question
a rectangle ahas an area of a^2 -16q units find its sides
98
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! ;) 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 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. 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 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 given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units By Pythagoras, if the sum of squares of two sides is equal to the square of the largest side then the triangle is said to be right-angled


Here given sides of the triangle are 7, 24 and 25


Largest side = 25


252 = 625  Sum of squares of remaining two sides 72 + 242 = 49 + 576 = 625


∴ 72 + 242 = 252 Therefore triangle having sides 7,24 and 25 is a right-angled triangle area of a semi-circle= 77 m2 we are given`16 cot x=12` .we are asked to find the following `(sin x-cos x)/(sin x+cos x)` we know that: `cot x= "base"/"perpendicular" ` → "base"=3 → "perpendicular"=4 → `"hypotenuse"= sqrt(("perpendicular")^2+("base")^2)` → `"hypotenuse"=sqrt(16+9)` →`"hypotenuse"=5` now we have `16 cot x=12` `cot x=12/16` `cot x=3/4`, we know sin x=`"perpendicular"/"hypotenuse" and cos x= "base"/"hypotenuse"` now we find `(sin x- cos x)/(sin z+cos x)` = `(4/5-3/5)/(4/5+3/5)` =`(1/5)/(7/5)` =`1/7` 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