side AD Mark J on side BC Choose positions such that: HI=GI (making $△ AHI ≅ △ DGI$ possible) $EJ = FJ$ (making $Δ BEJ ≅ △ CFJ$ possible) Step 4: The height of the rectangle $= h$ (height of the trapezium) Step 5: The length of rectangle $= \frac{( a + b )}{2}$ where $a$ and $b$ are parallel sides This ensures: Area of rectangle $= h \times \frac{( a + b )}{2}$ = Area of trapezium

Practical construction:

Draw the trapezium ABCD Calculate required rectangle length $= \frac{( AB + CD )}{2}$ Mark points H and E on AB such that the central portion has this length.

Draw perpendiculars to get rectangle EFGH.

Verify that the triangular pieces outside match those inside.

This construction beautifully demonstrates area conservation through dissection!

6. Using the idea of converting a trapezium into a rectangle of equal area and vice versa, construct a trapezium of area $144 cm^{2}$ . Sol. Area of the trapezium $= \frac{1}{2} \times (10 + 8) \times 16$ $= 144 cm^{2}$ Area of rectangle $= 16 \times 9 = 144 cm^{2}$

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

Sol. Here total area of hexagon $= 6 \times \frac{3}{4} a^{2} = \frac{3 3}{2} a^{2}$

Equilateral triangle:

Area $= \frac{3}{4} a^{2}$

Rhombus:

Area $= 2 \times \frac{3}{4} a^{2} = \frac{3}{2} a^{2}$

Trapezium:

Remaining Area = Total Area - Triangle Area - Rhombus Area $= \frac{6 3}{4} - \frac{3}{4} a^{2} - \frac{3}{2} a^{2}$ $= \frac{3 3}{4} a^{2}$

Ratio of area

= Triangle : Rhombus : Trapezium $= \frac{3}{4} a^{2} : \frac{3}{2} a^{2} : \frac{3 3}{4} a^{2} = 1 : 2 : 3$

8. ZYXW is a trapezium with $ZY ∥ WX . A$ is the midpoint of XY . Show that the area of the trapezium ZYXW is equal to the area of $Δ ZWB .$ .

Sol. $∠ ZAY = ∠ BAX$ (Vertically opposite angles)

A Y = A X (∵ A

is mid point of

X Y)

∴ ∠ YZB = ∠ XBZ

( ∵ ZY || XB alternate interior angles are equal)

∠ ZYA = ∠ BXA

( ∵ they are alternate interior angles) So

△ ZAY ≅ △ BAX

By the AAS congruence. So, Area of

Δ ZAY =

Area of

Δ BAX

Thus, Area of trapezium ZYXW = Area of triangle ZWB

5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 7

Visual Derivations: The solutions don't just provide formulas; they show how a parallelogram can be transformed into a rectangle or a trapezium into triangles, making the math intuitive.
Practical Contexts: These types of problem-based tasks relate to real-world scenarios, such as Rangoli patterns and measuring land with regional units (bigha, cent). Learning how to do this contributes to knowledge of daily life.
Step-by-Step Polygon Analysis: The strategies include step-by-step methods for calculating the area of irregularly shaped pieces of land using a "field book" approach. The students are taught to use these strategies systematically to find an area for any irregular shaped piece of land.
Critical Thinking with Area: Through "Math Talk" activities, students explore how shapes can be compressed and expanded while maintaining the same area, fostering deep geometric insight.
Enhanced Calculation Accuracy: Working through various worked examples can help students to identify common problems and differences between the 'slant height' and the 'vertical height' as used in calculating areas.
Academic and Competitive Readiness: This chapter is part of the CBSE curriculum (core) and is an essential component in many Olympiads where geometric measurement and properties have a high frequency of occurrence.

Frequently Asked Questions

What concepts are covered in NCERT Solutions for Class 8 Maths Chapter 7 Area?

How do NCERT Solutions for Class 8 help students understand area formulas?

Why is the topic of area important in mathematics?

What kinds of shapes are discussed in this chapter?

Can NCERT Solutions for Class 8 help students improve problem-solving skills?

How can students score better in area-based questions?

Are practical applications of area explained in this chapter?

Do NCERT Solutions for Class 8 include answers for all textbook exercises?

Why should students practice mensuration topics regularly?

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Frequently Asked Questions

What concepts are covered in NCERT Solutions for Class 8 Maths Chapter 7 Area?

How do NCERT Solutions for Class 8 help students understand area formulas?

Why is the topic of area important in mathematics?

What kinds of shapes are discussed in this chapter?

Can NCERT Solutions for Class 8 help students improve problem-solving skills?

How can students score better in area-based questions?

Are practical applications of area explained in this chapter?

Do NCERT Solutions for Class 8 include answers for all textbook exercises?

Why should students practice mensuration topics regularly?

Are these NCERT Solutions for Class 8 based on the latest CBSE syllabus?

Join ALLEN!

NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 7 Area

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 7 Solutions

2.0Key Concepts covered in Class 8 Maths : Area

3.0NCERT Solutions for Class 8 Maths Chapter 7 : All Exercises

4.0Detailed Class 8 Maths Chapter 7 Area - NCERT Solutions

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5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 7