Question

The length of the sides forming right angle of a right triangle are 5x cm and (3x-1)cm. If the area

of the triangle is 60 sq.cm. Find its hypotenuse. ​

Answer 1

Answer:

Hypotenuse of the triangle is 17 cm.

Step-by-step explanation:

Given :-

• The length of the sides forming right angle of a right triangle are 5x cm and (3x-1) cm.
• Area of the triangle is 60 cm².

To find :-

• Hypotenuse of the triangle.

Solution :-

Let the hypotenuse of the triangle be y cm.

And,

let the perpendicular of the triangle be 5x cm and the base of the triangle be (3x-1) cm.

Formula used :

${\boxed{\sf{Area\:of\: right\: triangle=\dfrac{1}{2}\times\:base\times\: perpendicular}}}$

According to the question ,

$\to \sf \: \dfrac{1}{2} \times (3x - 1) \times 5x = 60 \\ \\ \to \sf \: (3x - 1)5x = 120 \\ \\ \to \sf \: 15x^{2} - 5x = 120 \\ \\ \to \sf \: 5(3 {x}^{2} - x) = 120 \\ \\ \to \sf \: 3 {x}^{2} - x = 24 \\ \\ \to \sf \: 3 {x}^{2} - x - 24 = 0 \\ \\ \to \sf \: (x - 3)(3x + 8) = 0$

Either,

x - 3 = 0

→ x = 3

Or,

3x+8 = 0

→ x = -8/3 [ Impossible]

Therefore,

• Perpendicular = 5×3 = 15 cm
• Base = (3×3-1) = 8 cm

Now find the hypotenuse by using Pythagoras Theorem.

Hypotenuse² = Perpendicular ² + Base²

→ y ² = 15² + 8²

→ y ² = 225+64

→ y=√289

→ y = 17

Therefore, the hypotenuse of the triangle is 17 cm.

Answer 2

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Given :-

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

Area of the triangle is 60 cm².

Area of triangle = $\dfrac{1}{2}$×base × height

⠀⠀⠀⠀⠀= $\dfrac{1}{2}$× 5x (3x - 1 )

According to the question,

⠀⠀⠀⠀15$\dfrac{x}{2}$ - 5x = 120

⠀⠀⠀or, 3 $\dfrac{x}{2}$ - x - 24 = 0

⠀⠀⠀or, 3 $\dfrac{x}{2}$ - 9x + 8x - 24 = 0

⠀$\bold{<strong><u>or</u></strong><strong><u>,</u></strong> 3x ( x - 3 ) + 8 ( x - 3 ) = 0}$

$\bold{<strong><u>o</u></strong><strong><u>r</u></strong><strong><u>,</u></strong> ( x - 3 ) ( 3x + 8 ) = 0}$

⛬ x = 3, x = $\dfrac{8}{3}$

length can't be negative, so x = 3

$\bold{AB = 5 × 3}$

$\bold{ = 15cm, BC}$

$\bold{= 3x - 1 = 9 - 1}$

$\bold{= 8cm}$

⛬ AC $\implies \sf{ \sqrt{15² + 8²}}$

$\implies \sf{\sqrt{225+64}}$

$\implies \sf{ \sqrt{289}}$= 17cm

$\large{\boxed{\bold{\purple{Hence\: hypotenuse = 17cm}}}}$

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