Question

The lengths of two sides of a right triangle containing the right angle differ by 2cm . If the area of the triangle is 24 cm square , verify this area by using heron's formula .

Answer 1

$\huge\mathcal\pink{Solution:-}$

✧Let one side be = x i.e base

then other side will be y i.e height

Their difference is 2

❥x - y = 2

❥y = x - 2.....

❥Area = ½ × b × h = 24

❥½ × x × ( x - 2 ) = 24

❥x² - 2x - 48 = 0

❥x² - 8x + 6x - 48 = 0

❥x × ( x - 8 ) + 6 × ( x - 8 ) = 0

❥( x + 6 ) × ( x - 8 ) = 0

❥x = - 6 or 8 cm

✧side cannot be in negative.... so x = 8 cm

y = x - 2 = 6 cm

Hypotenuse = √ ( 8² + 6² )

❥= √ ( 100 )

❥= 10 cm

So perimeter is.....

✧Perimeter = x + y + hypotenuse

❥= 8 + 6 + 10

❥= 24 cm......

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

The length of the two sides of a right triangle containing the right angle differ by 2 cm. if the area of the triangle is 24cm sq, find the perimeter of the triangle.

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

let the 2 sides be x , x -2

${\bold{\blue{\boxed{\bf{ Given }}}}}$

• triangle is a right angle triangle
• area = 24cm^2
• sides = x , x - 2

${\bold{\blue{\boxed{\bf{Find }}}}}$

• 3rd side
• perimeter

${\bold{\blue{\boxed{\bf{solution}}}}}$

${\bold{\green{\boxed{\bf{area \: of \: triangle = \dfrac{1}{2} \times base \times height}}}}}$

$\sf\implies 24 = \dfrac{1}{2} \times x \times x-2$

$\sf\implies 48 = x^2 - 2x$

$\sf\implies x^2 + 2x - 48 = 0$

mid term split

$\sf\implies x^2 + 8x - 6x - 48 = 0$

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

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

x - 6 = 0 | x + 8 = 0

x = 0 + 6| x = 0 - 8

x. = 6 | x = -8

• side cannot be negative

therefore x = 6cm , x+2 = 6 + 2 = 8cm

2 sides of triangle = 6cm , 8cm

• though triangle is a right angle triangle
• applying Pythagoras theoram

${\bold{\blue{\boxed{\bf{{hypo.}^2 = {height}^2 - {base}^2 }}}}}$

$\sf\longrightarrow hypo^2 = 6^2 + 8^2$

$\sf\longrightarrow hypo^2 = 36 + 64$

$\sf\longrightarrow hypo^2 = 100$

$\sf\longrightarrow hypo = \sqrt{100}$

$\sf\longrightarrow hypo = 10cm$

${\bold{\blue{\boxed{\bf{perimeter = AB + BC + CA }}}}}$

$\sf\implies perimeter = 6 + 8 + 10$

$\sf\implies perimeter = 24cm$

${\bold{\blue{\boxed{\bf{perimeter = 24cm}}}}}$