Question

M is midpoint of QR. Angle PRQ=90° Prove that PQ square = 4PM square - 3PR square

Answer 1

Heya !!

Here's your answer.. ⬇⬇

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⚫ Given :- In ∆PRQ, angle R = 90°
M is midpoint of RQ.
RM = MQ = 1/2 RQ

⚫ To Prove :- PQ^2 = 4PM^2 - 3PR^2

⚫ Proof :- In ∆PRM, angle PRM = 90°
By pythagoras theorem...,

PM^2 = PR^2 + RM^2
PM^2 = PR^2 + ( 1/2 RQ )^2 --( M is midpoint )
PM^2 = PR^2 + RQ^2/4
PM^2 = ( 4PR^2 + RQ^2 )/4
4PM^2 = 4PR^2 + RQ^2
RQ^2 = 4PM^2 - 4PR^2 --- ( 1 )

In ∆PRQ, angle PRQ = 90°
By pythagoras theorem..,

PQ^2 = PR^2 + RQ^2
PQ^2 = PR^2 + ( 4PM^2 - 4PR^2 ) --( From eq.1 )
PQ^2 = PR^2 + 4PM^2 - 4PR^2
PQ^2 = 4PM^2 - 3PR^2

Hence proved PQ^2 = 4PM^2 - 3PR^2
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Hope it helps..
Thanks (:

Answer 2

Hi friend, Harish here.

Here is  your answer:

Given that,

1) ΔPRQ , right angled at R.

2) M is the mid point of the side QR. ( RM = QM )

To Prove,

(PQ)² = 4(PM)² - 3(PR)²

Solution:

We know that, In ΔPQR:

(PQ)² = (PR)² + (RQ)²   (Pythagorean Theorem).

Here RQ = 2 RM. Then,

⇒ $(PQ)\² = (PR)\² + (2RM)\²$

⇒ $(PQ)\² = (PR)\² + 4(RM)\²$   

⇒ $4(RM)\² = (PQ)\² - (PR)\²$

⇒ $(RM)^{2} = \frac{(PQ)^{2} - (PR)^{2}}{4}$   - (i)

Now, In ΔPRM,

$(PM)\² = (PR)\² + (RM)\²$  - (ii)

Now substitute (i) in (ii)

Then, 

⇒ $(PM)\² = (PR)\² + ( \frac{(PQ)^{2} - (PR)^{2}}{4})$

⇒ $(PM)\² = ( \frac{4(PR)\² +(PQ)^{2} - (PR)^{2}}{4})$

⇒ $4(PM)\² = [(4(PR)\² - (PR)\²) + (PQ)^{2}]$

⇒ $4(PM)\² = (3(PR)\²+ (PQ)^{2})$

⇒ $(PQ)^{2} = 4(PM)^{2} - 3(PR)^{2}$

Hence proved.
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Hope my answer is helpful to you.