In NQR , NR ^ 2 = QN ^ 2 + QR ^ 2 —(1)But QN = 1 2 QP So, QN ^ 2 = 1 4 QP ^ 2 –(2)Substituting (2) in (1), #160;we get NR ^ 2 = ...

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Standard VII

Mathematics

Pythagoras Theorem

4 RN2= PQ2 + ...

Question

$4 R N^{2} = P Q^{2} + 4 Q R^{2}$

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Solution

In $△ N Q R$ ,
${N R}^{2} = {Q N}^{2} + {Q R}^{2}$ ---(1)

But $Q N = \frac{1}{2} Q P$

So, ${Q N}^{2} = \frac{1}{4} {Q P}^{2}$ --(2)

Substituting (2) in (1), we get

${N R}^{2} = \frac{1}{4} {Q P}^{2} + {Q R}^{2}$

$4 {N R}^{2} = {Q P}^{2} + {4 Q R}^{2}$

Hence proved.

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Similar questions

4 R N^{2} = P Q^{2} + 4 P R^{2}

State whether the above statement is true or false.

Q. Evaluate the following:
(1) 3p

q^{3} \times

4qr
(2) -2

a^{3} b \times 3 a b^{2} c

Angles Q and R of a ΔPQR are 25° and 65°.

Write which of the following is true:

(i) PQ² + QR²= RP²

(ii) PQ² + RP²= QR²

(iii) RP² + QR²= PQ²

Question 6
Angles Q and R of a $Δ$ PQR are $25^{\circ}$ and $65^{\circ}$ .
Write which of the following is true:
(i) $P Q^{2} + Q R^{2} = R P^{2}$
(ii) $P Q^{2} + R P^{2} = Q R^{2}$
(iii) $R P^{2} + Q R^{2} = P Q^{2}$

Q. Simplify:

p^{2} q - p q^{2} - p q + r^{2}

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4RN2=PQ2+4QR2

In △NQR,NR2=QN2+QR2 ---(1)But QN=12QPSo, QN2=14QP2 --(2)Substituting (2) in (1), we getNR2=14QP2+QR24NR2=QP2+4QR2Hence proved.

$4 R N^{2} = P Q^{2} + 4 Q R^{2}$

In $△ N Q R$ ,
${N R}^{2} = {Q N}^{2} + {Q R}^{2}$ ---(1)

But $Q N = \frac{1}{2} Q P$

So, ${Q N}^{2} = \frac{1}{4} {Q P}^{2}$ --(2)

Substituting (2) in (1), we get

${N R}^{2} = \frac{1}{4} {Q P}^{2} + {Q R}^{2}$

$4 {N R}^{2} = {Q P}^{2} + {4 Q R}^{2}$

Hence proved.