area of triangle formed by the vertices `z, omega z and z+omega z` is `4/sqrt(3)`, `omega` is complex cube roots of unity then `|z|` is (a) `1` (b) `4/3` (c) `3/4` (d) `4/(sqrt3)`
solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer
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 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! ;) 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: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle 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. 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 given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b 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 in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2