a square lawn has a 2m wide path surrounding it if the area of the path is 136 square - math - perimeter and area
a square lawn has a 2m wide path surrounding it if the area of
the path is 136 square m, find the area of the lawn please give the
ans with detailed steps
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let the length of path be a m


total area ![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/ceaxujzx7923478049123545928%20png)


area of lawn = total area â€“ area of
path

![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/1800204/2012_12_04_21_09_56/mathmlequation2002748073868866733%20png)


area of lawn


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/dwqusyms3311632565229211681%20png)
you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) 18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is given:volume of given cube = 64cm3side = a cmvolume = (a)364 = (a)3a = 4cmtwo cubes are joined together.length of resultant cube = 4+4 = 8cmbreadth = 4cmheight = 4cm(a) lsa of cuboid = 2(lh+hb)= 2(8(4) + 4(4))= 2(32+16)= 2(48) = 96cm2⇒ lsa of cuboid = 66m2(b)tsa of cuboid = 2(lh+hb+lb)= 2(8(4) + 4(4) + 8(4))= 2(32+16+32)⇒ tsa of cuboid = 2(80) = 160m2 given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 area of trapezium = area of trapezium = = therefore = 24 cm given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region the straight distance between two place on a map can be measured by a strip of paper and divider.use a ruler to measure the distance between the two places. if the line is quite curved, use a string to determine the distance, and then measure the string. find the scale for the map you're going to use. it might be a ruler bar scale or a written scale, in words or numbers. Correct option is A) 9.378 m2


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result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits) area of a semi-circle= 77 m2