Prove that the sum of squares of the sides of a rectangle is equal to the sum of squares of its diagonals.
(N/A) Let $ABCD$ be a rectangle with sides $AB = CD = l$ and $BC = DA = b$. Let the diagonals be $AC$ and $BD$.
In a rectangle,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle ABC$. By the Pythagoras theorem:
$AC^2 = AB^2 + BC^2 = l^2 + b^2$
Similarly,in the right-angled triangle $\triangle BCD$:
$BD^2 = BC^2 + CD^2 = b^2 + l^2$
Sum of the squares of the diagonals:
$AC^2 + BD^2 = (l^2 + b^2) + (b^2 + l^2) = 2l^2 + 2b^2$
Sum of the squares of the sides:
$AB^2 + BC^2 + CD^2 + DA^2 = l^2 + b^2 + l^2 + b^2 = 2l^2 + 2b^2$
Thus,the sum of the squares of the sides is equal to the sum of the squares of the diagonals.
In a rectangle,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle ABC$. By the Pythagoras theorem:
$AC^2 = AB^2 + BC^2 = l^2 + b^2$
Similarly,in the right-angled triangle $\triangle BCD$:
$BD^2 = BC^2 + CD^2 = b^2 + l^2$
Sum of the squares of the diagonals:
$AC^2 + BD^2 = (l^2 + b^2) + (b^2 + l^2) = 2l^2 + 2b^2$
Sum of the squares of the sides:
$AB^2 + BC^2 + CD^2 + DA^2 = l^2 + b^2 + l^2 + b^2 = 2l^2 + 2b^2$
Thus,the sum of the squares of the sides is equal to the sum of the squares of the diagonals.
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