Question

The sides containing the right triangle differ in length by 7cm . if the hypotenuse js 13 cm long, determine the lengths of the two sides of the triangle​

Answer 1

Answer:

Let x cm be the one of the sides, then (x−2)cm be another side.

Area of triangle =24cm  

2

 (given)

We know, Area of triangle =  

2

1

​  

(Base×height)

⇒24=  

2

1

​  

×x×(x−2)

⇒48=x  

2

−2x

or x  

2

−2x−48=0

Solving above equation, we have

(x+6)(x−8)=0

x=−6 or x=8

Since length measure cannot be negative, so neglect x=−6

One side =8cm

Another Side =x−2=8−2=6cm

Apply Pythagoras theorem:

Hypotenuse  

2

= Base  

2

+ Perpendicular  

2

 

Hypotenuse  

2

=  

(8  

2

+6  

2

)

​  

 

Hypotenuse =  

100

​  

=10

Therefore, perimeter of triangle = Sum of all the sides =(6+8+10)cm=24cm

Step-by-step explanation:

Answer 2

Solution :

• Let the shorter sides of the given triangle be x cm then the length of the longer side will be (x + 7) cm.
• Hypotenuse of right angle ∆ = 13 cm

◖According to the Question now :

By using Phythagoras theorem we get :

⇏ (Hypotenuse)² = (Perpendicular)² + (base)²

⇏ (13)² = (x + 7)² + (x)²

⇏ 169 = x² + 2(7)(x) + (7)² + x²

⇏ 169 = x² + 14x + 49 + x²

⇏ 169 = 2x² + 14x + 49

⇏ 2x² + 14x + 49 - 169 = 0

⇏ 2x² + 14x - 120 = 0

Dividing all the eqⁿ by 2 we get :

⇏ x² + 7x - 60 = 0

Now, by using splitting middle term we get :

⇏x² + 12x - 5x -60= 0

⇏x(x + 12) - 5(x + 12) = 0

⇏(x + 12) (x - 5) = 0

⇏x = -12 or x = 5

Hence, ignoring negative value we get x as 5.

◖ Length of the remaining two sides of ∆ are :

⇏ Length of shorter side of ∆ = x

⇏ Length of shorter side of ∆ = 5 cm

⇏ Length of longer side of ∆ = x + 7

⇏ Length of longer side of ∆ = 5 + 7

⇏ Length of longer side of ∆ = 12 cm