Question

prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

Answer 1

$\Huge\textbf{\red{Solution}}$

$\textbf{\red{Given}}$

$\textbf{A rhombus ABCD with diagonals AC}$

$\textbf{and BD intersecting at a point O}$

To prove : sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

$\mathbf{=> AB^2 + BC^2 + CD^2 + AD^2 = AD^2 + BD^2}$

$\textbf{\green{Proof : }}$

$\bf\textrm{We know that}$

$\textbf{sides are rhombus are equal}$

$\mathbf{\bold{AB = BC = CD = AD}}$

$\bf\textsf{Also we know that diagonals of a rhombus}$$\bf\textsf{intersect each other at right angles}$

$\bf\textsf{Therefore}$

$\mathbf{∠ AOB = ∠ BOC = ∠COD = ∠AOD = 90°}$

AO = CO = 1/2 AC ------------(1)

BO = DO = 1/2 BD------------(2)

Now ∆ AOB is a right angled triangle

$\bf\textsf{Using Pythagoras theorem}$

$\mathbf{h^2 = p^2 + b^2}$

$\mathbf{AB^2 = OA^2 + OB^2}$

$\bf {AB}^{2} =( \dfrac{1}{2} {AC})^{2} + ( \dfrac{1}{2} {BD})^{2} ----from \: 1 \: and \: 2$

$\bf{AB}^{2} = \dfrac{ {AC}^{2} }{4} + \dfrac{ {BD}^{2} }{4}$

$\bf {AB}^{2} = \dfrac{ {AC}^{2} + {BD}^{2} }{4}$

$\bf 4{AB}^{2} = {AC}^{2} + {BD}^{2}$

$\bf{AB}^{2} + {AB}^{2} + {AB}^{2} + {AB}^{2} = {AC}^{2} + {BD}^{2}$

$\small\textbf{(SINCE SIDES \: OF \: A \: RHOMBUS \: ARE \: EQUAL \: )}$

$\mathbf{AB^2+ BC^2 + CD^2 + AD^2 = AD^2 + BD^2}$

$\Huge\mathrm{Thanks}$

Have a colossal day ahead

$\Huge\mathfrak{\blue{Be BrainlY}}$

Answer 2

I hope this is helpful