NCTB Class 7 Math Chapter Three Exercise 3 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 3 “Measurement” Exercise 3 Solution.

Board |
NCTB |

Class |
7 |

Subject |
Math |

Chapter |
3 |

Chapter Name |
“Measurement” |

Exercise |
3 Solution |

**Exercise:- 3**

**(1) Solution:-** 1sq.foot = (c) 929sqcm.

**(2) Solution:-** If the length of one edge of a cube is 3m then total surface area of the cube – (a) 5sqm.

Therefore, Area of one surface= (3×3) = 9sqm.

Therefore, Total area= (6×9) = 9sqm.

**(3) Solution:-** Let, breadth of garden = x m.

Therefore, length of garden= 3x m.

Therefore, 2 (3x+x) = 400

=> 4x = 200

=> x = 200/4

= 50

Therefore, Breadth = 50m.

Length= (3×50)m

= 150m (c)

**(4) Solution:-** Area = (150×50)sqm

(d) = 7500sqm.

**(5) Solution:-** meaning of deci in Latin – (b) One tenth.

**(6) solution:-** Perimeter of land = 2 (20+15)

= (b) 70m.

**(7) solution:**– Without work way, length of land

= {20 – (2+2)}m

= 16m.

Breadth of land

= {15- (2+2)}

= 11m.

Therefore, Area= (16×11) sqm

= 176sqm.

**(8) (a) Solution:-** 40390cm= 0.4039km.

**(8) (b) Solution:-** 75m 250mm

= 75m+ 250mm

= 75m+ 250/100m [Because, 1m= 1000mm]

= 75+0.5

= 75025m

= 75.25/1000

= 0.07525km.

**(9) Solution:-** 5.37 decimetre = (5.37x 10)m

= 53.4m

Therefore, 1m= 10decm.

Therefore, 53.7m = 537decm.

**(10) (a) Solution:-** Area of triangle

= ½ x .10x 6

= 30sqm.

**(10) (b) Solution:-** Area of triangle

= ½ x 25x14cm.

= 175sqcm.

**(11) Solution:-** Let, breadth of a rectangular plot x m. and length = 3x m.

Therefore, perimeter= 2 (3x+x)

= 2x4x

= 8x m.

Therefore, 8x= 1000

=> x= 1000/8

X = 125

Therefore, breadth= 125m.

Therefore, length= (3×125)m.

= 375m.

Therefore, length 375m and breadth 125m.

**(12) Solution:**– Perimeter of rectangular park

= 2 ( l+ b)

= 2 (100+50)m.

= (2×150)m.

= 300m.

Therefore, 1m cost to fencing is 100TK.

Therefore, 300m cost to fencing in (300×100) TK.

= 30000TK.

**(13) Solution:-** Given, base of parallelogram = 40m

Therefore, height of parallelogram = 50m.

Therefore, Area= (40×50) = 2000sqm.

(**14) Solution:-** Area of cube = (4x40sqm

= 16sqm.

Therefore, Total surface area of cube

= (16×6)

= 96sqm.

**(15) Solution:-** Produces potatoes

= 500kg 700gm

= (500×1000) + 700gm

= 5007000gm.

Joseph, 1 piece of land produces 500 700gm potatoes.

Therefore, 11 piece of land produces (11×500 700) Gm potatoes

= 5507700gm

= 5507000+7.00gm

= 5507000/1000+ 7000gm

= (5507kg+700gm)

**(16) Solution:-** In 16 area land produces 28 metric ton paddy

Therefore, 1 area land produces 28/16 metric ton paddy.

= 7/4 metric ton paddy.

= 1 – ¾ metric ton.

**(17) Solution:-** 1 month = 30days.

In 30 days produces rod = 200000m.t.

Therefore, 1 days produces rod = 200000/30 m.t

= 666 – 2/3m.t

= 666m.ton + 2/3x 1000kg

= 666m.ton +666- 2/3kg

= 666m.ton+666kg +2/3x 1000gm

= 666m.ton+ 666kg+666-2/3gm.

**(18) Solution:-** In 1 day sell 20kg 400gm.

= 20 400gm.

Therefore, in 30 day sell = (20400×30) gm

= 612 000gm

= 612kg.

**(19) Solution:-** in 1 piece of land produces 20 kg 850gm mustard = 20850gmmustard.

Therefore, in 7 piece of land produces = (20850×7) mustard.

= 145950gm.

= (145000+950)gm.

= 145kg 950gm.

**(20) Solution:-** Volume of mug = 1500cu.cm.

= 1500/1000 list

= 1.5 lit.

Therefore, 1.5 lit in 1 mug

Therefore, 1 lit in 1/1.5 mug

Therefore, 270 lit in 270/1.5 mug

= 180mug.

**(22) Solution:-** In 1 day requires milk 125 lit

Therefore, 30 day requires mil (30×1.255) lit

= 37.50lit.

Now, 1 lit of milk cost 52TK.

Therefore, 37.50 lit of milk cost (37.50×52) TK.

= 1650TK.

**(23) Solution:-**

**(24)** **Solution:-** Let, the breadth of room X m.

Therefore, length = (3xx) 3xm.

Therefore, area = (3Xx) sqm.

= 3x^{2}sqm.

Therefore, 7.50TK – 1 sqm.

1TK. — 1/7.50 sqm.

Therefore, 1102.50TK. – 1×1102.50/7.50 sqm

= 147sqm.

Therefore, Area of room 147sqm

Therefore, 3x^{2} = 147

=> x^{2} = 147/3

= 49

=> x **= √**49

= 7

Therefore, Breadth is 7m out length (7×3) = 21m.

**(25) Solution:- **

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