NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 7: Area

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• Figure it Out: Page 150
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In Class 8 Ganita Prakash Part 2 Chapter 7, Cutting and rearranging shapes, a method called dissection that also appears in the Sulba Sutras, then produces the formulas for the parallelogram, rhombus and trapezium. Along the way, the chapter shows why perimeter cannot measure area, uses reflections to find shortest paths and ends with real-life units, from square inches and acres to local measures such as bigha and katha.
Get here the Solutions and Answers of Figure it Out at Page 150.

Answer:
(i) Height of the rectangle of area 21 in² = 21 ÷ 7 = 3 in.
Height of the rectangle of area 28 in² = 4 + 3 = 7 in.
Its width = 28 ÷ 7 = 4 in.
Width of the rectangle of area 35 in² = 3 + 4 = 7 in.
Its height = 35 ÷ 7 = 5 in.
Height of the rectangle of area 14 in² = 5 + 2 = 7 in.
Therefore, the missing sidelength = 14 ÷ 7 = 2 in.

(ii) Width of the rectangle of area 29 m² = 29 ÷ 4 = 7.25 m.
Width of the rectangle of area 11 m² = 11 ÷ 4 = 2.75 m.
Area of the lower rectangle = 50 − 29 = 21 m².
Its width is also 7.25 m.
Therefore, its height = 21 ÷ 7.25 = 84/29 m.
The missing sidelengths are 7.25 m, 2.75 m and 84/29 m.

(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?

Answer:
(i) We need the length and width of both the outer rectangle ABCD and the park EFGH.
I take the outer rectangle to be 24 m long and 16 m wide, and the park to be 20 m long and 12 m wide.
Area of the outer rectangle = 24 × 16 = 384 m².
Area of the park = 20 × 12 = 240 m².
Area of the path = 384 − 240 = 144 m².
If the outer length and width are L and B, and the park’s length and width are l and b, then
Area of the path = LB − lb.

(ii) The widths of the path alone are not enough. We also need the length and width of the park.
I take the park’s length as 20 m, its width as 12 m, and the path’s width as 2 m along each side.
The two strips beside the longer sides, including the four corners, each have dimensions 24 m × 2 m.
The other two strips each have dimensions 12 m × 2 m.
Area of the path = 2 × 24 × 2 + 2 × 12 × 2
= 96 + 48
= 144 m².
For a park of length l and width b, with a path of uniform width w,
Area of the path = 2w(l + 2w) + 2bw
= 2w(l + b) + 4w².
If the widths on the left, right, top and bottom are p, q, r and s respectively, then
Area of the path = (l + p + q)(b + r + s) − lb.

(iii) No, the area of the path does not change.
The areas of the outer rectangle and the park remain the same.
Since area of the path = area of the outer rectangle – area of the park, the area of the path remains unchanged.

Answer:
We need the widths of the two paths.
I take each path to be 1 m wide.
Area of the path along the 14 m side = 14 × 1 = 14 m².
Area of the path along the 12 m side = 12 × 1 = 12 m².
The 1 m × 1 m square where the paths cross is counted twice.
So, we subtract its area once.
Area of the crosspath = 14 + 12 − 1 = 25 m².
If the widths of the paths along the 14 m and 12 m sides are x m and y m respectively, then
Area of the crosspath = (14x + 12y − xy) m².

[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?
Answer:
The width of the tube is 1 unit.
In the bent tube in the hint, the two strips each measure 5 units × 1 unit.
The 1 unit × 1 unit square at the bend is counted twice.
Area of the bent tube = 5 × 1 + 5 × 1 − 1 × 1
= 9 square units.
So, the straight tube of width 1 unit must have length 9 units.

For the spiral tube, the lengths of the nine strips are 20, 20, 20, 15, 15, 10, 10, 5 and 5 units.
Sum of their areas = (20 + 20 + 20 + 15 + 15 + 10 + 10 + 5 + 5) × 1
= 120 square units.
There are 8 bends. At each bend, a 1 unit × 1 unit square has been counted twice.
Area of the spiral tube = 120 − 8 × 1
= 112 square units.

Answer:
Let the original sidelength of the square be a units.
Its area = a² square units.
The diagonal divides the square into two equal areas. So, region 3 has area a²/2.
The other half is divided into two triangles with equal bases and the same height.
So, regions 1 and 2 each have area a²/4.

When the sidelength is doubled, the new area of the square = (2a) × (2a) = 4a².
The new areas are:
Region 1: 4a²/4 = a².
Region 2: 4a²/4 = a².
Region 3: 4a²/2 = 2a².
Thus, the area of each region becomes 4 times its original area.
The increase in each region is 3 times its original area or 300%.

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.
Answer:
I drew two slanting perpendicular lines through the centre of a cardboard square and cut along them. I marked the four pieces I, II, III and IV.
I turned the pieces and arranged them as shown below. The four right angles formed at the intersection of the cuts became the four corners of the larger square.

The pieces form a larger square with a square hole inside. The total area of the four pieces remains equal to the area of the original square.

Get here the Solutions and Answers of Figure it Out at Page 157.

Answer:
Area of a triangle = 1/2 × base × height.
(i) In ∆ABC, BC = 4 cm and AE = 3 cm.
Area of ∆ABC = 1/2 × 4 × 3 = 6 cm².

(ii) In ∆DEF, EF = 5 cm and DN = 3.2 cm.
Area of ∆DEF = 1/2 × 5 × 3.2 = 8 cm².

(iii) In ∆ANT, AT = 3 cm and AN = 4 cm.
Area of ∆ANT = 1/2 × 3 × 4 = 6 cm².

Answer:
Taking BC as the base, the corresponding height is AX.
Area of ∆ABC = 1/2 × BC × AX
= 1/2 × 6 × 4
= 12 square units.

Taking AC as the base, the corresponding height is BY.
12 = 1/2 × 8 × BY
12 = 4 × BY
BY = 12 ÷ 4 = 3 units.
Therefore, the altitude BY is 3 units.

Answer:
In the isosceles triangle SUB, the altitude SE divides the base UB into two equal parts.
So, UE = EB.
Triangles SEU and SEB have equal bases and the same height SE.
Therefore, their areas are equal.
Area of ∆SEU = 24 square units.
Area of ∆SUB = 24 + 24 = 48 square units.

Answer:
Take a rectangle ABCD. Extend AB to E so that BE = AB. Join DE.

Triangle DAE has base AE = 2 × AB and height AD.
Area of ∆DAE = 1/2 × AE × AD
= 1/2 × 2 × AB × AD
= AB × AD.
Area of rectangle ABCD = AB × AD.
Therefore, ∆DAE and rectangle ABCD have equal areas.

Answer:
Take a triangle ABC with base BC = b and altitude AX = h.
Draw a rectangle PQRS of length b and width h/2.

Area of ∆ABC = 1/2 × b × h.
Area of rectangle PQRS = b × h/2 = 1/2 × b × h.
Therefore, the rectangle and the triangle have equal areas.

(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?

Answer:
Let the sidelength of each square be s units. Let DH meet AB at P.

DA = BH and AD is parallel to BH.
The right triangles DAP and HBP also have equal angles at P.
So, they are congruent.
Therefore, AP = PB = s/2.
The red region is ∆DCH.
DC = s and CH = 2s.
Area of the red region = 1/2 × s × 2s = s².
The blue region is ∆DAP.
Area of the blue region = 1/2 × s × s/2 = s²/4.
Thus, the red region has the same area as one square and the blue region has one-fourth of that area.

(i) Area of the blue region = 49 ÷ 4 = 12.25 square units.

(ii) Let the area of each square be S square units.
S + S/4 = 180
5S/4 = 180
S = 180 × 4/5 = 144.
Therefore, the area of each square is 144 square units.

Answer:
Join NY.

Since XN = NZ, triangles XNY and YNZ have equal bases and the same height from Y.
So, area of ∆XNY = 1/2 × area of ∆XYZ.

Since XM = MY, triangles XMN and YMN have equal bases and the same height from N.
So, area of ∆XMN = 1/2 × area of ∆XNY.

Therefore,
Area of ∆XMN = 1/2 × 1/2 × area of ∆XYZ
= 1/4 × area of ∆XYZ.
The required fraction is 1/4.

Answer:
Let H represent the house and T the water tank.
Reflect T across the nearer riverbank to get T′.
Join H to T′. Let this line meet the riverbank at P.

The shortest path is H → P → T.
For any point Q on the riverbank, QT = QT′ because T′ is the reflection of T.
So, the length of H → Q → T equals the length of H → Q → T′.
The straight line HT′ is the shortest path from H to T′.
Therefore, Gopal should walk straight from his house to P, collect water there and then walk straight to the tank.

Get here the Solutions and Answers of Figure it Out at Page 160.

Answer:
The diagonal AC divides quadrilateral ABCD into ∆ABC and ∆ADC.
Area of ∆ABC = 1/2 × AC × BM
= 1/2 × 22 × 3
= 33 cm².

Area of ∆ADC = 1/2 × AC × DN
= 1/2 × 22 × 3
= 33 cm².
Area of quadrilateral ABCD = 33 + 33 = 66 cm².

Answer:
Area of rectangle ABCD = 18 × 10 = 180 cm².
The two unshaded parts are ∆AEF and ∆EBC.
Area of ∆AEF = 1/2 × AE × AF
= 1/2 × 10 × 6
= 30 cm².

Area of ∆EBC = 1/2 × EB × BC
= 1/2 × 8 × 10
= 40 cm².
Area of the shaded region = 180 − 30 − 40 = 110 cm².

Answer:
I would measure:
(i) The length of one side, a.
(ii) The perpendicular distance from the centre to that side, h.

On joining the centre O to all six vertices, the regular hexagon is divided into six equal triangles.
Area of each triangle = 1/2 × a × h.
Area of the hexagon = 6 × 1/2 × a × h = 3ah.

Answer:
Let the width of the rectangle be b. Let the heights of the upper and lower blue triangles be h₁ and h₂.
Both triangles have a base of length b.
Also, h₁ + h₂ is the height of the rectangle.
Area of the blue region = 1/2 × b × h₁ + 1/2 × b × h₂
= 1/2 × b × (h₁ + h₂).

Area of the rectangle = b × (h₁ + h₂).
Therefore, the blue region is 1/2 of the total area of the rectangle.

Answer:
Let the quadrilateral be ABCD.

1. Draw a diagonal AC inside the quadrilateral.
2. Mark E as the midpoint of AB and F as the midpoint of AD.
3. Join EC and FC. The quadrilateral AECF is the required quadrilateral.

Since AE = EB, triangles AEC and BEC have equal bases and the same height from C.
So, area of ∆AEC = 1/2 × area of ∆ABC.
Similarly, AF = FD, so area of ∆AFC = 1/2 × area of ∆ADC.
Area of quadrilateral AECF = area of ∆AEC + area of ∆AFC
= 1/2 × (area of ∆ABC + area of ∆ADC)
= 1/2 × area of quadrilateral ABCD.

Get here the Solutions and Answers of Figure it Out at Page 162.

(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, and which has the minimum perimeter?

Answer:
(i) All the parallelograms have equal areas because they have the same base and the same height.
Each has a base of 5 grid units and a height of 3 grid units.
Area of each parallelogram = 5 × 3 = 15 square grid units.

(ii) Their perimeters are not all equal.
Perimeter of a parallelogram = 2 × (base + adjacent side).
Their bases are equal, but their sloping sides have different lengths. Among these figures, (g) has the longest sloping side and appears to have the maximum perimeter. Figure (a) has the shortest sloping side and appears to have the minimum perimeter.

Answer:
Area of a parallelogram = base × corresponding height.
(i) Area = 7 × 4 = 28 cm².

(ii) Area = 5 × 3 = 15 cm².

(iii) Taking the side of length 5 cm as the base, the corresponding height is 4.8 cm.
Area = 5 × 4.8 = 24 cm².

(iv) Taking the side of length 2 cm as the base, the corresponding height is 4.4 cm.
Area = 2 × 4.4 = 8.8 cm².

Answer:
Taking SR as the base,
Area of parallelogram PQRS = SR × QM
= 12 × 6
= 72 cm².

Taking PS as the base, QN is the corresponding height.
So, PS × QN = 72.
⇒ 7.6 × QN = 72
⇒ QN = 72 ÷ 7.6
= 720/76
= 180/19 cm.
Therefore, QN = 180/19 cm, or approximately 9.47 cm.

Answer:
The rectangle has the greater area.
Taking the side of length 5 cm as the base, the rectangle has a height of 4 cm.
Area of the rectangle = 5 × 4 = 20 cm².

The slanting parallelogram has the same base, but its perpendicular height is less than its side of length 4 cm.
So, its area is less than 5 × 4, that is, less than 20 cm².

Answer:
Method 1: Using the same base and height
Choose a longest side BC of the triangle as its base. Draw AD perpendicular to BC.
Let BC = b and AD = h.
Draw perpendiculars to BC at B and C. Draw a line through A parallel to BC to complete rectangle PQCB.
Area of ∆ABC = 1/2 × b × h.
Area of rectangle PQCB = b × h.
Therefore, the rectangle has twice the area of the triangle.

Method 2: Using two identical triangles
Take two identical triangular cut-outs. Turn one through a half-turn and join it to the other along a matching side to form a parallelogram.
Cut off a triangular corner of the parallelogram and move it to the other end to form a rectangle.
The rectangle has the combined area of the two identical triangles.
So, its area is twice the area of the original triangle.

Answer:
Take a triangular cut-out ABC.
Choose a longest side BC as the base, so that the perpendicular from A meets BC at D.
Let BC = b and AD = h.

1. Mark E and F as the midpoints of AB and AC. Join EF. It is parallel to BC and passes halfway up the triangle.
2. Let EF meet AD at G. Cut along EF and then cut the top triangle along AG.
3. Turn ∆AEG through a half-turn about E and ∆AGF through a half-turn about F.
4. Place these pieces in the two corners beside trapezium EBCF. They complete rectangle XYCB.

The rectangle has length b and breadth h/2.
Area of the rectangle = b × h/2 = 1/2 × b × h.
This is the area of the original triangle.
Also, no piece is removed or overlapped during the rearrangement.

[Hint: Show that triangles ∆ADB and ∆ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]
Answer:
Cut the isosceles triangle ABC along its height AD.
Since AB = AC, AD is common and both angles at D are right angles, ∆ADB and ∆ADC are congruent.
Also, BD = DC.

Turn one cut-out over if needed.
Rearrange the two right-angled triangles so that their equal sloping sides AB and AC meet along a diagonal of a rectangle. Their perpendicular sides form the boundary of the rectangle.

If BC = b and AD = h, the rectangle has sides b/2 and h.
Area of the rectangle = b/2 × h = 1/2 × b × h.
Thus, the rectangle and the original triangle have equal areas.

Answer:

1. Cut the rectangle along a diagonal. We get two congruent right-angled triangles.
2. Turn one piece over if needed. Join one pair of their equal perpendicular sides, keeping the other pair on a straight line on opposite sides of the joining point.
3. The two equal hypotenuses become the equal sides of an isosceles triangle.

For example, let the rectangle have sides ℓ and b. Join the sides of length b to form AD and place the sides of length ℓ along BD and DC.

The isosceles triangle has base 2ℓ and height b.
Area of the triangle = 1/2 × 2ℓ × b = ℓb.
This is equal to the area of the rectangle.

Answer:
Let the common sidelength be s and the height of the equilateral triangle be h.
The height h is less than the sloping side s. So, h < s.
Area of the square = s × s = s².
Area of one equilateral triangle = 1/2 × s × h.
Since h < s, this is less than 1/2 × s², and hence less than s².
Therefore, the square has a greater area than one equilateral triangle.

Area of two identical equilateral triangles = 2 × 1/2 × s × h = s × h.
Since h < s, we get s × h < s × s = s².
Therefore, the square also has a greater area than the two equilateral triangles together.

Get here the Solutions and Answers of Figure it Out at Page 169.

Answer:
Area of a rhombus = 1/2 × product of its diagonals
= 1/2 × 20 × 15
= 150 cm².
Therefore, the area of the rhombus is 150 cm².

Answer:

1. Take a rectangle ABCD of length 2a and breadth b.
2. Mark M and N as the midpoints of AB and DC respectively.
3. Cut along MN, AN and BN. We get four congruent right-angled triangles, each with perpendicular sides a and b.
4. Rearrange these four triangles with their right angles meeting at O, as shown.

The four outer sides are the hypotenuses of the congruent triangles, so they are equal.
Thus, PQRS is a rhombus.
All four pieces are used without gaps or overlaps.
Therefore, the rhombus and the rectangle have equal areas.

Answer:
Area of a trapezium = 1/2 × height × sum of the parallel sides.
(i) The parallel sides are 10 ft and 7 ft, and the perpendicular distance between them is 16 ft.
Area = 1/2 × 16 × (10 + 7)
= 8 × 17
= 136 ft².

(ii) The parallel sides are 24 m and 36 m, and the height is 14 m.
Area = 1/2 × 14 × (24 + 36)
= 7 × 60
= 420 m².

(iii) The parallel sides are 14 in and 6 in, and the perpendicular distance between them is 10 in.
Area = 1/2 × 10 × (14 + 6)
= 5 × 20
= 100 in².

(iv) The parallel sides are 12 ft and 18 ft, and the height is 8 ft.
Area = 1/2 × 8 × (12 + 18)
= 4 × 30
= 120 ft².

Answer:

1. Take an isosceles trapezium ABCD with AB ∥ DC and DC longer than AB.
2. Mark I and J as the midpoints of AD and BC respectively.
3. Draw IG and JF perpendicular to DC.
4. Cut off triangles DGI and CFJ.
5. Turn triangle DGI through 180° about I. It fits into the space AHI above I.
6. Turn triangle CFJ through 180° about J. It fits into the space BEJ above J.

The pieces form rectangle HEFG. Since we have only rearranged the pieces, the rectangle has the same area as the trapezium.

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.]
Answer:

1. Find the midpoints I and J of AD and BC respectively.
2. Through I, draw a line perpendicular to DC. Let it meet DC at G and the line AB at H.
3. Through J, draw another line perpendicular to DC. Let it meet DC at F and the line AB at E. Extend AB where necessary.

Thus, E, F, G and H are the required vertices. HE ∥ GF, and HG and EF are perpendicular to these sides. Hence, EFGH is a rectangle.
Now, AI = DI because I is the midpoint of AD.
Also, ∠AIH = ∠DIG, as they are vertically opposite angles.
Since AB ∥ DC, ∠HAI = ∠GDI.
Therefore, ∆AHI ≅ ∆DGI by ASA congruence.
Similarly, ∆BEJ ≅ ∆CFJ.

The two triangular parts removed from the trapezium have the same areas as the two triangular parts added to form the rectangle. Therefore, trapezium ABCD and rectangle EFGH have equal areas.

Answer:

1. Draw a rectangle EFGH with HE = GF = 12 cm and HG = EF = 12 cm.
2. Mark A and B on HE so that HA = EB = 2 cm. Then AB = 12 − 2 − 2 = 8 cm.
3. Mark I and J as the midpoints of HG and EF respectively. Join AI and BJ.
4. Cut off triangles AHI and BEJ. Turn them through 180° about I and J respectively. They form triangles DGI and CFJ outside the two sides of the rectangle.

We get trapezium ABCD. Here, DG = FC = 2 cm.
DC = DG + GF + FC
= 2 + 12 + 2
= 16 cm.
Its height is 12 cm.

Area of the trapezium = 1/2 × 12 × (8 + 16)
= 6 × 24
= 144 cm².
Thus, a trapezium with parallel sides 8 cm and 16 cm and height 12 cm is the required trapezium.

Answer:
Join the centre of the regular hexagon to all its vertices. The hexagon is divided into six congruent equilateral triangles.

The trapezium contains 3 of these triangles.
The equilateral triangle contains 1 of these triangles.
The rhombus contains 2 of these triangles.
Therefore,
Area of trapezium : Area of equilateral triangle : Area of rhombus = 3 : 1 : 2.

Answer:
Since A is the midpoint of XY, YA = AX.
In triangles YAZ and XAB,
∠YAZ = ∠XAB, as they are vertically opposite angles.
∠ZYA = ∠BXA, as ZY ∥ XB and these are alternate interior angles.
YA = AX.
Therefore, ∆YAZ ≅ ∆XAB by ASA congruence.
Hence, area of ∆YAZ = area of ∆XAB.

Adding the area of quadrilateral ZAXW to both sides, we get
Area of ∆YAZ + Area of ZAXW = Area of ∆XAB + Area of ZAXW.
Therefore, area of trapezium ZYXW = area of ∆ZWB.

Rebuild these formulas instead of memorising them; each one starts from a rectangle. A triangle is half its enclosing rectangle; a parallelogram becomes a rectangle by moving one triangle; a rhombus is two triangles on one diagonal; and two copies of a trapezium form a parallelogram.
Always use the perpendicular height, not a slanted side, which is why a 5 cm by 4 cm parallelogram has less area than a rectangle with the same sides. The dissection answers in these Area Class 8 solutions come from the Sulba Sutras, which also gave us Chapter 2’s theorem. Area completes Ganita Prakash Part 2; for Part 1 and every other chapter, visit our NCERT Solutions for Class 8 Maths.