Question

A wire of length 72 cm is bent to form a right-angled triangle of hypotenuse 30 cm. Find the lengths of the
other two sides.​

Answer 1

Answer:

24cm and 18 cm

Step-by-step explanation:

Let their length be 'x' and 'y'.

Total length of the wire would not change.  As now, it is a triangle.

∴ total length = perimeter of Δ

                 72 = x + y + 30

            42 - x = y          ...(1)

As this is a right angled triangle:  using Pythagoras theorem,

⇒ hypotenuse² = x² + y²

⇒ 30² = x² + (42 - x)²        {from(1)}

⇒ 900 = x² + 1764 + x² - 84x

⇒ 0 = 2x² + 864 - 84x

⇒ 0 = x² + 432 - 42x

⇒ 0 = x² - 18x - 24x + 432

⇒ 0 = x(x - 18) - 24(x - 18)

⇒ 0 = (x - 18)(x - 24)

  Hence, x = 18 or 24

when x = 18, y = 24

          x = 24, y = 18

Answer 2

Given : A wire of length 72 cm is bent to form a right-angled triangle of hypotenuse 30 cm.

Exigency to find : The Length of two other sides .

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❍ Let's Consider the Length of other two sides be a & b respectively.

Given that ,

• A wire of length 72 cm is bent to form a right-angled triangle .

⠀⠀⠀⠀So ,

• The Perimeter of Right angled triangle is 72 cm

⠀⠀⠀⠀&

• One Side of a Triangle or Hypotenuse is of 30 cm .

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Triangle)} \:: a + b + c }\bigg\rgroup$

⠀⠀⠀⠀ Here A , B & C are three sides of Triangle .

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf 72 = 30 + a + b$

$\qquad:\implies \sf 72 - 30 = a + b$

$\qquad:\implies \sf 42 = a + b$

$\qquad:\implies \bf 42 - a = b \:or \:\: Side_2 \:\:\qquad \longrightarrow \:\:Eq.1$

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$\dag\:\:\bf{ By\:Pythagoras \:Theorem \:\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Hypotenuse^2 = Side_1\:^2 + Side_2\:^2 }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \sf :\implies 30^2 = a^2 + b^2$

$\qquad \sf :\implies 900 = a^2 + b^2$

$\qquad \sf :\implies 900 = a^2 + (42- a ) ^2\qquad \qquad [ From\:\:: Eq.1 ]$

$\qquad \sf :\implies 900 = a^2 + (42 - a)^2$

• Algebraic Indentity = (a- b)² = a² + b² - 2ab

⠀By Using this Algebraic indentity :

$\qquad \sf :\implies 900 = a^2 + (42 - a)^2$

$\qquad \sf :\implies 900 = a^2 + 42^2 - 2 \times 42 \times a + a^2$

$\qquad \sf :\implies 900 = a^2 + 1764 + a^2 - 84a$

$\qquad \sf :\implies 900 = 2a^2 + 1764 - 84a$

$\qquad \sf :\implies 0 = 2a^2 + 1764-900 - 84a$

$\qquad \sf :\implies 0 = 2a^2 + 864 - 84a$ [ Canceling each term by 2 ]

$\qquad \sf :\implies 0 = a^2 + 432 - 42a$

$\qquad \sf :\implies 0 = a^2 -42a + 432$

$\qquad \sf :\implies 0 = a^2 - 18a - 24a + 432$

$\qquad \sf :\implies 0 = a( a - 18)-24 ( a - 18 )$

$\qquad \sf :\implies 0 = (a -18) (a - 24)$

$\qquad \longmapsto \frak{\underline{\purple{\:a = 24\: or\:18 }} }\bigstar$

⠀⠀⠀⠀⠀Therefore,

• When ,

⠀⠀⠀⠀⠀Side 1 = 18 , Side 2 = 24

• Or ,

⠀⠀⠀⠀⠀Side 1 = 24 , Side 2 = 18 .

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Length \:of\:two\:sides\:of\:Right-\:Angled \:Triangle \:is\:\bf{24\:cm\:\:\&\:\:18cm}}}}$

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