Since rhombus is also a kind of parallelogram. ∴ Area of rhombus = Base x Altitude = 6 x 4 = 24 cm² Also Area of rhombus = 1/2d₁d₂ ⇒ 24=1/25x8xd₂ ⇒ 24=4d₂ ⇒ d₂=24/4=6cm Hence, the length of the other diagonal is 6 cm. Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to:Read more
Since rhombus is also a kind of parallelogram.
∴ Area of rhombus = Base x Altitude
= 6 x 4 = 24 cm²
Also Area of rhombus = 1/2d₁d₂
⇒ 24=1/25x8xd₂ ⇒ 24=4d₂ ⇒ d₂=24/4=6cm
Hence, the length of the other diagonal is 6 cm.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Given: d₁ =7.5 cm and d₂=12 cm We know that, Area of rhombus = 1/2xd₁xd₂=1/2x7.5x12=45cm² Hence, area of rhombus is 45 cm². Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Given: d₁ =7.5 cm and d₂=12 cm
We know that,
Area of rhombus = 1/2xd₁xd₂=1/2×7.5×12=45cm²
Hence, area of rhombus is 45 cm².
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Here h₁ = 13 m, h₂ = 8 m and AC = 24 m Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC = 1/2bxh₁+1/2bxh₂ = 1/2b(h₁+h₂) = 1/2x24x(13+8)=1/2x24x21=252m² Hence, required area of the field is 252 m². Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to: https://www.Read more
Here h₁ = 13 m, h₂ = 8 m and AC = 24 m
Area of quadrilateral ABCD
= Area of ∆ABC + Area of ∆ADC
= 1/2bxh₁+1/2bxh₂
= 1/2b(h₁+h₂)
= 1/2x24x(13+8)=1/2x24x21=252m²
Hence, required area of the field is 252 m².
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m ∵ Perimeter of trapezium ABCD = AB + BC + CD + DA ⇒ 120 = AB + 48 + 17 + 40 ⇒ 120 = AB = 105 ⇒ AB = 120 – 105 = 15 m Now Area of the field = 1/2x(BC+AD)xAB = 1/2 x (48+40)x15=1/2x88x15=660m² Hence, area of the field ABCD is 660 m² Class 8Read more
Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD = AB + BC + CD + DA
⇒ 120 = AB + 48 + 17 + 40 ⇒ 120 = AB = 105
⇒ AB = 120 – 105 = 15 m
Now Area of the field = 1/2x(BC+AD)xAB
= 1/2 x (48+40)x15=1/2x88x15=660m²
Hence, area of the field ABCD is 660 m²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Let the length of the other parallel side be b. Length of one parallel side (a) = 10 am and height (h) = 4 cm Area of trapezium = 1/2 (a+b)xh ⇒ 34=1/2(10+b)x4 ⇒ 34 = (10+b)x2 ⇒ 34=20+2b ⇒ 34-20=2b ⇒ 14=2b ⇒ 7=b ⇒ b=7 Hence, the another required parallel side is 7 cm. Class 8 Maths Chapter 11 ExercisRead more
Let the length of the other parallel side be b.
Length of one parallel side (a) = 10 am and height (h) = 4 cm
Area of trapezium = 1/2 (a+b)xh
⇒ 34=1/2(10+b)x4 ⇒ 34 = (10+b)x2
⇒ 34=20+2b ⇒ 34-20=2b
⇒ 14=2b ⇒ 7=b ⇒ b=7
Hence, the another required parallel side is 7 cm.
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Since rhombus is also a kind of parallelogram. ∴ Area of rhombus = Base x Altitude = 6 x 4 = 24 cm² Also Area of rhombus = 1/2d₁d₂ ⇒ 24=1/25x8xd₂ ⇒ 24=4d₂ ⇒ d₂=24/4=6cm Hence, the length of the other diagonal is 6 cm. Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to:Read more
Since rhombus is also a kind of parallelogram.
∴ Area of rhombus = Base x Altitude
= 6 x 4 = 24 cm²
Also Area of rhombus = 1/2d₁d₂
⇒ 24=1/25x8xd₂ ⇒ 24=4d₂ ⇒ d₂=24/4=6cm
Hence, the length of the other diagonal is 6 cm.
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 diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Given: d₁ =7.5 cm and d₂=12 cm We know that, Area of rhombus = 1/2xd₁xd₂=1/2x7.5x12=45cm² Hence, area of rhombus is 45 cm². Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Given: d₁ =7.5 cm and d₂=12 cm
We know that,
Area of rhombus = 1/2xd₁xd₂=1/2×7.5×12=45cm²
Hence, area of rhombus is 45 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/
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Here h₁ = 13 m, h₂ = 8 m and AC = 24 m Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC = 1/2bxh₁+1/2bxh₂ = 1/2b(h₁+h₂) = 1/2x24x(13+8)=1/2x24x21=252m² Hence, required area of the field is 252 m². Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video for more answers vist to: https://www.Read more
Here h₁ = 13 m, h₂ = 8 m and AC = 24 m
Area of quadrilateral ABCD
= Area of ∆ABC + Area of ∆ADC
= 1/2bxh₁+1/2bxh₂
= 1/2b(h₁+h₂)
= 1/2x24x(13+8)=1/2x24x21=252m²
Hence, required area of the field is 252 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/
Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m ∵ Perimeter of trapezium ABCD = AB + BC + CD + DA ⇒ 120 = AB + 48 + 17 + 40 ⇒ 120 = AB = 105 ⇒ AB = 120 – 105 = 15 m Now Area of the field = 1/2x(BC+AD)xAB = 1/2 x (48+40)x15=1/2x88x15=660m² Hence, area of the field ABCD is 660 m² Class 8Read more
Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD = AB + BC + CD + DA
⇒ 120 = AB + 48 + 17 + 40 ⇒ 120 = AB = 105
⇒ AB = 120 – 105 = 15 m
Now Area of the field = 1/2x(BC+AD)xAB
= 1/2 x (48+40)x15=1/2x88x15=660m²
Hence, area of the field ABCD is 660 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/
The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Let the length of the other parallel side be b. Length of one parallel side (a) = 10 am and height (h) = 4 cm Area of trapezium = 1/2 (a+b)xh ⇒ 34=1/2(10+b)x4 ⇒ 34 = (10+b)x2 ⇒ 34=20+2b ⇒ 34-20=2b ⇒ 14=2b ⇒ 7=b ⇒ b=7 Hence, the another required parallel side is 7 cm. Class 8 Maths Chapter 11 ExercisRead more
Let the length of the other parallel side be b.
Length of one parallel side (a) = 10 am and height (h) = 4 cm
Area of trapezium = 1/2 (a+b)xh
⇒ 34=1/2(10+b)x4 ⇒ 34 = (10+b)x2
⇒ 34=20+2b ⇒ 34-20=2b
⇒ 14=2b ⇒ 7=b ⇒ b=7
Hence, the another required parallel side is 7 cm.
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/