Question

The hypotense of a right angled triangle is 5 cm and the other two sides differ by 1 cm .find the length of the sides of the triangle

Answer 1

★Given :-

• A right triangle whose hypotenuse is 5 cm
• Other two sides are differ by 1 cm

★To find :-

The length of two sides of triangle

★ Solution:-

Let the one side be x

then the other side be x+1

• According to Pythagoras formula

$\boxed{\sf Hypotenuse^2=Base^2+ perpendicular^2}$

Put the values in the formula !

→ (5)²=(x)²+(x+1)²

→ 25= x²+x²+2x+1

→ 25-1=2x²+2x

→ 0=2x²+2x-24

or

→ 2(x²+x-12)=0

→ x²+(4-3)x-12=0/2

→ x²+4x-3x-12=0

→ x(x+4)-3(x+4)=0

→ (x+4)(x-3) =0

→ x= (-4) or 3

Sides can't be negative ,So we take the positive value .

So value of x = 3 and (x+1)= 4

The length of other two sides of triangle is 3cm and 4cm.

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

The hypotenuse of a right angled triangle = 5cm

The other sides of the triangle differ by 1

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

The lengths of the sides of the triangle.

${\green{\underline{\underline{\bold{Solution:-}}}}}$

Given,

Sum of other two sides differ by 1

Let one of the sides be x

The other side = x + 1

By using Pythagores theorem:-

(Base)² +(Height)² = (Hypotenuse)²

$\implies \bf {x}^{2} + {(x + 1)}^{2} = {5}^{2}$

$\implies \bf {x}^{2} + {x}^{2} + 1 + 2x = 25$

$\implies \bf2 {x}^{2} + 2x + 1 = 25$

$\implies \bf 2{x}^{2} + 2x + 1 - 25 = 0$

$\implies \bf2 {x}^{2} + 2x - 24 = 0$

Divide whole equation by 2

$\implies \bf {x}^{2} + x - 12 = 0$

$\bf By factorisation:-$

$\implies \bf {x}^{2} + 4x - 3x - 12 = 0$

$\implies \bf x(x + 4) - 3(x + 4) = 0$

$\implies \bf(x + 4)(x - 3) = 0$

$\implies \bf x = - 4 \: ,\: \: 3$

Since the sides of a triangle cannot be negative

The sides of the triangle are 4cm and 3cm