Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
Let the breadth of floor be 'b' m. Then, length of the floor is 'l = (b + 25)' Area of the rectangular floor = l x b = (b + 25) × b According to the question, (b + 15) (b + 8) = (b + 25) × b b 2 + 8 b + 15 b + 120 = b 2 + 25 b 2b = 120 b = 60 m. l = b + 25 = 60 + 25 = 85 m. Area of the floor = 85 × 60 = 5100 sq.m.
Area of equilateral triangle = ?3a2/4 = x ......(i)And perimeter = 3a = y ? a = y/3 ....(ii)Now, Putting the value of a from Eq. (ii) in Eq. (i). we get?3 (y/3)2/4 = x ? x = ?3 x y2/36? x = y2/3?3x = y2/12?312?3 x = y2On squaring both sides, we get y4 = 432x2
The man takes 3600 s for 14.4 km The man will take 88 s for 14.4 x (88/3600) = 352/1000 km = 352 m Now, circumference of circular field = 352 m ? 2?r = 352 m2 x (22/7) x r = 352 ? r = 56 m Therefore, area of the field = ?r2= (22/7) x 56 x 56 = 8 x 22 x 56 m2= 9856 sq m.
Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
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