Question

the lengths of two sides of a right trianngle containing right angle are 5x cm and 3x-1 cm .If it's area is 60 cm^2, find it's perimeter
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Answer 1

Solution :

Let the right triangle be ABC.

AB = 5x cm

BC = ( 3x - 1 ) cm

Given, Area of right triangle = 60 $\mathsf{{cm} ^{2}}$

Area of triangle = $\mathsf{\dfrac{1}{2}\:{\times{Base{\times{Height}}}}}$

60 = $\mathsf{\dfrac{1}{2}\:{\times{( 5x ){\times{( 3x \:-: 1 )}}}}}$

60 × 2 = $\mathsf{15{x} ^{2}\:-\:5x}$

$\mathsf{15{x} ^{2}\:-\:5x\:-\:120\:=\:0}$

Dividing by 5,

$\mathsf{3{x} ^{2}\:-\:x\:-\:24\:=\:0}$

$\mathsf{3{x} ^{2}\:-\:9x\:+\:8x\:-\:24\:=\:0}$

3x ( x - 3 ) + 8( x - 3 ) = 0

( 3x + 8) ( x - 3 ) = 0

x = 3, - 8/3

Since, the value of x = 3 is admissible as the second value is negative which will not be admissible.

Now,

AB = 5 ( 3) cm = 15 cm.

↪️ AB = 15 cm.

BC = ( 3×3 - 1 ) = 9 - 1 = 8 cm.

↪️ BC = 8 cm.

Now, Using PYTHAGORA'S theorem,

AC = $\sqrt{\mathsf{{AB} ^{2}\:+\:{BC}^{2}}}$

AC = $\sqrt{\mathsf{{15} ^{2}\:+\:{8}^{2}}}$

AC = $\sqrt{\mathsf{225\:+\:64}}$

AC = $\sqrt{\mathsf{289}}$

↪️ AC = 17 cm.

Now, Perimeter of right triangle ABC = AB + BC + AC

Perimeter = ( 15 + 8 + 17) cm.

Perimeter = 40 cm.