Sides of a triangle are 7, 24 and 25. Determine whether the triangle is right-angled triangle or not
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
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 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. 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 let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved answer: k=7 consider the parallelogram abcd with diagonals ac and bd as shown and they intersect at right angles at oa parallelogram is a square if its adjacent angles are 90° and adjacent sides are equalso we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a squareconsider δbad and δcdaac = bd … givenab = dc … opposite sides of a parallelogramad is the common sidetherefore, δbad ≅ δcda … sss test for congruency⇒ ∠bad = ∠cda …corresponding angles of congruent triangles …(i)as it is given that abcd is parallelogram⇒ ∠bad + ∠cda = 180° … sum of adjacent angles of a parallelogram is 180°using equation (i)⇒ ∠cda + ∠cda = 180°⇒ 2 × ∠cda = 180°⇒ ∠cda = 90°⇒ ∠bad = 90°thus, adjacent angles are 90° … (iii)consider δaob and δaod∠aob = ∠aod … both 90° because given that diagonals intersect at right anglesod = ob … diagonals of a parallelogram bisect each otherao is the common sidetherefore, δaob ≅ δaod … sas test for congruency⇒ ab = ad … corresponding sides of congruent trianglesthus, adjacent sides are equal … (iv)from statements (iii) and (iv) we can conclude that parallelogram abcd is a squaretherefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square. answer: `1/221`
solution: n/a fleming’s right hand rule is used to find the direction of the? correct 3 3/7 days