Question

In a right triangle ABC is right-angled at B If P and Q are points on the sides AB and BC respectively then

• A. $A{Q}^2+C{P}^2=2(A{C}^2+P{Q}^2)$
• B. $2(A{Q}^2+C{P}^2)=A{C}^2+P{Q}^2$
• C. $A{Q}^2+C{P}^2=A{C}^2+P{Q}^2$
• D. $AQ+CP=\frac{1}{2}(AC+PQ)$

Answer 1

The correct option is C $A{Q}^2+C{P}^2=A{C}^2+P{Q}^2$

From figure,

In $△ABC$,

From pythagorous theorem,

$A{B}^2+B{C}^2=A{C}^2$ ..........(1)

Similarly,

In $△ABQ$,

$A{B}^2+B{Q}^2=A{Q}^2$ ..............(2)

And,

In $△PBQ$,

$P{B}^2+B{Q}^2=P{Q}^2$ ..............(3)

and,

In $△PBC$,

$P{B}^2+B{C}^2=C{P}^2$ ................(4)

Now from Eq. (1) ,(2) , (3) & (4),

We get,

$A{Q}^2+C{P}^2=A{C}^2+P{Q}^2$