Question

The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse wll be 9√5 cm. How long are the legs of the triangle?

Answer 1

 

SOLUTION :

Given : Hypotenuse of a right triangle is 3√10 cm.

Let the length of  smaller leg be x cm and the larger leg be y cm.

By using Pythagoras theorem,

H² = B² + P²

x² + y²  = (3√10)²

x² + y² = 9×10

x² + y² = 90

y² = 90 - x² . ………….(1)

Given if the smaller leg is tripled and the longer leg is doubled, new hypotenuse will be 9√5 cm

Smaller leg = 3x  

longer leg = 2y

By using Pythagoras theorem,

H² = B² + P²

(3x)² + (2y)² = (9√5)²

9x² + 4y²  = 81×5

9x² + 4y²  = 405

9x² + 4(90 - x²) = 405

[From eq 1]

9x² + 360 - 4x² = 405

9x² - 4x² + 360 - 405 = 0

5x² - 45 = 0

5x² = 45

x² = 45/5

x² = 9

x = √9

x = ±3

Since, length of a side can't be negative, so x ≠ - 3  

Therefore, x = 3

Length of  smaller leg be 3 cm

On putting x = 3 in eq 1,

y² = 90 - x²

y² = 90 - 3²

y² = 90 - 9

y² = 81

y = √81

y = ± 9

Since, length of a side can't be negative, so y ≠ - 9  

Therefore, y = 9

Length of  larger leg be 9 cm

Hence, the length of  smaller leg be 3 cm and the larger leg be 9 cm.

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Answer 2

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Hii dear here is your answer

Let the Sides of the Triangle be a,b and c.
By the Pythagoras Theorem
a^2+b^2=c^2
Given relation- a^2+b^2= (3√10)^2=90……(1)
Now, smaller leg, i.e 'a' is tripled, thus new side is 3a
The other leg is doubled, i.e 2b.
Now given relation,
(3a)^2+(2b)^2 = (9√5)^2
9a^2+4b^2= 405……….(2)
By eq.(1) and eq.(2)
There are 2 variables and 2 equations.Thus, they can be solved by many methods, the simplest ones being Substitution or Elimination.
We will use Substitution.
Eq.1 can also be written as → a^2=90-b^2
Substituting this value in the second equation
9(90-b^2) + 4b^2= 405
810–9b^2 + 4b^2= 405
Thus -5 b^2 = -405
b^2= 81
b=9cm
Thus a^2 + 9^2 = 90
a^2= 90–81=9
Thus a=3 cm.

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