If P and Q are points on sides CA and CB respectively of - ABC right angled at C- then AQ 2- BP 2- -A- AB 2- PQ 2B- AC 2- P C 2C- AB 2- PQ 2D- CQ 2- CB 2

The correct option is C AB^2 + nbsp; PQ^2 nbsp; In right angled triangle nbsp;ACQ, AQ^2 = nbsp; AC^2 + nbsp; CQ^2 ....(1) In right angled triang ...

145

You visited us 145 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Pythagoras Theorem

If P and Q ar...

Question

If P and Q are points on sides CA and CB respectively of ΔABC right angled at C, then ${A Q}^{2}$ + ${B P}^{2} = ?$

{C Q}^{2}

{C B}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{{A B}^{2}}{{P Q}^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{A C}^{2}

{P C}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{A B}^{2}

{P Q}^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D ${A B}^{2}$ + ${P Q}^{2}$

In right- angled triangle ACQ,

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

In right- angled triangle PCB,

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

In right- angled triangle ABC,

${A B}^{2} = {A C}^{2} + {B C}^{2} . . . . . . . (3)$

In right- angled triangle PQC,

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

Adding (1) and (2)

$\Rightarrow {A Q}^{2}$ + ${B P}^{2}$ = ( ${A C}^{2}$ + ${C Q}^{2}$ ) + ( ${P C}^{2}$ + ${C B}^{2}$ )

$\Rightarrow {A Q}^{2}$ + ${B P}^{2}$ = ( ${A C}^{2}$ + ${B C}^{2}$ ) + ( ${P C}^{2}$ + ${Q C}^{2}$ )

$\Rightarrow {A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$

Suggest Corrections

Similar questions

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$

Q. If

P

and

Q

are point on sides

C A

and

C B

respectively of

Δ A B C

, right angled at

C

, prove that

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

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$ [4 MARKS]

Q. In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

(a) AQ² + CP² = 2(AC²+ PQ²)
(b) 2(AQ² + CP²) = AC² + PQ²
(c) AQ²+ CP² = AC² + PQ²
(d)

AQ + CP = \frac{1}{2} (AC + PQ)

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE²+ BD² = AB² + DE²

Join BYJU'S Learning Program

If P and Q are points on sides CA and CB respectively of ΔABC right angled at C, then AQ2 + BP2= ?

If P and Q are points on sides CA and CB respectively of ΔABC right angled at C, then ${A Q}^{2}$ + ${B P}^{2} = ?$