Prove that sum of square of diagonal of parallelogram is equal to sum of squares...

Sum of square of diagonal of parallelogram is equal to sum of squares of its sides is proved

Solution:

The diagram for this sum is attached below

In parallelogram ABCD, AB = CD, BC = AD

Draw perpendiculars from C and D on AB as shown.

In right angled Δ AEC,

$\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AE}^{2}+\mathrm{CE}^{2}[\mathrm{By} \text { Pythagoras theorem }]} \\\\ {\Rightarrow \mathrm{AC}^{2}=(\mathrm{AB}+\mathrm{BE})^{2}+\mathrm{CE}^{2}} \\\\ {\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2}+2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2} \rightarrow(1)}\end{array}$

From the figure CD = EF (Since CDFE is a rectangle)

But CD= AB

⇒ AB = CD = EF

Also CE = DF (Distance between two parallel lines)

ΔAFD ≅ ΔBEC (RHS congruence rule)

⇒ AF = BE

Consider right angled ΔDFB

$\begin{array}{l}{\mathrm{BD}^{2}=\mathrm{BF}^{2}+\mathrm{DF}^{2}[\mathrm{By} \text { Pythagoras theorem }]} \\\\ {=(\mathrm{EF}-\mathrm{BE})^{2}+\mathrm{CE}^{2}[\text { since DF }=\mathrm{CE}]} \\\\ {=(\mathrm{AB}-\mathrm{BE})^{2}+\mathrm{CE}^{2}[\text { Since } \mathrm{EF}=\mathrm{AB}]} \\\\ {\Rightarrow \mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2}-2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2} \rightarrow(2)}\end{array}$

Add (1) and (2),

$\mathrm{AC}^{2}+\mathrm{BD}^{2}=\left(\mathrm{AB}^{2}+\mathrm{BE}^{2}+2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2}\right)+\left(\mathrm{AB}^{2}+\mathrm{BE}^{2}-2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2}\right)$

$\begin{array}{l}{=2 \mathrm{AB}^{2}+2 \mathrm{BE}^{2}+2 \mathrm{CE}^{2}} \\\\ {\mathrm{AC}^{2}+\mathrm{BD}^{2}=2 \mathrm{AB}^{2}+2\left(\mathrm{BE}^{2}+\mathrm{CE}^{2}\right) \rightarrow(3)}\end{array}$

$\begin{array}{l}{\text { From right angled } \Delta \mathrm{BEC}, \mathrm{BC}^{2}=\mathrm{BE}^{2}+\mathrm{CE}^{2}[\mathrm{By}\text { Pythagoras theorem] } } \\\\ {\text { Hence equation }(3) \text { becomes, } \mathrm{AC}^{2}+\mathrm{BD}^{2}=2 \mathrm{AB}^{2}+2 \mathrm{BC}^{2}} \\\\ {=\mathrm{AB}^{2}+\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{BC}^{2}} \\\\{=\mathrm{AB}^{2}+\mathrm{CD}^{2}+\mathrm{BC}^{2}+\mathrm{AD}^{2}}\end{array}$

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

Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Learn more about parallelogram

Prove that sum of squares of sides of a parallelogram is equal to sum of squares of its diagonals.

brainly.in/question/2686535

Questions

Answers & Comments

Sum of square of diagonal of parallelogram is equal to sum of squares of its sides is proved

Add an Answer

what is pinacole pinacolone rearrangement

Mods bohot fast ho gaye h .... koi nahi apan answer bhi denge or spam bhi kareng...

What are function of placenta ?...

What is midpoint formula (come on Be fst and give answr)v...

(Who rc17) explain function of galvanometer...

What is scale factor...

What are mammary glands...

What is compensation point? Explain in brief...

Hey everyone What is Rowlatt Act?...

Who was witer of hind swaraj...

Recommend Questions

Helpful Links

Helpful Social