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 there be n sides of the polygon. Then it has n vertices. The total number of straight lines obtained by joining n vertices by talking 2 at a time is nC2 These nC2 lines also include n sides of polygon. Therefore, the number of diagonals formed is nC2 - n. Thus, nC2 - n = 44? [n(n - 1)/2] - n = 44? ( n2 - 3n) / 2 = 44? n2 - 3n = 88 ? n2 - 3n - 88 = 0 ?(n - 11) (n + 8) = 0 ? n = 11
a1 = 68/4 = 17 cmand a2 = 60/4 = 15 cm [ where a1 and a2 are sides]According to the question,Area of the third square = [(17)2 - (15)2 ] = (17 + 15) (17 - 15) = 32 x 2 = 64 sq cmLet a3 = Side of the third square.According to the question, (a3)2 = 64 sq cm ? a3 = ?64 = 8 cm? Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.
Original area = (22/7) x 9 x 9 cm2New area = (22/7) x 7 x 7 cm2? Decrease = 22/7 x [(9)2 -(7)2] cm2=(22/7) x 16 x 2 cm2Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %= 39.5 %
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