In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB²

Question

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB²

KnowledgeBoat · Accepted Answer

(i) In right angle triangle ACQ, By pythagoras theorem we get, ⇒ AQ² = AC² + CQ² Multiplying both sides by 9 we get, ⇒ 9AQ² = 9AC² + 9CQ² ⇒ 9AQ² = 9AC² + (3CQ)² .......(i) Given, BQ : CQ = 1 : 2 Substituting above value in (i) we get, ⇒ 9AQ² = 9AC² + (2BC)² ⇒ 9AQ² = 9AC² + 4BC². Hence, proved that 9AQ² = 9AC² + 4BC². (ii) In right angle triangle BPC, By pythagoras theorem we get, ⇒ BP² = BC² + CP² Multiplying both sides by 9 we get, ⇒ 9BP² = 9BC² + 9CP² ⇒ 9BP² = 9BC² + (3CP)² .......(ii) Given, AP : PC = 1 : 2 Substituting above value in (ii) we get, ⇒ 9BP² = 9BC² + (2AC)² ⇒ 9BP² = 9BC² + 4AC². Hence, proved that 9BP² = 9BC² + 4AC². (iii) In right angle triangle ABC, By pythagoras theorem we get, ⇒ AB² = AC² + BC² ......(iii) Adding resultants from part (i) and (ii) we get, 9AQ² + 9BP² = 9AC² + 4BC² + 9BC² + 4AC² 9(AQ² + BP²) = 13AC² + 13BC² 9(AQ² + BP²) = 13(AC² + BC²) 9(AQ² + BP²) = 13AB² [From (iii)]. Hence, proved that 9(AQ² + BP²) = 13AB².

Mathematics

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB²

Pythagoras Theorem

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Mathematics

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that(i) 9AQ2 = 9AC2 + 4BC2(ii) 9BP2 = 9BC2 + 4AC2(iii) 9(AQ2 + BP2) = 13AB2

Pythagoras Theorem

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In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB²

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8PT² = 3PR² + 5PS².

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