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Class 7 R S Aggarwal Maths Solutions

Contents

20. Mensuration 20A, 20B, 20C

 

EXERCISE -20A MENSURATION

 

Q1. L = 24.5m breadth = 18m.

Area of rectangle = l x b = 24.5x 18 = 441m2.

Area of rectangle = l x b = 12. 5 x 8 = 10 m2.

Q2. Let ABCD be the rectangular plot. Then, AB = 48m & ac = 50m.

Let BC = x m. From right triangle ABC, we have, www.rsmaths99.com

AC2 = AB2 + BC2 => 502 = 482 + X2 => X2 = 2500 2304 = 196.

  X = 14 m. BC = 14m. Hence, the area of the plot = (48 x 14)m2 = 672m2.

 

Q3. Let the sides be 4y & 3y, then, 4y x 3y = 1728 m2. 12y2 = 1728.

Y2 =

The sides are 4y = 4 x 12 = 48m. and 3y = 3 x 12 = 36m.

Total Fencing = 2(48 + 36)m = 2 x 84 = 168m.

Cost of fencing = 168 x 30 = Rs. 5040. www.rsmaths99.com

Q4. Given, L= 64 m, we know, area of rectangle = l x b = 3584 = 64 x b.

B = Now, perimeter = 2(64 + 56) = 2 x 120 = 240 m x 5 = 1200m.

Since, in 6000m he takes time = 60min.

In 1 min =

In 1200 m =

Q5. Area = 15 x 14 = 600m2. Area of stone = 6dm x 5 dm = 30 dm2.

Since, 1m2 = 100dm2. Then, 600dm2 = 60000dm2.

No. of stone required = www.rsmaths99.com

Q6. Area = 13 x 9 = 117m2. And 75 cm =

Q7. Given, length of room = l = 15m. Let breadth of room = bm.

Breadth of carpet = 75m =

Length of carpet =

Now, area of room = area of carpet. => 15 x b = =

Q9. Given, length of hall = l = 10m www.rsmaths99.com

Breadth of hall = b = 10m. length of hall = l = 5m.

Length of largest pole = = . = = 15m.

 

Q10. Area of square = a2 = (8.5)2m= 8.5 x 8.5 = 72.25 m2.

 

Q11. Area of the square =

=

Q12. 16200 m2 = = => x2 = 32400 => x = = 180 m.

 

Q13. We know, 1 hectare = 10000 m2. Then, www.rsmaths99.com

Area of square = . Now, area of square =

5000 = y = = 100 m.

Q14. Area of square = a2sq. => 6084 = a2. a = 78 x 4 = 312 x 4 = 1248 m.

 

Q18. Given, Area of four walls of room = 77m2.

  2h(l + b) = 77. => 2h(7.5 + 3.5) = 77. => 2h x 11 = 77.

 

  h = =www.rsmaths99.com

 

Q19. Given, Area of four walls= 120 m2.

Also length = 2 breadth => l = 26. => height = h = 4m.

Now, area of 4 walls = 120 m2. => 2h(l + b) = 120.

  2 x 4 [2b + b]= 120. => 3b = www.rsmaths99.com

  B = l = 10m. area of floor = l x b = 10 x 5 = 50m2.

 

Q20. Area to be painted = area of four walls (area of 2 doors + area of 2 windows)

  2h (l + b) (2 x 1.5 x 1 + 2 x 2 x 1)

=> 2 x 3.4 (8.5 + 6.5) - (3 + 4 ). => 6.8 x 15 7 = 102 7 = 95.

Cost of painting = Rs. 95 x 160 = Rs. 15200.

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Q4.

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EF = 38 + 2 x 2.5 = 38 + 5 = 43m. FG = 25 + 5 = 30m.

Area ABCD = 38 x 25 = 950m2. Area of EFGH = 43 X 30 = 1290m2.

Area of path = 1290 950 = 340 x 120 = Rs.40800.

Q5.

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EF = 9.5 + 2 x 1.25 = 9.5 + 2.50 = 12m. FG = 6 + 2.50 = 8.50m.

Area of EFGH = 12 x 8.50 = 102m2. Area ABCD = 9.5 x 6 = 57m2.

Area of verandah = 102 57 = 45m2.

Cost of cementing the floor of verandah = 45 x 80 = Rs.3600.

 

Q6. Area of ABCD = (2m 80cm)2 = (2.80m)2 = 280cm2.

EF = 280cm + 2 x 30 cm = 280 + 60 cm = 340cm. Area of EFGH = 340cm2 .

Increase in area = (3402 - 2802)cm2 =115600 78400 = 37200cm2 = 3.72m2.

Area of enlarged flower bed = 115600cm2 = 11.56m2.

 

Q7. Perimeter of park = 2 x (l + b)= 2 x (2y + y) = 2 x 3y = 6y = 240m. => y =

Length = 2 x y = 2 x 40 = 80m. breadth = y = 40 m.

Area of ABCD = 80 x 40 = 3200m2. www.rsmaths99.com

EF = 80 2 x 2m = 80 4 = 76m. FG = 40 4 = 36m.

Area of EFGH = 76 X 36 = 2736m2. Area of path = 3200 2736 = 464m2.

Cost of paving the path = 464 x 80 = Rs.37120.

 

Q8. Area of ABCD = (22 x 15 5)m = 341m2. EF = 22 2 x 0.75 = 22 1.5 = 20.5m.

FG = 15.5 1.5 = 14m.

Area of EFGH =area uncovered = 20.5 x 14 = 287m2.

Area of the carpet = 341 287 = 54m2.

Now, area = 287m2 and breadth = 82cm. Then, area = l x b = l x

L =

Cost for 1m = Rs. 60 . www.rsmaths99.com

Cost for 350m = 60 x 350 = Rs.21000.

 

Q9. Area of path = 165m2. Let AB = y m. Then EF = (y + 2.5 +2.5)m = (y + 5 )m.

Area of the square ABCD = AB2 = ( y2) m2.

Area of square EFGH = EF2 = (y + 5)2 m2.

Area of path = area of EFGH - Area of the ABCD .

= {(y + 5)2 - y2} m2. = (Y2 + 10Y + 25 Y2) m2. = (25 + 10Y) m2

25 + 10Y = 165 => 10Y = 165 25 = 140. => Y =

Area of lawn = (14)2m2 = 14 x 14m2 = 196m2.

 

Q10. Given, area of path = 305m2. Let AB = 5y , BC = 2y.

Area of ABCD = 5y x 2y = 10y2 m2.

EF = 2y + 5 = (5y + 5)m. FG = (2y + 5) m .

Area of EFGH = (5y + 5) (2y + 5) m = 10y2 + 35 y + 25.

Area of path = Area of EFGH - Area of ABCD. www.rsmaths99.com

= 10y2 + 35 y + 25 10y2 = 35y + 25.

35y + 25 = 305. => 35y = 280 => y =

AB = l = 5 x 8 = 40m. and BC = b = 2 x 8 = 16m.

 

Q11.

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EF = 70m, FG = 5m, LK =50, JK = 5m.

Area of path = Area EFGH + area IJKL area MNOP.

= 70 x 5 + 50 x 5 5 x 5 = 350m2 + 250m2 25 = 575m2.

Cost of constructing the road at Rs. 120 pre m2 = 600 x 120 = Rs. 69000.

 

Q12.

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AB = 115m, BC = 2m, EF = 64m, EH = 2.5m.

Area of path = Area ABCD + area EFGH - area MNOP

= (115 X2) + (64 X 2.5) + (2.5 X 2)

= 230 + 160 5 = 385 m2. Cost of gravelling the roads = 385 x 60 = Rs.23100.

 

Q13.

Area of ABCD = 50 x 40 = 2000m2. EF= 50m, FG = 2m, LK = 40m, JK = 2.5m.

Area of the road = area EFGH + area IJKL area MNOP.

= (50 x 2) + (40 x 2.5) + (2 x 2.5) = 100 + 100 5= 195m2.

Remaining portion of the field = 2000 195 = 1805m2.

 

Q14. I] EF = 43 2 x 1.5 = 43 3 = 40m. FG = 27 2 x 1 = 27 2 = 25m.

Area EFGH = 40 X 25 = 1000m2. Area ABCD = 43 x 27 = 1161m2

Area of shaded region = 1161 1000 = 161m2.

Ii] EF = 40m, FG = 3m, LK = 40m, JK = 2m, AB = 40m, BC = 40m.

Area of ABCD = 40 x 40 = 1600m2. Area of EFGH = 40 x 3 = 120m2.

Area of IJKL = 40 X 2 = 80m2. Area of MNOP = 2 X 3 = 6m2.

Area of unshaded region = area EFGH + area IJKL area MNOP.

= 120 + 80 6 = 194m2.

Area of shaded region = area ABCD area of unshaded region.

= 1600 194 = 1406m2.

Q15. Area of ABCD = 24 x 19 = 456m2. www.rsmaths99.com

In area EFCD : DC = 24m, CF = 19 16.5 = 2.5m. area = 2.4 x 2.5 = 60m2.

In area HBCG : BC = 19m, HB = 4m. Area = 4 x 2.5 = 10m2.

Area of shaded region = area EFGH + area HBCG area OFCG.

= 60 + 76 -10 = 126m2.

 

Q16. In area ANMC : AB = 15cm, CM = 3cm. Area ANMC = 15 x 3 = 45cm2.

In MDLK : MD = 12 3 = 9cm, DL = 3cm. Area MDLK = 9 x 3 = 27cm2.

In FEBN : NB = 12 3 = 9cm, EB = 3cm. (since, AB = CD). Area = 9 x 3 = 27cm2.

In area GHIJ : GH = 5cm, HI = 3cm. Area GHIJ = 5 x 3 = 15 cm2.

Total area = area ANMC + area MDLK + area FEBN + area GHIJ.

= 45 + 27 + 27 + 15 = 114cm2. www.rsmaths99.com

 

Q17. In area ABCP : AB = 3.5m, AP = 0.5m. Area ABCP = 3.5 x 0.5 = 1.75m2.

In area ODEN : OD = 3.5 (0.5 + 0.5) = 3.5 -1 = 2.5m. DE = 0.5m.

Area ODEN = 2.5 X 0.5 = 1.25m2.

In area MFGL: MF = OD (0.5 +0.5)= 2.5 1 = 1.5m. FG = 0.5m.

Area MFGL = 1.5 x 0.5 = 0.75m2.

In area KHIJ : KH = MF -1= 1.5 -1 = 0.5m. HI = 0.5m.

Area KHIJ =0.5 x 0.5 = 0.25. www.rsmaths99.com

Total area = area ABCP + area ODEN + area MFGL + area KHIJ

= 1.75 + 1.25 + 0.75 + 0.25 = 4m2.

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EXERCISE 20C AREA OF PARALLELOGRAM AND RHMBUS

1.      Area of ||gm ABCD = 2 x (area of triangle ABC)

2.      Area of ||gm ABCD = (base x height) sq units.

3.      Area of rhombus = b x h

4.      Area of rhombus =

 

Q1. Area of ||gm ABCD = (base x height) sq units.

= 32 x 16.5 = 528cm2.

Q2. Base = 1m60cm = 1.60m & height = 75 cm = 0.75m.

Area of ||gm ABCD = 1.60 x 0.75 = 1.2m2.

Q3. b = 14dm = 140cm, h = 6.5dm = 65cm. (since, 1dm = 10cm)

I] Area of ||gm = b x h = 140 x 65 = 9100cm2. www.rsmaths99.com

Ii] b = 14dm = 140cm = h = 6.5dm = 65cm = 0.65m.

Area of ||gm = b x h = 1.40 x 0.65 = 0.91m2.

 

Q4. Given, AB = 15cm. area = 54cm2. Area = b x h => 54 = 15 x h.

h =

 

Q5. Given, b= 18cm, area = 153. Area of ||gm = b x h

=> 153 = 18 x h. => h = www.rsmaths99.com

Q6. Area of ||gm ABCD =b x h = AB x AL = 18 x 6.4 = 15.2cm2.

Area of ||gm ABCD =b x h = BC x AM.

115.2 = 12 x AM. AM =

 

Q7. Let the required height be h cm. Then, h x 8 = 15 x 4. => h =

 

Q8. Area of ||gm = b x h. => 108 = y x => 108 x 3 = y2.

Y = = = 2 x 9 = 18cm.

Base = 18cm, height = www.rsmaths99.com

 

Q9. area of the ||gm = b x h

512 = 2y x y => 512 = 2 y2. y2 = y = = 16 cm.

B = 2 x 16 = 32 cm. and h = y = 16 cm.

 

Q10. I] Area of rhombus = b x h = 12 x 7.5 cm = 90 cm2.

Ii] 2dm = 20 cm. then, Area of rhombus = b x h = 20 x 12.6 = 252 cm2.

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Q11. Area of rhombus =

= = 224 cm2.

Ii] 8dm = 80cm, then, 8dm5cm = 85 cm, and 5dm 6cm = 56 cm.

Area =

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Q12. Let ABCD be the rhombus in which AB = 20 cm and AC = 24 cm.

We know that the diagonals of a rhombus bisect each other at right angles.

OA = = = and LAOB = 90o.

OB = = = 16 cm.

BD = 2 x OB = 2 x 16 = 32 cm.

Area of rhombus ABCD =

= 24 x 32 =

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Q13. Area = 148.8 cm2. Let y be the another diagonal.

Area of rhombus =

148.8 = => 148.8 = 9.6y y = 15.5 cm.

Hence the another diagonal = 15.5 cm.

 

Q14. Area ABCD = 119 cm2. Perimeter = 56 cm.

4 (side)= 56 cm. [ Perimeter = 4(side)]

  Side =

Now, area of the Rhombus = side x height.

119 = 14 x h. h =

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Q15. Let base = y . area of the rhombus = y x 17.5.

441= y x 17.5 y = 25.2 cm.

 

Q16. Area of triangle =

Area of rhombus =

204.6 =

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