Question

the sides of a right angled triangle is x plus 1 and x plus 3 and length of its hypotenuse is 10cm find the area of the triangle​

Answer 1

Answer:

The area of the right triangle is 24 cm².

Step-by-step-explanation:

Let the base of the right triangle be x + 1.

And the height of the right triangle be x + 3.

We have given that,

The length of the hypotenuse is 10 cm.

Now, by Pythagoras theorem,

( Hypotenuse )² = ( Base )² + ( Height )²

⇒ ( 10 )² = ( x + 1 )² + ( x + 3 )²

By using ( a + b )² = a² + 2ab + b², we get,

⇒ 100 = x² + 2 * x * 1 + 1² + x² + 2 * x * 3 + 3²

⇒ 100 = x² + 2x + 1 + x² + 6x + 9

⇒ x² + x² + 2x + 6x + 1 + 9 = 100

⇒ 2x² + 8x + 10 = 100

⇒ 2x² + 8x + 10 - 100 = 0

⇒ 2x² + 8x - 90 = 0

By dividing each term by 2, we get,

⇒ x² + 4x - 45 = 0

⇒ x² + 9x - 5x - 45 = 0

⇒ x ( x + 9 ) - 5 ( x + 9 ) = 0

⇒ ( x + 9 ) ( x - 5 ) = 0

⇒ ( x + 9 ) = 0 OR ( x - 5 ) = 0

⇒ x + 9 = 0 OR x - 5 = 0

⇒ x = - 9 OR x = 5

By using x = - 9, we get,

Base = x + 1

⇒ Base = - 9 + 1

⇒ Base = - 8

As length of base of right triangle can't be negative, x = - 9 is unacceptable.

∴ x = 5

By using x = 5, we get,

Base = x + 1

⇒ Base = 5 + 1

⇒ Base = 6 cm

Now,

Height = x + 3

⇒ Height = 5 + 3

⇒ Height = 8 cm

Now, we have to find the area of triangle.

We know that,

Area of triangle = ( Base * Height ) / 2

⇒ Area of triangle = ( 6 * 8 ) / 2

⇒ Area of triangle = 6 * 8 ÷ 2

⇒ Area of triangle = 6 * 4

⇒ Area of triangle = 24 cm²

∴ The area of the right triangle is 24 cm².

Answer 2

$\Large \underline{ \underline{ \text{Question:}}}$

• The sides of a right angled triangle is 'x + 1' and 'x + 3' and length of its hypotenuse is 10cm. Find the Area of the Triangle.

$\Large \underline{ \underline{ \text{Solution:}}}$

Given that,

• $\text{The Base of Right-Angled} \: \triangle \: (B) = x + 1,$

• $\text{The Perpendicular of Right-Angled} \: \triangle \: (P) = x + 3,$

And,

• $\text{The Hypotenuse of Right-Angled} \: \triangle \: (H) = 10cm$

According to Pythagoras theorem, The sum of Squares of Perpendicular and Base is equals to The Square of the Hypotenuse. As,

• $\boxed{ {P}^{2} + {B}^{2} = { H}^{2} }$

Substituting the given values and Finding the value of 'x',

$:\longrightarrow {(x + 3)}^{2} + {(x + 1)}^{2} = { (10)}^{2} \\ \\ [ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} ] \\ \\ :\longrightarrow {x}^{2} + 2(3)x + {3}^{2} + {x}^{2} + 2(x) + {1}^{2} = 100 \\ \\ :\longrightarrow {x}^{2} + 6x + 9 + {x}^{2} + 2x + 1 = 100 \\ \\ :\longrightarrow 2 {x}^{2} + 8x + 10 = 100 \\ \\ :\longrightarrow 2( {x}^{2} + 4x + 5) = 2(50) \\ \\ :\longrightarrow {x}^{2} + 4x + 5 = 50 \\ \\ :\longrightarrow {x}^{2} + 4x - 45 = 0$

We have,

• ${x}^{2} + 4x - 45 = 0$

By Middle - term Spitting method,

Finding the value of 'x',

$:\longrightarrow {x}^{2} + 4x - 45 = 0 \\ \\ :\longrightarrow {x}^{2} + 9x - 5x - 45 = 0 \\ \\ :\longrightarrow x(x + 9) - 5(x + 9) = 0 \\ \\ :\longrightarrow (x - 5)(x + 9) = 0 \\ \\ :\longrightarrow x = 5 \: \text{and} \: - 9$

Here,

• 'x' can be 5 and -9.

Substituting the value of 'x' by - 9,

$:\longrightarrow B = - 9 + 1, \\ \\ :\longrightarrow B = - 8cm$

As Length can't be negative.

So, - 9 can't be the value of 'x'.

Substituting the value of 'x' by 5,

• $:\longrightarrow B = x + 1 \\ \\ :\longrightarrow B = 5 + 1 \\ \\ :\longrightarrow B = 6cm$

And,

• $:\longrightarrow P= x + 3\\ \\ :\longrightarrow P= 5 + 3 \\ \\ :\longrightarrow P = 8cm$

As we know,

• $\boxed{{\text{Area}}_{(\text{Right-Angled}\triangle)} = \frac{1}{2} \times B \times P }$

Substituting the values and Finding the Area of ∆,

$:\longrightarrow {\text{Area}}_{(\text{Right-Angled}\triangle)} = \frac{1}{2} \times (6cm) \times (8cm) \\ \\ :\longrightarrow {\text{Area}}_{(\text{Right-Angled}\triangle)} = 3cm \times 8cm \\ \\ :\longrightarrow \boxed{ {\text{Area}}_{(\text{Right-Angled}\triangle)} = 24 {cm}^{2} }$

Therefore,

• The Area of Triangle is 24cm².

$\\ \Large \underline{ \underline{ \text{Required Answer:}}}$

• The Area of Triangle is 24cm².