Question

5.BR & AP are medians of triangle POR right angled at Q Prove that 4BR + 4AP = 5PR​

Answer 1

Question:

BR and AP are medians of triangle PQR right angled at Q. Prove that 4 BR² + 4 AP² = 5 PR²

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\rm{Given:}}}}$

• ΔPQR with ∠Q = 90°, BR and AP are medians.

$\Large{\underline{\underline{\rm{To\:Prove:}}}}$

• 4 BR² + 4 AP² = 5 PR²

$\Large{\underline{\underline{\rm{Proof:}}}}$

→ Since BR and AP are medians,

  AQ = 1/2 × RQ-----(1)

  QB = 1/2 × PQ----(2)

→ Consider Δ APQ

→ By Pythagoras theorem,

  AP² = PQ² + AQ²-----(3)

→ Substitute 1 and 2 in 3

  AP² = PQ² + (1/2 × RQ)²

  AP² = (2 QB)² + 1/4 × RQ²

  AP² = 4 QB² + 1/4 × RQ²-------(4)

→ Consider ΔQBR

→ By Pythagoras theorem,

  RB² = QB² + RQ²-----(5)

→ Add equation 4 and 5

  AP² + BR² =  4 QB² + 1/4 × RQ² + QB² + RQ²

  AP² + BR² = 5 QB² + (RQ² + 4 RQ²)/4

  AP² + BR² = 5 QB² + (5 RQ²)/4

  AP² + BR² = (20 QB² + 5 RQ²)/4

→ Multiply the whole equation by 4

  4 ( AP² + BR²) = 20 QB² + 5 RQ²-------(6)

→ Now consider Δ QPR

→ By pythagoras theorem,

  PR² = PQ² + QR²

  QR² = PR² - PQ²

  QR² = PR² - ( 2 QB)² -------∵ PQ = 2 QB

  QR² = PR² - 4 QB²--------(7)

→ Substitute equation 7 in 6

   4 ( AP² + BR²) = 20 QB² + 5 (PR² - 4 QB²)

   4 BR² + 4 AP² = 20 QB² + 5 PR² - 20 QB²

→ Cancelling 20 QB²

   4 BR² + 4 AP² = 5 PR²

→ Hence proved.

 

Answer 2

Question :-

BR and AP are the medians of traingle PQR right angled at Q. Prove that, 4BR² + 4AP² = 5PR²

Answer :-

Given,

PQR is a right angled ∆ ,right angled at Q.

So, < Q = 90°

PQ is the perpendicular whereas QR is the hypotenuse. BR and AP are medians.

★ Concept :-

Here the concept of Pythagoras Theorem is used where,

Hypotenuse² = Base² + Perpendicular²

Also, here the concept of median is used where, median bisects the opposite side of the triangle.

» To prove :- 4BR² + 4AP² = 5PR²

★Solution :-

Since, its given that ,

BR and AP are medians.

So, from figure,

• AQ = ½ (RQ)

• QB = ½ (PQ)

Now by applying, Pythagoras Theorem right angled ∆ APQ,

✒ AP² = PQ² + AQ²

✒ AP² = (2QB)² + (½ RQ)²

(since AQ = ½(RQ), and QB = ½PQ)

✒ AP² = 4QB² + ¼ RQ² .. (i)

Now by applying Pythagoras Theorem in right angled ∆ QBR, we get,

✒ RB² = QB² + RQ² ... (ii)

Adding equations (i) and (ii), we get,

✒ AP² + RB² = 4QB² + ¼ RQ² + QB² + RQ²

✒ AP² + RB² = 5QB² + ¼ { RQ² + 4RQ²}

✒ AP² + RB² = 5QB² + ¼ (5RQ²)

✒ AP² + RB² = ¼(20QB² + 5RQ²)

✒ ¼(4AP² + 4RB²) = ¼(20QB² + 5RQ²)

✒ 4(AP² + RB²) = 20QB² + 5RQ² ...(iii)

By applying Pythagoras Theorem in triangle QPR, we get,

▶ PR² = PQ² + QR²

▶QR² = PR² - PQ²

▶ QR² = PR² - (2QB)²

▶ QR² = PR² - 4QB² ... (iv)

From equations (iii) and (iv) , we get,

▶ 4(AP² + RB²) = 20QB² + 5(PR² - 4QB²)

▶ 4AP² + 4RB² = 20QB² + 5PR² - 20QB²

▶ 4AP² + 4RB² = 5PR²

On rearranging, we get,

▶ 4BR² + 4AP² = 5PR²

Hence proved.

★More to know :-

• Median of the traingle, bisects the opposite side from which vertex its propagating.

• Also median divides the triangle into two halfs.

• Pythagoras Theorem was first used somewhere in early AD period, which from then is continued till today.