New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Perimeter and Area Chapter 14C Solution
(1) (a) Area = (17 X 10) cm2
= 170 cm2
(b) Area = (5 X 8) m2
= 40 m2
(c) Area = (200 X 120) m2
= 24000 m2
(d) Area = 152 m2
= 225 m2
(e) Area = 6.32 km2
= 39.69 km2
(2) (a) Breadth = 90 sqcm/15 cm
= 6 cm
Perimeter = 2 X (15+6) cm
= 2 X 21 cm
= 42 cm
(b) Breadth = 117 sq cm/13 cm
= 9 cm
Perimeter = 2 X (13+9) cm
= 2 X 22
= 44 cm
(3) (a) Length = 42 sq cm/6 cm
= 7 cm
Perimeter = 2 X (7+6) cm
= 2 X 13 cm
= 26 cm
(b) Length = 480 sqcm/20 cm
= 24 cm
Perimeter = 2 X (24+20) cm
= 2 X 44 cm
= 88 cm
(4) (a) Side2 = 49 cm2
Or, side = √49cm2
= 7 cm
Perimeter = (4 X 7) cm
= 28 cm
(b) Side2 = 144 cm2
Or, side = √144cm2
= 12 cm
∴ Perimeter = (4 X 12) cm
= 48 cm
(c) Side2 = 256 cm2
Or, side = √256cm2
= 16 cm
∴ Perimeter = (4 X 16) cm
= 54 cm
(d) Side2 = 900 cm2
Or, side = √900cm2
= 30 cm
∴ Perimeter = (4 X 30) cm
= 120 cm
(5) Area of the carpet = (5 X 4) m2
= 20 cm
Area of the square room = 72 m2
= 49 m2
Uncovered area of the floor = (49 – 20) m2
= 29 m2
Ans – uncovered area of the floor is 29 m2
(6)
Area of the lawn PQRS = (10 X 8) m2
= 80 m2
BC = AD = (10 + 1 + 1) m
= 12 m
AB = CD = (8 + 1 + 1) m
= 10 m
Area of the lawn along with path ABCD = (12 X 10) m2
= 120 m2
∴ Area of the path = (120 – 80) m2
= 40 m2
Ans – Area of the path is 40 m2
(7) Area of the picture = (30 X 24) cm2
= 720 cm2
Length of the picture along with card = 30 + (2 X 3) cm
= 30 + 6
= 36
Breadth of the picture along with card = 24 X (2X3) cm
= 24 + 6 cm
= 30 cm
∴ Area of the card = (36 X 30) cm2
= 1080 cm2
∴ The area of the margin = (1080 – 720) cm2
= 360 cm2
Ans – The area of the margin is 360 cm2
(8) Area of the courtyard = (40 X 24) m2
= 960
Length of the courtyard without path = 40 – (3 X 2) m
= 40 – 6 m
= 34 m
Breadth of the courtyard with path = 24 – (3X2) m
= 24 – 6 m
= 18 m
∴ Area of the courtyard without = (34X18) m2
= 612
∴ Area of the path = (960 – 612) m2
= 348 m2
Ans – Area of the path is 348 m2
(9) 1 hector = 10000 sq meter
6 hector = 60000 sq meter
Area of each garden = (40 X 15) m2
= 600 m2
No of gardens can be made out = 60000/600
= 100
Ans – 100 gardens can be made out of this land.
(10) Area of the bricks = (22 X 15) cm2
= 330 cm2
Area of the court = (3000 X 2200) cm2
= 6.600,000 cm2
∴ N. of bricks will be required to pave = 6600000/330
= 20,000
Ans – 20,000 bricks is required to pave the court.
(11) Area of the (25 X 30) cm2
= 750 cm2
Area of the floor room = (900 X 600) cm2
= 540000 cm2
No. of tile to cover the floor = 540000/750
= 720
∴ Total cost = RS (720 X 200)
= RS 144000
Ans – Total cost of covering the floor RS 144000
(12) Area of four walls = 2 (7+5) X3 m2
= 24 X 3 m2
= 72 m2
Ans – Total area of the room is 72 m2
(13) Area of the four wall = 2 (8+6) X 4 m2
= 28 X 4 m2
= 112
Area of roll = (0.8 X 5) m
= 4 m
No. of rolls require = 112/4 = 28
∴ Cost of preparing the room = RS (28 X 120)
= RS 3360
(14) Area of the room (5 X 4.8) m2
= 24 m2
Width of the carpet = 1.2 m
Length of 5m can cover the area of the room = 1.2 X 5 = 6m2
No. of carpet of same length needed to cover the entire room = 24/6
= 4 carpets of 5 m long
Total 4 carpet of 5 m long = (5 X 4) m
= 20 m
∴ Cost of 20m carpet = RS (50 X 20)
= RS 3000
Ans – It would cost RS 3000 to carpet a room.
(15) (a)
BG = 5cm, EG = 6cm
Area of ABGE = 5 X 6 cm2
= 30 cm2
GC = DF = (7 – 5) cm
= 2 cm
DC = FG = 2 cm
Area of GCDF = 22 cm2
= 4 cm2
∴ Area of A B G C D F F = (30 + 4) cm2
= 34 cm2 (Ans)
(b)
AB = FP = (3 + 3 + 3) cm
= 9 cm
Area of ABPF = (10 X 9) cm2
= 90 cm2
SP = EQ = (3+3) cm
= 6 cm
SE = PQ = 4 cm
Area of PQES = (6 X 4) cm2
= 24 cm2
RQ = DC = 3 cm
RD = QC 4 cm
Area of QCDR (3 X 4) cm2
= 12 cm
∴ Area of A B P Q C D E F = (90 + 24 + 12) cm2
= 126 cm2 (Ans)
(c)
XY = UT = 6 cm
XU = YT = (9-7) cm
= 2 cm
Area of YTUX = (6 X 2) cm2
= 12 cm2
PW = XV = 9 cm
PX = WV = 4 cm
Area of XVYP = (9 X 4) cm2
= 36 cm2
YS = QR = 12 cm
YQ = SR = 5 cm
Area of QRSY = (12 X 5) cm2
= 60 cm2
Area of the figure P Q R S T U V W = (12 + 35 + 60) cm2
= 108 cm2 (Ans)
(16 (a) Area of the inner unshaded part
= (2 X 7) cm2
= 14 cm2
Area of whole figure along with shaded part
– (15 X 6) cm2
= 90 cm2
∴ Area of the shaded part = (90 – 14) cm2
= 76 cm2 (Ans)
(b) Length of the inner unshaded part = 20 – (4 X 2) cm
= 20 – 8 cm
= 12 cm
Breadth of the total part including shaded part
= 16 – (3+5) cm
= 16 – 8 cm
= 8 cm
∴ Area of the inner unshaded part
= (12 X 8) cm2
= 96 cm2
∴ Area of the total part = (20 X 16) cm2
= 320 cm2
∴ Area of the shaded part = (320 – 96) cm2
= 224 cm2 (Ans)
(c)
Area of shaded part ABGH = (3 X 6) cm2
= 18 cm2
Area of shaded part CDEF = (3 X 6) cm2
= 18 cm2
IL = JK = (5+3+5) cm
= 13 cm
IJ = LK = 4 cm
∴ Area of IJKL = (13 X 4) cm2 = 52 cm2
∴ Area of total shaded part = (18 + 18 + 52) cm2
= 88 cm2 (Ans)
(d)
Area of PQCB = (2 X 4) cm2
= 8 cm2
Area of NOED = (2 X 4) cm2
= 8 cm2
AF = RG = (5+2+4+2+5) cm
= 18 cm
AR = FG = 3 cm
Area of AFGR = (18 X 3) cm2
= 54 cm2
Area of JKRI = (3 X 4) cm2
= 12 cm2
Area of LMGH = (3 X 4) cm2
= 12 cm2
∴ Area of total shaded part = (8+8+54+12+12) cm2
= 94 cm2 (Ans)
(17) Area of one flower bed = (5 X 4) m2
= 20 cm2
Each flower bed has 2 m path between them
So, length of the garden = (5+5+2) m
= 12 m
=Breadth of the garden = (4+4+2) m
= 10 m
∴ Area of the garden = (12 X 10) m2
= 120 m2
The area of the grass region
= Area of the garden – area of 4 flower beds
= 120 – (4 X 20)
= 120 – 80
= 40 m2
The length of water square = (12 + 3 + 3) m
= 18 m
The breadth of region with garden path = (10+3+3) m
= 16 m
∴ The area of the garden with gravel path = (18 X 16) m2
= 288 m2
∴ Area of the gravel path = (288 – 120) m2
= 168 m2
Ans – Total area of grassy region is 40 m2
Area of the gravel path is 168 m2.
