determine whether the triangle having sides (a – 1) cm, cm and (a + 1) cm is a right angled triangle.
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 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 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 correct option is a) right electrode will increase 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! ;) fleming’s right hand rule is used to find the direction of the? a - ii, b - iv, c - i, d - iii answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a correct a, b and c