Question

9
Show that in rhombus the sum of the
square of its sides is equal to the Sum
its diagonals​

Answer 1

Appropriate Question :-

Show that, in rhombus the sum of the square of its sides is equal to the sum of square of its diagonals.

Solution :-

Let us consider a rhombus ABCD in which diagonals AC and BD intersects at O.

We know,

• In rhombus, all sides are equal.

So,

• Let AB = BC = CD = DA = x.

And

• Diagonals bisect each other at right angles.

So,

$\red{\rm :\longmapsto\:\bf \: OA = OC = \dfrac{1}{2}AC} \\ \sf \: \blue{and} \\ \red{\rm :\longmapsto\:\bf \: OB = OD = \dfrac{1}{2}BD}$

Now,

• In right triangle OAB,

By using Pythagoras Theorem,

$\red{\rm :\longmapsto\:\bf \: {OA}^{2} + {OB}^{2} = {AB}^{2} }$

$\rm :\longmapsto\: {\bigg(\dfrac{AC}{2}\bigg)}^{2} + {\bigg(\dfrac{BD}{2}\bigg)}^{2} = {x}^{2}$

$\rm :\longmapsto\:\dfrac{ {AC}^{2} }{4} + \dfrac{ {BD}^{2} }{4} = {x}^{2}$

$\rm :\longmapsto\: {AC}^{2} + {BD}^{2} = {4x}^{2}$

$\rm :\longmapsto\: {AC}^{2} + {BD}^{2} = {x}^{2} + {x}^{2} + {x}^{2} + {x}^{2}$

$\red{\rm :\longmapsto\: {AC}^{2}+{BD}^{2}={AB}^{2} +{BC}^{2} +{CD}^{2}+{DA}^{2}}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. Pythagoras Theorem :-

• This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

• This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

• This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

• If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.