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Area Exercise 7.2 Class 8 Maths – Chapter 7 NCERT Solutions

NCERT Class 8 Maths Term 2 Chapter 7, Area Exercise 7.2, detailed solutions based on the updated syllabus are provided below. Prepared by the ALLEN expert team, these solutions give clear, step-by-step explanations for practice and revision.In Exercise 7.2 we deal with areas of triangles and unknown altitudes, converting rectangles and triangles into figures of equal area, and shaded regions and area relationships based on midpoints.

1.0Download Class 8 Maths Chapter 7 Ex 7.2 NCERT Solutions PDF

Download the Exercise 7.2 NCERT Solutions PDF to follow the solutions to the area-based questions and refer to them whenever you need to revisit the exercise.

Chapter 7 : Area Exercise 7.2

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.2 Class 8 Maths (Ganita Prakash Term 2)

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)

3.0Key Concepts of Chapter 7 Area Exercise 7.2

Calculating triangle areas from the given base and height.
Using the relationship between the areas of triangles AXC, AXB, and ABC to obtain altitude BY.
Using the equal-area halves of an isosceles triangle to obtain the area of triangle SUB.
Transforming a rectangle into an equal-area triangle using the Śulba-Sūtras construction.
Transforming a triangle into an equal-area rectangle using its base and half of its height.
Working out shaded areas formed by three identical squares and the area fraction of triangle XMN within triangle XYZ.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.2

Calculates the areas of triangles using their given base and height.
Uses related triangle areas to determine the unknown altitude BY.
Applies the equal-area halves of an isosceles triangle to calculate the area of triangle SUB.
Follows the Śulba-Sūtras construction for converting a rectangle into an equal-area triangle.
Converts a triangle into a rectangle with the same area using its base and height.
Calculates shaded regions formed by three identical squares and determines the area fraction of triangle XMN within triangle XYZ.

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

Exercise 7.2

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 7.2 Class 8 Maths – Chapter 7 NCERT Solutions

NCERT Class 8 Maths Term 2 Chapter 7, Area Exercise 7.2, detailed solutions based on the updated syllabus are provided below. Prepared by the ALLEN expert team, these solutions give clear, step-by-step explanations for practice and revision.In Exercise 7.2 we deal with areas of triangles and unknown altitudes, converting rectangles and triangles into figures of equal area, and shaded regions and area relationships based on midpoints.

1.0Download Class 8 Maths Chapter 7 Ex 7.2 NCERT Solutions PDF

Download the Exercise 7.2 NCERT Solutions PDF to follow the solutions to the area-based questions and refer to them whenever you need to revisit the exercise.

Chapter 7 : Area Exercise 7.2

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.2 Class 8 Maths (Ganita Prakash Term 2)

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)

3.0Key Concepts of Chapter 7 Area Exercise 7.2

Calculating triangle areas from the given base and height.
Using the relationship between the areas of triangles AXC, AXB, and ABC to obtain altitude BY.
Using the equal-area halves of an isosceles triangle to obtain the area of triangle SUB.
Transforming a rectangle into an equal-area triangle using the Śulba-Sūtras construction.
Transforming a triangle into an equal-area rectangle using its base and half of its height.
Working out shaded areas formed by three identical squares and the area fraction of triangle XMN within triangle XYZ.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.2

Calculates the areas of triangles using their given base and height.
Uses related triangle areas to determine the unknown altitude BY.
Applies the equal-area halves of an isosceles triangle to calculate the area of triangle SUB.
Follows the Śulba-Sūtras construction for converting a rectangle into an equal-area triangle.
Converts a triangle into a rectangle with the same area using its base and height.
Calculates shaded regions formed by three identical squares and determines the area fraction of triangle XMN within triangle XYZ.

Frequently Asked Questions

How is the area of a triangle calculated in Exercise 7.2?

How is altitude BY obtained in Exercise 7.2?

How is the area of triangle SUB calculated?

What equal-area transformations are included in Exercise 7.2?

What is the area relationship between triangles XMN and XYZ?

Join ALLEN!

Area Exercise 7.2 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.2 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.2 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.2

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.2

Frequently Asked Questions

How is the area of a triangle calculated in Exercise 7.2?

How is altitude BY obtained in Exercise 7.2?

How is the area of triangle SUB calculated?

What equal-area transformations are included in Exercise 7.2?

What is the area relationship between triangles XMN and XYZ?

Join ALLEN!

Area Exercise 7.2 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.2 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.2 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.2

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.2