the length of a rectangle is 20 cm more than its breadth.if the perimeter is 100 cm find the dimensions of the rectangle and find its area
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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! ;) total cost of fencing = rs 9600
rate of fencing = rs 24
then, perimeter of field = cost/rate = 9600/24 = 400mts
let the length and breadth of the rectangular field be 5x and
3x
then perimeter= 2(l+b)
= 2(5x+3x)
= 2*8x = 16x
but the perimeter of the given field is 400 m
so, 16x = 400
x = 400/16 = 25
length of the field = 5*25mts = 125mts
breadth of the field = 3*25mts = 75mts 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 the dimensions of base of a rectangular box of greatest volume that can be constructed from 100 cm square of hard board sheet, if the box is three times as long as it is wide, are \_\_\_\_\_\_. the sides of the triangle are 6 cm, 8 cm and 10 cm. the longest side is 10 cm. (10)2 = 100 ….(i) now, the sum of the squares of the other two sides will be, (6)2 + (8)2 = 36 + 64 = 100 ….(ii) (10)2 = (6)2 + (8)2 ….. from (i) and (ii) by the converse of pythagoras theorem: the given sides form a right angled triangle. the distance to be covered to go one round along perimeter of this quadrilateral is12 cm + 3 cm + 5.6 cm + 19 cm consider the following figure:two circles with centres q and r touch each other externally at x. another circle, with centre p, touches the other two circles internally at s and t. if pq = 8 cm, qr = 11 cm and pr = 9 cm, find the radius of circle r. area of a semi-circle= 77 m2 hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t.