Question

the lengths of the two sides of a right triangle containing the right angle differ by 2 cm. if the area of the triangle is 24cm^2, find the perimeter of the triangle.​

Answer 1

Answer:

Area of triangle = 24 cm²

Let the height of the right angle triangle be x cm and base be y cm.

It is given that, the lengths of the two sides of a right triangle containing the right angle differ by 2 cm :]

Therefore,

➳ x - y = 2 cm

➳ x = 2 + y .........[Equation (i)]

As it is said in question that we need to find the area of triangle. So, to Calculate area of triangle we have to use below given formula :]

➳ Area of Triangle = ½ × base × height

Now,using this formula we will calculate the area of triangle :]

➳ 24 = ½ * y * (2 + y)

➳ 48 = 2y + y²

➳ y² + 2y - 48 = 0

➳ y² + 8y - 6y - 48 = 0

➳ y(y + 8) - 6(y + 8) = 0

➳ (y + 8) (y - 6) = 0

➳ y = -8 or 6

Side of the triangle cannot be negative. Therefore, y = 6 cm

Now, Putting the value y = 6 in equation (i) we get,

➳ x = 2 + y

➳ x = 2 + 6

➳ x = 8 cm

Therefore,

Base of triangle = y = 6 cm

Height of the triangle = x = 8 cm

Now, we will find the third side of a right angle triangle by using the Pythagoras theorem.

➳ (Hypotenuse)² = (Oneside)² + (Other side)²

➳ Hypotenuse² = 8² + 6²

➳ Hypotenuse² = 64 + 36

➳ Hypotenuse² = 100

➳ Hypotenuse = 10 cm

Hence, the third side is 10 cm.

Now, we can calculate the perimeter of triangle :

➳ Perimeter of triangle = 8 + 6 + 10

➳ Perimeter of triangle = 24 cm

Therefore, the perimeter of triangle is 24 cm

Answer 2

Given :-

A right triangle whose sides differ = 2 cm

Area of the triangle = 24 cm²

To Find :-

The sides of the triangle.

The perimeter of the triangle.

Solution :-

We know that,

• b = Base
• h = Height
• p = Perimeter

Given that,

Right triangle whose sides differ = 2 cm

Area of the triangle = 24 cm²

According to the question,

Let x and (x - 2) be one of the sides and the other side respectively.

From the formula,

$\underline{\boxed{\sf Area \ of \ triangle= \dfrac{1}{2} \: (Base) \times (Height)}}$

Substituting their values, we get

$\sf \longrightarrow 24=\dfrac{1}{2} \times x \times (x-2)$

$\sf \longrightarrow 48=x^{2}-2x$

$\sf Or \ x^{2}-2x-48=0$

By solving them, we get

$\sf (x+6)(x-8)=0$

$\sf x=-6 \ or \ x=8$

Since length measure cannot be negative,

So neglect x = −6

One side = 8 cm

∴ Another Side = $\sf x-2=8-2=6 \ cm$

By applying Pythagoras theorem,

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

Substituting their values, we get

$\sf Hypotenuse^{2}=\sqrt{(8)^{2}+(6)^{2}}$

$\sf Hypotenuse^{2}=\sqrt{100} =10$

Perimeter of triangle = Sum of all the sides

(6 + 8 + 10) cm = 24 cm

Therefore, the perimeter of the triangle is 24 cm