if a square and a Rombus stand on the same base,prove that the area of the square is greater than that of the rombus.
RoshanRox:
I think the question is wrong?
nope
its in my book
you are of which class
9th
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let the side of the square be "a"
let the diagonals be 2d' and 2d"
area of square "a square (a2)
using Pythagoras theorem
d' ka square +d" ka square =a ka square
ar. of rhombu a2= 1/2 ×2d'×2d" =2d'd"
(d'-d")=s' ka sq. + d" ka sq. -2 d' d"
area of square = positive quantity + area of rhombus
therefore area of sq. is greater than area of rh.
let the diagonals be 2d' and 2d"
area of square "a square (a2)
using Pythagoras theorem
d' ka square +d" ka square =a ka square
ar. of rhombu a2= 1/2 ×2d'×2d" =2d'd"
(d'-d")=s' ka sq. + d" ka sq. -2 d' d"
area of square = positive quantity + area of rhombus
therefore area of sq. is greater than area of rh.
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