Here, AC = 22 cm, BM = 3 cm, DN = 3 cm Area of quadrilateral ABCDF = Area of ∆ABC + Area of ∆ADC = 1/2 x AC x BM + 1/2 x AC x DN = 1/2 x 22 x 3 + 1/2 x 22 x 3 = 3 x 11 + 3 x 11 = 33 + 33 = 66 cm² Thus, the area of quadrilateral ABCD is cm². Class 7 Maths Chapter 11 Exercise 11.4 for more answers visRead more
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm
Area of quadrilateral ABCDF = Area of ∆ABC + Area of ∆ADC
= 1/2 x AC x BM + 1/2 x AC x DN
= 1/2 x 22 x 3 + 1/2 x 22 x 3
= 3 x 11 + 3 x 11
= 33 + 33
= 66 cm²
Thus, the area of quadrilateral ABCD is cm².
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm Area of shaded portion = Area of rectangle ABCD – (Area of ∆FAE + area of ∆EBC) = (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC) = (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10) = 180 – (30 + 40) = 180 – 70 = 110 cm² (ii) Here, SR = SU +Read more
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion
= Area of rectangle ABCD – (Area of ∆FAE + area of ∆EBC)
= (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC)
= (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10)
= 180 – (30 + 40)
= 180 – 70
= 110 cm²
(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS – Area of ∆QPT – Area of ∆TSU – Area of ∆UQR
= (SR x QR) – 1/2 x PQ x PT – 1/2 x ST x SU – 1/2
= 20 x 20 – 1/2 x 20 x 10 – 1/2 x 10 x 10 – 1/2 x 20 x 10
= 400 – 100 – 50 – 100
= 150 cm²
Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5m And radius of the circular flower bed = 2m (i) Area of the whole land = length x breadth = 10 x 5 = 50 m² (ii) Area of flower bed = πr² = 3.14 x 2 x 2 = 12.56 m² (iii) Area of lawn excluding the area of the flower bed = area ofRead more
Length of rectangular lawn = 10 m,
breadth of the rectangular lawn = 5m
And radius of the circular flower bed = 2m
(i) Area of the whole land = length x breadth
= 10 x 5 = 50 m²
(ii) Area of flower bed = πr²
= 3.14 x 2 x 2 = 12.56 m²
(iii) Area of lawn excluding the area of the flower bed
= area of lawn – area of flower bed
= 50 – 12.56
= 37.44 m²
(iv) The circumference of the flower bed = 2πr
= 2 x 3.14 x 2 = 12.56 m
Radius of pipe = 4 cm Wrapping cord around circular pipe = 2πr = 2 x 3.14 x 4 = 25.12 cm Again, wrapping cord around a square = 4 x side = 4 x 4 = 16 cm Remaining cord = Cord wrapped on pipe – Cord wrapped on square = 25.12 – 16 = 9.12 cm Thus, she has left 9.12 cm cord. Class 7 Maths Chapter 11 ExeRead more
Radius of pipe = 4 cm
Wrapping cord around circular pipe = 2πr
= 2 x 3.14 x 4 = 25.12 cm
Again, wrapping cord around a square = 4 x side
= 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16
= 9.12 cm
Thus, she has left 9.12 cm cord.
(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and EF = 90 m and KL = 3 m and KN = 3 m Area of roads = Area of PQRS + Area of EFGH – Area of KLMN [∵ KLMN is taken twice, which is to be subtracted] = PS x PQ + EF x EH – KL x KN = (60 x 3) + (90 x 3) – (3 x 3) = 180 + 270 – 9 = 441 m² (ii) The cost of 1m²Read more
(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and
EF = 90 m and KL = 3 m and KN = 3 m
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (60 x 3) + (90 x 3) – (3 x 3)
= 180 + 270 – 9
= 441 m²
(ii) The cost of 1m² constructing the roads = ₹110
The cost of 441m² constructing the roads = ₹110 x 441 = ₹48,510
Therefore, the cost of constructing the roads = ₹48,510
Find the area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM⊥ AC, DN ⊥AC.
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm Area of quadrilateral ABCDF = Area of ∆ABC + Area of ∆ADC = 1/2 x AC x BM + 1/2 x AC x DN = 1/2 x 22 x 3 + 1/2 x 22 x 3 = 3 x 11 + 3 x 11 = 33 + 33 = 66 cm² Thus, the area of quadrilateral ABCD is cm². Class 7 Maths Chapter 11 Exercise 11.4 for more answers visRead more
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm
Area of quadrilateral ABCDF = Area of ∆ABC + Area of ∆ADC
= 1/2 x AC x BM + 1/2 x AC x DN
= 1/2 x 22 x 3 + 1/2 x 22 x 3
= 3 x 11 + 3 x 11
= 33 + 33
= 66 cm²
Thus, the area of quadrilateral ABCD is cm².
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
In the following figures, find the area of the shaded portions:
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm Area of shaded portion = Area of rectangle ABCD – (Area of ∆FAE + area of ∆EBC) = (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC) = (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10) = 180 – (30 + 40) = 180 – 70 = 110 cm² (ii) Here, SR = SU +Read more
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion
= Area of rectangle ABCD – (Area of ∆FAE + area of ∆EBC)
= (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC)
= (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10)
= 180 – (30 + 40)
= 180 – 70
= 110 cm²
(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS – Area of ∆QPT – Area of ∆TSU – Area of ∆UQR
= (SR x QR) – 1/2 x PQ x PT – 1/2 x ST x SU – 1/2
= 20 x 20 – 1/2 x 20 x 10 – 1/2 x 10 x 10 – 1/2 x 20 x 10
= 400 – 100 – 50 – 100
= 150 cm²
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (i) the area of the whole land. (ii) the area of the flower bed. (iii) the area of the lawn excluding the area of the flower bed. (iv) the circumference of the flower bed.
Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5m And radius of the circular flower bed = 2m (i) Area of the whole land = length x breadth = 10 x 5 = 50 m² (ii) Area of flower bed = πr² = 3.14 x 2 x 2 = 12.56 m² (iii) Area of lawn excluding the area of the flower bed = area ofRead more
Length of rectangular lawn = 10 m,
breadth of the rectangular lawn = 5m
And radius of the circular flower bed = 2m
(i) Area of the whole land = length x breadth
= 10 x 5 = 50 m²
(ii) Area of flower bed = πr²
= 3.14 x 2 x 2 = 12.56 m²
(iii) Area of lawn excluding the area of the flower bed
= area of lawn – area of flower bed
= 50 – 12.56
= 37.44 m²
(iv) The circumference of the flower bed = 2πr
= 2 x 3.14 x 2 = 12.56 m
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? ( Take π = 3.14)
Radius of pipe = 4 cm Wrapping cord around circular pipe = 2πr = 2 x 3.14 x 4 = 25.12 cm Again, wrapping cord around a square = 4 x side = 4 x 4 = 16 cm Remaining cord = Cord wrapped on pipe – Cord wrapped on square = 25.12 – 16 = 9.12 cm Thus, she has left 9.12 cm cord. Class 7 Maths Chapter 11 ExeRead more
Radius of pipe = 4 cm
Wrapping cord around circular pipe = 2πr
= 2 x 3.14 x 4 = 25.12 cm
Again, wrapping cord around a square = 4 x side
= 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16
= 9.12 cm
Thus, she has left 9.12 cm cord.
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find: (i) the area covered by the roads. (ii) the cost of constructing the roads at the rate of ₹110 per m².
(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and EF = 90 m and KL = 3 m and KN = 3 m Area of roads = Area of PQRS + Area of EFGH – Area of KLMN [∵ KLMN is taken twice, which is to be subtracted] = PS x PQ + EF x EH – KL x KN = (60 x 3) + (90 x 3) – (3 x 3) = 180 + 270 – 9 = 441 m² (ii) The cost of 1m²Read more
(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and
EF = 90 m and KL = 3 m and KN = 3 m
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (60 x 3) + (90 x 3) – (3 x 3)
= 180 + 270 – 9
= 441 m²
(ii) The cost of 1m² constructing the roads = ₹110
The cost of 441m² constructing the roads = ₹110 x 441 = ₹48,510
Therefore, the cost of constructing the roads = ₹48,510
Class 7 Maths Chapter 11 Exercise 11.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/