Here two of given figures (I) and (II) are similar in dimensions. And also figures (III) and (IV) are similar in dimensions. ∴ Area of figure (I) = Area of trapezium = 1/2(a+b)xh = 1/2(28+20)x4=1/2x48x4=96cm² Also Area of figure (II) = 96cm² Now Area of figure (III) = Area of trapezium = 1/2(a+b)xhRead more
Here two of given figures (I) and (II) are similar in
dimensions.
And also figures (III) and (IV) are similar in dimensions.
∴ Area of figure (I) = Area of trapezium = 1/2(a+b)xh
= 1/2(28+20)x4=1/2x48x4=96cm²
Also Area of figure (II) = 96cm²
Now Area of figure (III) = Area of trapezium = 1/2(a+b)xh
= 1/2(24+16)x4=1/2x40x4=80cm²
Also Area of figure (IV) = 80 cm²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
First way: By Jyoti’s diagram, Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP = 1/2(AP+BC) x CP+1/2 (ED + AP) x DP = 1/2(30+15) x CP+1/2 (15 + 30) x DP = 1/2(30+15) x (CP+DP) = 1/2x45xcd= 1/2x45x15=337.5 m² Second way: By Kavita’s diagram Here, a perpendicular AM drawn to BE. AMRead more
First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= 1/2(AP+BC) x CP+1/2 (ED + AP) x DP
= 1/2(30+15) x CP+1/2 (15 + 30) x DP
= 1/2(30+15) x (CP+DP)
= 1/2x45xcd= 1/2x45x15=337.5 m²
Second way: By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
AM = 30 – 15 = 15 m
Area of pentagon = Area of ∆ABE + Area of square BCDE
=1/2x15x15+15×15=112.5+225.0=337.5 m²
Hence, total area of pentagon shaped park = 337.5 m².
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Given: Octagon having eight equal sides, each 5 m. Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively. Now Area of two trapeziums = 2x1/2Read more
Given: Octagon having eight equal sides, each 5 m.
Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and
perpendicular sides are 11 m and 4 m respectively and third figure
is rectangle having length and breadth 11 m and 5 m respectively.
Now Area of two trapeziums = 2×1/2(a+b)xh
= 2×1/2(11+5)x4=4×16=64cm²
And Area of rectangle = length x breadth
= 11 x 5 = 55m²
∴ Total area of octagon = 64 + 55 = 119 m²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Given: Perpendicular distance (h) = 100 m Area of the trapezium shaped field = 10500 m² Let side along the road be x m and side along the river = 2x m ∴ Area of the trapezium field = 1/2(a+b)x7 ⇒ 10500 = 1/2 (x+2x)x100 ⇒ 10500=3x50 ⇒ 3x=10500/50 ⇒ x=10500/50x3 ⇒ x=70m Hence, the side along the riverRead more
Given: Perpendicular distance (h) = 100 m
Area of the trapezium shaped field = 10500 m²
Let side along the road be x m and side along the river = 2x m
∴ Area of the trapezium field = 1/2(a+b)x7
⇒ 10500 = 1/2 (x+2x)x100 ⇒ 10500=3×50
⇒ 3x=10500/50 ⇒ x=10500/50×3 ⇒ x=70m
Hence, the side along the river = 2x = 2 x 70 = 140 m.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Here, d₁ = 45 cm and d₂ = 30 cm ∵ Area of one tile = 1/2d₁d₂=1/2x45x30=675cm² ∴ Area of 3000 tiles = 675 x 3000 = 2025000 cm² = 2025000/10000 = 202.50 m2 [∵ 1m²=10000 cm²] ∵ Cost of polishing the floor per sq. meter = ₹ 4 ∴ Cost of polishing the floor per 202.50 sq. meter = 4 x 202.50 = ₹ 810 Hence,Read more
Here, d₁ = 45 cm and d₂ = 30 cm
∵ Area of one tile = 1/2d₁d₂=1/2x45x30=675cm²
∴ Area of 3000 tiles = 675 x 3000 = 2025000 cm²
= 2025000/10000 = 202.50 m2 [∵ 1m²=10000 cm²]
∵ Cost of polishing the floor per sq. meter = ₹ 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 x 202.50 = ₹ 810
Hence, the total cost of polishing the floor is ₹ 810.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is same.
Here two of given figures (I) and (II) are similar in dimensions. And also figures (III) and (IV) are similar in dimensions. ∴ Area of figure (I) = Area of trapezium = 1/2(a+b)xh = 1/2(28+20)x4=1/2x48x4=96cm² Also Area of figure (II) = 96cm² Now Area of figure (III) = Area of trapezium = 1/2(a+b)xhRead more
Here two of given figures (I) and (II) are similar in
dimensions.
And also figures (III) and (IV) are similar in dimensions.
∴ Area of figure (I) = Area of trapezium = 1/2(a+b)xh
= 1/2(28+20)x4=1/2x48x4=96cm²
Also Area of figure (II) = 96cm²
Now Area of figure (III) = Area of trapezium = 1/2(a+b)xh
= 1/2(24+16)x4=1/2x40x4=80cm²
Also Area of figure (IV) = 80 cm²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?
First way: By Jyoti’s diagram, Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP = 1/2(AP+BC) x CP+1/2 (ED + AP) x DP = 1/2(30+15) x CP+1/2 (15 + 30) x DP = 1/2(30+15) x (CP+DP) = 1/2x45xcd= 1/2x45x15=337.5 m² Second way: By Kavita’s diagram Here, a perpendicular AM drawn to BE. AMRead more
First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= 1/2(AP+BC) x CP+1/2 (ED + AP) x DP
= 1/2(30+15) x CP+1/2 (15 + 30) x DP
= 1/2(30+15) x (CP+DP)
= 1/2x45xcd= 1/2x45x15=337.5 m²
Second way: By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
AM = 30 – 15 = 15 m
Area of pentagon = Area of ∆ABE + Area of square BCDE
=1/2x15x15+15×15=112.5+225.0=337.5 m²
Hence, total area of pentagon shaped park = 337.5 m².
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Given: Octagon having eight equal sides, each 5 m. Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively. Now Area of two trapeziums = 2x1/2Read more
Given: Octagon having eight equal sides, each 5 m.
Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and
perpendicular sides are 11 m and 4 m respectively and third figure
is rectangle having length and breadth 11 m and 5 m respectively.
Now Area of two trapeziums = 2×1/2(a+b)xh
= 2×1/2(11+5)x4=4×16=64cm²
And Area of rectangle = length x breadth
= 11 x 5 = 55m²
∴ Total area of octagon = 64 + 55 = 119 m²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Given: Perpendicular distance (h) = 100 m Area of the trapezium shaped field = 10500 m² Let side along the road be x m and side along the river = 2x m ∴ Area of the trapezium field = 1/2(a+b)x7 ⇒ 10500 = 1/2 (x+2x)x100 ⇒ 10500=3x50 ⇒ 3x=10500/50 ⇒ x=10500/50x3 ⇒ x=70m Hence, the side along the riverRead more
Given: Perpendicular distance (h) = 100 m
Area of the trapezium shaped field = 10500 m²
Let side along the road be x m and side along the river = 2x m
∴ Area of the trapezium field = 1/2(a+b)x7
⇒ 10500 = 1/2 (x+2x)x100 ⇒ 10500=3×50
⇒ 3x=10500/50 ⇒ x=10500/50×3 ⇒ x=70m
Hence, the side along the river = 2x = 2 x 70 = 140 m.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m ² is ₹ 4.
Here, d₁ = 45 cm and d₂ = 30 cm ∵ Area of one tile = 1/2d₁d₂=1/2x45x30=675cm² ∴ Area of 3000 tiles = 675 x 3000 = 2025000 cm² = 2025000/10000 = 202.50 m2 [∵ 1m²=10000 cm²] ∵ Cost of polishing the floor per sq. meter = ₹ 4 ∴ Cost of polishing the floor per 202.50 sq. meter = 4 x 202.50 = ₹ 810 Hence,Read more
Here, d₁ = 45 cm and d₂ = 30 cm
∵ Area of one tile = 1/2d₁d₂=1/2x45x30=675cm²
∴ Area of 3000 tiles = 675 x 3000 = 2025000 cm²
= 2025000/10000 = 202.50 m2 [∵ 1m²=10000 cm²]
∵ Cost of polishing the floor per sq. meter = ₹ 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 x 202.50 = ₹ 810
Hence, the total cost of polishing the floor is ₹ 810.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/