Question

The perimeter of a right angled triangle is 72 cm and its area is 216 cm2. Find the sum of the lengths of its perpendicular sides.

Answer 1

$\sf\red{Given:-}$

• The perimeter of a right angled triangle is 72cm and its area is 216cm².

$\sf\red{To \ Find :-}$

• Sum of length of its perpendicular sides

$\sf\red{Solution :-}$

Let,

• Base = a
• Perpendicular = b
• Hypotenuse = c

Given that,

a + b + c = 72cm

→ a + b = 72 - c ....i)

ab/2 = 216cm²

→ ab = 432 ....ii)

Then value of AB + BC

c² = a² + b² [ Pythagoras theorem ]

________________________________

⟹ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

⟹ (72)² = c² + c² + 2ab + 2bc + 2ac

⟹ 5184 = 2c² + 2ab + 2c (b + a)

⟹ 5184 = 2c² + 2ab + 2c (72 - c)

⟹ 5184 = 2ab + 144c

⟹ 5184 = 2 × 432 + 144c

⟹ 5184 = 864 + 144c

⟹ 144c = 4320

⟹ c = 30

As in first equation,

➝ a + b = 72 - c

➝ a + b = 72 - 30

➝ a + b = 42

Hence, the sum of the lengths of perpendicular sides is 42.

Answer 2

$\large \underline \mathfrak \red{Question}$

↓↓↓↓↓↓↓↓↓↓

The perimeter of a right angled triangle is 72 cm and its area is 216 cm2. Find the sum of the lengths of its perpendicular sides.

$\large \underline \mathfrak \green{answer}$

Let perpendicular sides of the right triangle are a and b.

So if hypotenuse =h

${ \rm \: then \: h^2= a^2+b^2 }$

$\rm{now \: perimeter = h+a+b =72 cm}</p><p></p><p></p><p>[tex] \rm{→a+b =72-h …1}</p><p>$

\rm{→and \: area =0.5 ×a ×b = 216 cm^2

}

[/tex]

______________________________

$\rm{→a^2+b^2= h^2}$

$\rm{→or, (a+b)^2–2ab=h^2}$

$\rm{→or, (72-h)^2–2×432=h^2}$

$\rm{→or, 72^2–2×72×h+h^2 -864=h^2}$

$\rm{→or \: 5184–144h +h^2–864=h^2}$

$\rm{→or, 144h =4320}$

$\rm{→or , h=30}$

_____________________________

→→from (1)

$\rm{72-h=a+b , or, a+b =72–30=42}$

sum of the perpendicular sides

=>42 cm

$\rm \blue{@BrainlyBlockBusterBB}$