Diameter = 10 cm ∴ Radius = 10/2 = 5 cm According to question, Perimeter of figure = Circumference of semi-circle + diameter = πr + D = 22/7x5+10 = 110/7+10 = 110+70/7 = 180/7 = 25.71 cm Thus, the perimeter of the given figure is 25.71 cm. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vistRead more
Diameter = 10 cm
∴ Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D
= 22/7×5+10 = 110/7+10
= 110+70/7 = 180/7 = 25.71 cm
Thus, the perimeter of the given figure is 25.71 cm.
Diameter of the circular table cover = 1.5 m ∴ Radius of the circular table cover = 15/2 m Circumference of circular table cover = 2πr = 2x3.14x15/2 = 4.71 m Therefore the length of required lace is 4.71 m. Now the cost of 1 m lace = ₹ 15 Then the cost of 4.71 m lace = 15 x 4.71 = ₹ 70.65 Hence, theRead more
Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 15/2 m
Circumference of circular table cover = 2πr
= 2×3.14×15/2
= 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = ₹ 15
Then the cost of 4.71 m lace = 15 x 4.71
= ₹ 70.65
Hence, the cost of 4.71 m lace is ₹ 70.65.
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm Area of remaining sheet = Area of circular sheet – Area of removed circle = πR² - πr² = π(R²-r²) = π(4² - 3²) = π(16-9) = 3.14 x 7 = 21.98 cm² Thus, the area of remaining sheet is 21.98 cm². Class 7 Maths Chapter 11 ExerciseRead more
Radius of circular sheet (R) = 4 cm and
radius of removed circle (r) = 3 cm
Area of remaining sheet
= Area of circular sheet – Area of removed circle
= πR² – πr² = π(R²-r²)
= π(4² – 3²) = π(16-9)
= 3.14 x 7 = 21.98 cm²
Thus, the area of remaining sheet is 21.98 cm².
Diameter of the circular garden = 21 m ∴ Radius of the circular garden = 21/2 Now Circumference of circular garden = 2πr = 2x22/7x21/2 = 22 x 3 = 66 m The gardener makes 2 rounds of fence so the total length of the rope of fencing = 2x2πr = 2 x 66 = 132 m Since, the cost of 1 meter rope = ₹ 4 TherefRead more
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21/2
Now Circumference of circular garden = 2πr = 2×22/7×21/2
= 22 x 3 = 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2×2πr
= 2 x 66 = 132 m
Since, the cost of 1 meter rope = ₹ 4
Therefore, cost of 132 meter rope = 4 x 132 = ₹ 528
Circumference of the circular sheet = 154 m ⇒ 2πr = 154 m ⇒ r = 154/2π ⇒ r = 154x7/2x22 = 24.5 m Now Area of circular sheet = πr² = 22/7x 24.5X24.5 = 22 x 3.5 x 24.5 = 1886.5 m² Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m² respectively. Class 7 Maths Chapter 11 Exercise 11.3Read more
Circumference of the circular sheet = 154 m
⇒ 2πr = 154 m
⇒ r = 154/2π
⇒ r = 154×7/2×22 = 24.5 m
Now Area of circular sheet = πr² = 22/7x 24.5X24.5
= 22 x 3.5 x 24.5 = 1886.5 m²
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m² respectively.
(a) Area of circle = πr²=22/7x14x14 = 22 x 2 x 14 = 616 mm² (b) Diameter = 49 m ∴ radius = 49/2 = 24.5 m ∴ Area of circle = πr²=22/7x24.5x24.5 = 22 x 3.5 x 24.5 = 1886.5 m² (c) Area of circle = πr²=22/7x5x5 557/7 cm² Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://www.tiwariRead more
(a) Area of circle = πr²=22/7x14x14
= 22 x 2 x 14
= 616 mm²
(b) Diameter = 49 m
∴ radius = 49/2 = 24.5 m
∴ Area of circle = πr²=22/7×24.5×24.5
= 22 x 3.5 x 24.5
= 1886.5 m²
(a) Circumference of the circle = 2πr = 2x22/7x14 = 88 cm (b) Circumference of the circle = 2πr = 2x22/7x28 = 176 cm (c) Circumference of the circle = 2πr = 2x22/7x21 = 132 cm Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/mathsRead more
(a) Circumference of the circle = 2πr = 2×22/7×14 = 88 cm
(b) Circumference of the circle = 2πr = 2×22/7×28 = 176 cm
(c) Circumference of the circle = 2πr = 2×22/7×21 = 132 cm
In ∆ABC, AD = 6 cm and BC = 9 cm Area of triangle = 1/2 x base x height = 1/2 x BC x AD = 1/2 x 9 x 6 = 27 cm² Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE ⇒ 27 = 1/2 x 7.5 x CE ⇒ CE = 27x2/7.5 ⇒ CE = 7.2 cm Thus, height from C to AB i.e., CE is 7.2 cm. Class 7 Maths Chapter 11 ExerRead more
In ∆ABC, AD = 6 cm and BC = 9 cm
Area of triangle = 1/2 x base x height = 1/2 x BC x AD
= 1/2 x 9 x 6 = 27 cm²
Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE
⇒ 27 = 1/2 x 7.5 x CE
⇒ CE = 27×2/7.5
⇒ CE = 7.2 cm
Thus, height from C to AB i.e., CE is 7.2 cm.
In right angles triangle BAC, AB = 5 cm and AC = 12 cm Area of triangle = 1/2 x base x height = 1/2 x AB x AC = 1/2 x 5 x 12 = 30 cm² Now, in ∆ABC Area of triangle ABC = 1/2 x BC x AD ⇒ 30 = 1/2 x 13 x AD ⇒ AD = 30x2/13 = 60/13 Class 7 Maths Chapter 11 Exercise 11.2 for more answers vist to: https:/Read more
In right angles triangle BAC, AB = 5 cm and AC = 12 cm
Area of triangle = 1/2 x base x height = 1/2 x AB x AC
= 1/2 x 5 x 12 = 30 cm²
Now, in ∆ABC
Area of triangle ABC = 1/2 x BC x AD
⇒ 30 = 1/2 x 13 x AD
⇒ AD = 30×2/13 = 60/13
Given: Area of parallelogram = 1470 cm² Base (AB) = 35 cm and base (AD) = 49 cm Since Area of parallelogram = base x height ⇒ 1470 = 35 x DL ⇒ DL = 1470/35 ⇒ DL = 42 cm Again, Area of parallelogram = base x height ⇒ 1470 = 49 x BM ⇒ BM = 1470/49 ⇒ BM = 30 cm Thus, the lengths of DL and BM are 42 cmRead more
Given: Area of parallelogram = 1470 cm²
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ DL = 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ BM = 1470/49
⇒ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Diameter = 10 cm ∴ Radius = 10/2 = 5 cm According to question, Perimeter of figure = Circumference of semi-circle + diameter = πr + D = 22/7x5+10 = 110/7+10 = 110+70/7 = 180/7 = 25.71 cm Thus, the perimeter of the given figure is 25.71 cm. Class 7 Maths Chapter 11 Exercise 11.3 for more answers vistRead more
Diameter = 10 cm
∴ Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D
= 22/7×5+10 = 110/7+10
= 110+70/7 = 180/7 = 25.71 cm
Thus, the perimeter of the given figure is 25.71 cm.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹15. ( Take π = 3.14)
Diameter of the circular table cover = 1.5 m ∴ Radius of the circular table cover = 15/2 m Circumference of circular table cover = 2πr = 2x3.14x15/2 = 4.71 m Therefore the length of required lace is 4.71 m. Now the cost of 1 m lace = ₹ 15 Then the cost of 4.71 m lace = 15 x 4.71 = ₹ 70.65 Hence, theRead more
Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 15/2 m
Circumference of circular table cover = 2πr
= 2×3.14×15/2
= 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = ₹ 15
Then the cost of 4.71 m lace = 15 x 4.71
= ₹ 70.65
Hence, the cost of 4.71 m lace is ₹ 70.65.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet ( Take π = 3.14)
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm Area of remaining sheet = Area of circular sheet – Area of removed circle = πR² - πr² = π(R²-r²) = π(4² - 3²) = π(16-9) = 3.14 x 7 = 21.98 cm² Thus, the area of remaining sheet is 21.98 cm². Class 7 Maths Chapter 11 ExerciseRead more
Radius of circular sheet (R) = 4 cm and
radius of removed circle (r) = 3 cm
Area of remaining sheet
= Area of circular sheet – Area of removed circle
= πR² – πr² = π(R²-r²)
= π(4² – 3²) = π(16-9)
= 3.14 x 7 = 21.98 cm²
Thus, the area of remaining sheet is 21.98 cm².
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it cost ₹4 per meter. ( Take π = 22/7)
Diameter of the circular garden = 21 m ∴ Radius of the circular garden = 21/2 Now Circumference of circular garden = 2πr = 2x22/7x21/2 = 22 x 3 = 66 m The gardener makes 2 rounds of fence so the total length of the rope of fencing = 2x2πr = 2 x 66 = 132 m Since, the cost of 1 meter rope = ₹ 4 TherefRead more
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21/2
Now Circumference of circular garden = 2πr = 2×22/7×21/2
= 22 x 3 = 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2×2πr
= 2 x 66 = 132 m
Since, the cost of 1 meter rope = ₹ 4
Therefore, cost of 132 meter rope = 4 x 132 = ₹ 528
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. ( Take π = 22/7)
Circumference of the circular sheet = 154 m ⇒ 2πr = 154 m ⇒ r = 154/2π ⇒ r = 154x7/2x22 = 24.5 m Now Area of circular sheet = πr² = 22/7x 24.5X24.5 = 22 x 3.5 x 24.5 = 1886.5 m² Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m² respectively. Class 7 Maths Chapter 11 Exercise 11.3Read more
Circumference of the circular sheet = 154 m
⇒ 2πr = 154 m
⇒ r = 154/2π
⇒ r = 154×7/2×22 = 24.5 m
Now Area of circular sheet = πr² = 22/7x 24.5X24.5
= 22 x 3.5 x 24.5 = 1886.5 m²
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m² respectively.
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Find the area of the following circles, given that: ( Take π = 22/7) (a) radius = 14 mm (b) diameter = 49 m (c) radius 5 cm
(a) Area of circle = πr²=22/7x14x14 = 22 x 2 x 14 = 616 mm² (b) Diameter = 49 m ∴ radius = 49/2 = 24.5 m ∴ Area of circle = πr²=22/7x24.5x24.5 = 22 x 3.5 x 24.5 = 1886.5 m² (c) Area of circle = πr²=22/7x5x5 557/7 cm² Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://www.tiwariRead more
(a) Area of circle = πr²=22/7x14x14
= 22 x 2 x 14
= 616 mm²
(b) Diameter = 49 m
∴ radius = 49/2 = 24.5 m
∴ Area of circle = πr²=22/7×24.5×24.5
= 22 x 3.5 x 24.5
= 1886.5 m²
(c) Area of circle = πr²=22/7x5x5
557/7 cm²
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Find the circumference of the circles with the following radius: ( Take π = 22/7) (a) 14 cm (b) 28 mm (c) 21 cm
(a) Circumference of the circle = 2πr = 2x22/7x14 = 88 cm (b) Circumference of the circle = 2πr = 2x22/7x28 = 176 cm (c) Circumference of the circle = 2πr = 2x22/7x21 = 132 cm Class 7 Maths Chapter 11 Exercise 11.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/mathsRead more
(a) Circumference of the circle = 2πr = 2×22/7×14 = 88 cm
(b) Circumference of the circle = 2πr = 2×22/7×28 = 176 cm
(c) Circumference of the circle = 2πr = 2×22/7×21 = 132 cm
Class 7 Maths Chapter 11 Exercise 11.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?
In ∆ABC, AD = 6 cm and BC = 9 cm Area of triangle = 1/2 x base x height = 1/2 x BC x AD = 1/2 x 9 x 6 = 27 cm² Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE ⇒ 27 = 1/2 x 7.5 x CE ⇒ CE = 27x2/7.5 ⇒ CE = 7.2 cm Thus, height from C to AB i.e., CE is 7.2 cm. Class 7 Maths Chapter 11 ExerRead more
In ∆ABC, AD = 6 cm and BC = 9 cm
Area of triangle = 1/2 x base x height = 1/2 x BC x AD
= 1/2 x 9 x 6 = 27 cm²
Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE
⇒ 27 = 1/2 x 7.5 x CE
⇒ CE = 27×2/7.5
⇒ CE = 7.2 cm
Thus, height from C to AB i.e., CE is 7.2 cm.
Class 7 Maths Chapter 11 Exercise 11.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
∆ABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also, find the length of AD.
In right angles triangle BAC, AB = 5 cm and AC = 12 cm Area of triangle = 1/2 x base x height = 1/2 x AB x AC = 1/2 x 5 x 12 = 30 cm² Now, in ∆ABC Area of triangle ABC = 1/2 x BC x AD ⇒ 30 = 1/2 x 13 x AD ⇒ AD = 30x2/13 = 60/13 Class 7 Maths Chapter 11 Exercise 11.2 for more answers vist to: https:/Read more
In right angles triangle BAC, AB = 5 cm and AC = 12 cm
Area of triangle = 1/2 x base x height = 1/2 x AB x AC
= 1/2 x 5 x 12 = 30 cm²
Now, in ∆ABC
Area of triangle ABC = 1/2 x BC x AD
⇒ 30 = 1/2 x 13 x AD
⇒ AD = 30×2/13 = 60/13
Class 7 Maths Chapter 11 Exercise 11.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Given: Area of parallelogram = 1470 cm² Base (AB) = 35 cm and base (AD) = 49 cm Since Area of parallelogram = base x height ⇒ 1470 = 35 x DL ⇒ DL = 1470/35 ⇒ DL = 42 cm Again, Area of parallelogram = base x height ⇒ 1470 = 49 x BM ⇒ BM = 1470/49 ⇒ BM = 30 cm Thus, the lengths of DL and BM are 42 cmRead more
Given: Area of parallelogram = 1470 cm²
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ DL = 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ BM = 1470/49
⇒ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Class 7 Maths Chapter 11 Exercise 11.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/