Question

The hypotenuse of a right angled triangle is 13 m long. If the base of the triangle is 7 m more than the other side, find the sides of the triangle.​

Answer 1

Given :-

• Length of hypotenuse = 13 m.
• Base of the triangle = 7 m more than the other side (altitude).

To Find :-

• Sides of the triangle.

Solution :-

Let:-

• One side (altitude) be (x) m

Then,

• Other side (base) = (7 + x) m

By Pythagoras theorem,

$\implies\rm{13^2=x^2+(7+x)^2}$

$\implies\rm{169=x^2+x^2+49+14x}$

$\implies\rm{2x^2+14x+49-169=0}$

$\implies\rm{2x^2+14x-120=0}$

$\implies\rm{x^2+7x-60=0}$

Splitting the middle term, we get,

$\implies\rm{x^2+12x-5x-60=0}$

$\implies\rm{x(x+12)-5(x+12)=0}$

$\implies\rm{(x+12)(x-5)=0}$

Therefore,

$\implies\rm{x=-12\:or\:x=5}$

Neglecting negative value, we get,

$\implies\rm{x=}\:\bold{5}$

Hence,

• One side (x) = 5 m.
• Other side (x + 7) = 12 m.

Answer 2

Given ,

• Hypotenuse are right angled triangle (h) = 13 m

• The base of triangle is 7 m more than the other side

Let ,

The perpendicular (other side) be " x "

We know that , in right angled triangle

$\boxed{ \sf{(hypotenuse)}^{2} = { (base)}^{2} + {(perpendicular)}^{2} }$

Thus ,

(13)² = (x + 7)² + (x)²

169 = (x)² + (7)² + 2(x)(7) + (x)²

120 = 2(x)² + 14x

2(x)² + 14x - 120 = 0

(x)² + 7x - 60 = 0

(x)² + 12x - 5x - 60 = 0

x(x + 12) - 5(x + 12) = 0

(x - 5)(x + 12) = 0

x = 5 or x = -12

Since , the length can't be negative

Thus ,

• Perpendicular = 5 m
• Base = 12 m