Question

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Class : 9

Subject : Maths

Chapter : Heron's Formula

$\large{\underline{\bf{QUESTION :}}}$

The perimeter of a right triangle is $\bf{60 \: cm }$. It's hypotenuse is $\bf{26 \: cm}$. Find the $\bf{other \: two \: sides}$ and the $\bf{area}$ of the triangle.

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Answer 1

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$<b>Solution:$

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Let a,b, c are the sides of a triangle .

Given that hypotenuse(c) = 26 cm.

a + b + c = 60 cm.

∴ a + b = 60 - c = 60 - 26 =

∴ a + b = 34 --------(1)

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In a right angled triangle c² = a² + b²

a² + b² = 262 = 676 ---------(2)

Now a² + ( 34 - a)² = 676.  [ from (1)]

⇒ a² + a² - 68a + 1156 = 676

⇒ 2a² - 34a + 480 = 0

⇒ a² - 17a + 240 = 0

⇒ a² - 10a - 24a + 240 = 0

⇒ a(a - 10) - 24(a - 10) = 0

∴ a = 10 , 24

but b = 34 - a then b = 24 , 10

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∴ sides of triangle are 10 , 24
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Answer 2

Step-by-step explanation:

Let a,b, c are the sides of a triangle .

Given that hypotenuse(c) = 26 cm.

a + b + c = 60 cm.

∴ a + b = 60 - c = 60 - 26 =

∴ a + b = 34 --------(1)

===================================

In a right angled triangle c² = a² + b²

a² + b² = 262 = 676 ---------(2)

Now a² + ( 34 - a)² = 676. [ from (1)]

⇒ a² + a² - 68a + 1156 = 676

⇒ 2a² - 34a + 480 = 0

⇒ a² - 17a + 240 = 0

⇒ a² - 10a - 24a + 240 = 0

⇒ a(a - 10) - 24(a - 10) = 0

∴ a = 10 , 24

but b = 34 - a then b = 24 , 10

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∴ sides of triangle are 10 , 24

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