## QN07: The altitude of a right Ais 7cm less than its base. If the hypotenuse is 13cm. find the other two sides (7cm)

Question

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## Answers ( )

## Solution

Let the base of right triangle be x the altitude be (x – 7)

Hypotenuse = 13 cm

(hypotenuse)²=(base)²+(perpendicular)²

(13)² = (x)² + (x – 7)²

169 = x² + x² + (7)² – 2 × x × 7

169 = 2x² + 49 – 14x

169 – 49 = 2x² – 14x

120 = 2x² – 14x

2x² – 14x – 120 = 0

2(x² – 7x – 60) = 0

x² – 7x – 60 = 0

x² + 7x – 12x – 60x = 0

x(x + 7) – 12(x + 7) = 0

(x + 7)(x – 12) = 0

______________

(x + 7) = 0

x = – 7

____________

(x – 12) = 0

x = 12

______________

Base = x = 12 m

Altitude = (x – 7) = (12 – 7) = 5m

GivenThe altitude of a right is 7cm less than its base. If the hypotenuse is 13cm.

Find outFind the other two sides

Solution★Letthebaseofrighttrianglebexthealtitudebe(x–7)AccordingtothePythagorastheorem(hypotenuse)²=(base)²+(perpendicular)²

➟ (13)² = (x)² + (x – 7)²

Applyidentity:(a–b)²=a²+b²–2ab➟ 169 = x² + x² + (7)² – 2 × x × 7

➟ 169 = 2x² + 49 – 14x

➟ 169 – 49 = 2x² – 14x

➟ 120 = 2x² – 14x

➟ 2x² – 14x – 120 = 0

➟ 2(x² – 7x – 60) = 0

➟ x² – 7x – 60 = 0

Splittingmiddleterm➟x² + 7x – 12x – 60x = 0

➟ x(x + 7) – 12(x + 7) = 0

➟ (x + 7)(x – 12) = 0

Either➟ (x + 7) = 0

➟ x = – 7

Or➟ (x – 12) = 0

➟ x = 12

★

Length can’t be in negative★Hence,Base= x = 12 mAltitude= (x – 7) = (12 – 7) = 5m