**New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Perimeter and Area Chapter 14C Solution**

**(1) **(a) Area = (17 X 10) cm^{2}

= 170 cm^{2}

(b) Area = (5 X 8) m^{2}

= 40 m^{2 }

(c) Area = (200 X 120) m^{2}

= 24000 m^{2}

(d) Area = 15^{2} m^{2}

= 225 m^{2}

(e) Area = 6.3^{2} km^{2}

= 39.69 km^{2}

**(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) Side^{2} = 49 cm^{2}

Or, side = √49cm^{2}

= 7 cm

Perimeter = (4 X 7) cm

= 28 cm

(b) Side^{2} = 144 cm^{2}

Or, side = √144cm^{2}

= 12 cm

∴ Perimeter = (4 X 12) cm

= 48 cm

(c) Side^{2} = 256 cm^{2}

Or, side = √256cm^{2}

= 16 cm

∴ Perimeter = (4 X 16) cm

= 54 cm

(d) Side^{2} = 900 cm^{2}

Or, side = √900cm^{2}

= 30 cm

**∴ ****Perimeter = (4 X 30) cm **

**= 120 cm **

**(5) **Area of the carpet = (5 X 4) m^{2}

= 20 cm

Area of the square room = 7^{2} m^{2}

= 49 m^{2}

Uncovered area of the floor = (49 – 20) m^{2}

= 29 m^{2}

Ans – uncovered area of the floor is 29 m^{2}

**(6) **

Area of the lawn PQRS = (10 X 8) m^{2}

= 80 m^{2}

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) m^{2}

= 120 m^{2}

∴ Area of the path = (120 – 80) m^{2}

= 40 m^{2}

Ans – Area of the path is 40 m^{2}

**(7) **Area of the picture = (30 X 24) cm^{2}

= 720 cm^{2}

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) cm^{2}

= 1080 cm^{2}

∴ The area of the margin = (1080 – 720) cm^{2}

= 360 cm^{2}

Ans – The area of the margin is 360 cm^{2}

**(8) **Area of the courtyard = (40 X 24) m^{2}

= 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) m^{2}

= 612

∴ Area of the path = (960 – 612) m^{2}

= 348 m^{2}

Ans – Area of the path is 348 m^{2}

**(9) **1 hector = 10000 sq meter

6 hector = 60000 sq meter

Area of each garden = (40 X 15) m^{2}

= 600 m^{2}

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) cm^{2}

= 330 cm^{2}

Area of the court = (3000 X 2200) cm^{2}

= 6.600,000 cm^{2}

∴ 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) cm^{2}

= 750 cm^{2}

Area of the floor room = (900 X 600) cm^{2}

= 540000 cm^{2}

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 m^{2}

= 24 X 3 m^{2}

= 72 m^{2}

Ans – Total area of the room is 72 m^{2}

**(13) **Area of the four wall = 2 (8+6) X 4 m^{2}

= 28 X 4 m^{2}

= 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) m^{2}

= 24 m^{2}

Width of the carpet = 1.2 m

Length of 5m can cover the area of the room = 1.2 X 5 = 6m^{2}

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 cm^{2}

= 30 cm^{2}

GC = DF = (7 – 5) cm

= 2 cm

DC = FG = 2 cm

Area of GCDF = 2^{2} cm^{2}

= 4 cm^{2}

∴ Area of A B G C D F F = (30 + 4) cm^{2}

= 34 cm^{2} (Ans)

(b)

AB = FP = (3 + 3 + 3) cm

= 9 cm

Area of ABPF = (10 X 9) cm^{2}

= 90 cm^{2}

SP = EQ = (3+3) cm

= 6 cm

SE = PQ = 4 cm

Area of PQES = (6 X 4) cm^{2}

= 24 cm^{2}

RQ = DC = 3 cm

RD = QC 4 cm

Area of QCDR (3 X 4) cm^{2}

= 12 cm

∴ Area of A B P Q C D E F = (90 + 24 + 12) cm^{2}

= 126 cm^{2} (Ans)

(c)

XY = UT = 6 cm

XU = YT = (9-7) cm

= 2 cm

Area of YTUX = (6 X 2) cm^{2}

= 12 cm^{2 }

PW = XV = 9 cm

PX = WV = 4 cm

Area of XVYP = (9 X 4) cm^{2}

= 36 cm^{2}

YS = QR = 12 cm

YQ = SR = 5 cm

Area of QRSY = (12 X 5) cm^{2}

= 60 cm^{2}

Area of the figure P Q R S T U V W = (12 + 35 + 60) cm^{2}

= 108 cm^{2} (Ans)

**(16 **(a) Area of the inner unshaded part

= (2 X 7) cm^{2}

= 14 cm^{2}

Area of whole figure along with shaded part

– (15 X 6) cm^{2}

= 90 cm^{2}

∴ Area of the shaded part = (90 – 14) cm^{2}

= 76 cm^{2} (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) cm^{2}

= 96 cm^{2}

∴ Area of the total part = (20 X 16) cm^{2}

= 320 cm^{2}

∴ Area of the shaded part = (320 – 96) cm^{2}

= 224 cm^{2} (Ans)

(c)

Area of shaded part ABGH = (3 X 6) cm^{2}

= 18 cm^{2}

Area of shaded part CDEF = (3 X 6) cm^{2}

= 18 cm^{2}

IL = JK = (5+3+5) cm

= 13 cm

IJ = LK = 4 cm

∴ Area of IJKL = (13 X 4) cm^{2} = 52 cm^{2}

∴ Area of total shaded part = (18 + 18 + 52) cm^{2}

= 88 cm^{2} (Ans)

(d)

Area of PQCB = (2 X 4) cm^{2}

= 8 cm^{2}

Area of NOED = (2 X 4) cm^{2}

= 8 cm^{2}

AF = RG = (5+2+4+2+5) cm

= 18 cm

AR = FG = 3 cm

Area of AFGR = (18 X 3) cm^{2}

= 54 cm^{2}

Area of JKRI = (3 X 4) cm^{2}

= 12 cm^{2}

Area of LMGH = (3 X 4) cm^{2}

= 12 cm^{2}

∴ Area of total shaded part = (8+8+54+12+12) cm^{2}

= 94 cm^{2} (Ans)

**(17) **Area of one flower bed = (5 X 4) m^{2}

= 20 cm^{2}

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) m^{2}

= 120 m^{2}

The area of the grass region

= Area of the garden – area of 4 flower beds

= 120 – (4 X 20)

= 120 – 80

= 40 m^{2}

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) m^{2}

= 288 m^{2}

∴ Area of the gravel path = (288 – 120) m^{2}

= 168 m^{2}

Ans – Total area of grassy region is 40 m^{2}

Area of the gravel path is 168 m^{2}.