Question

The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm. If the area
of the triangle be 60 sq cm, calculate the length of the sides of the triangle.
1
[Hint: Area of a triangle = base x height.]
2​

Answer 1

Answer:

given:-sides are 5 x cm & (3x -1)cm.

area= 60 cm^2

we have to find the length of sides of triangle

Step-by-step explanation:

• area of triangle =1/2 × base × height
• 60cm^2 =1/2 × (3x-1)cm × 5xcm
• 60 =1/2 × 15x^2 -5x
• 60×2=15x^2-5x
• 15x^2 - 5x -120=0
• dividing whole equation by 5
• 3x^2 - 1x - 24=0
• 3x^2 - 9x + 8x -24 =0
• 3x(x-3) + 8(x-3)=0
• taking x-3 common then,
• (x-3)(3x+8)=0
• x-3=0
• x=3
• Hence,sides of ∆are 15cm and 8cm.

Answer 2

Given :-

The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm.

Area of the triangle = 60 cm²

To Find :-

The length of the sides of the triangle.

Solution :-

By the formula,

$\underline{\boxed{\sf Area \ of \ rectangle=\dfrac{1}{2} \times Base \times Height}}$

Given that,

Sides of a right-angled triangle = 5x and (3x - 1) cm

Area (a) = 60 cm²

Substituting their values,

$\sf 60=\dfrac{1}{2} \times 5x \times (3x-1)$

By transposing,

$\sf 5x(3x - 1) = 60 \times 2$

$\sf 5x(3x - 1) = 120$

$\sf x(3x - 1) = \dfrac{120}{5}$

$\sf 3x^2 - x = 24$

$\sf 3x^2 - x - 24 = 0$

$\sf 3x^2 + 8x - 9x - 24 = 0$

$\sf x(3x + 8) - 3(3x + 8)$

$\sf (x - 3)(3x + 8)$

$\sf x = 3 \ or \ \dfrac{-8}{3}$

Hence,

The value cannot be negative, then the value of 'x' would be 3.

Therefore, the sides would be

5x = 5 × 3

= 15 cm

(3x - 1) = [(3 × 3) - 1]

= 9 - 1 = 8 cm

Therefore, the sides of a right-angled triangle are 15 cm and 8 cm.

By Pythagoras theorem,

$\underline{\boxed{\sf Hypotenuse=a^2+b^2}}$

Substituting their values,

$\sf c^2 = 15^2 + 8^2$

$\sf c^2=225 + 64$

$\sf c^2= 289$

Finding h,

$\sf c=\sqrt{289}$

$\sf c=17 \ cm$

Therefore, the sides of the triangle are 8cm, 15cm, and 17cm respectively.