The perimeter of a right angled triangle is 70 units and its hypotenuse is 29 units we would like to find the length of the other sides. Pls, guys solve it, by quadratic equations!!!?

Question

Answer

Let the length of one of the sides be x and the other be y

given, hypotenuse = 29 units
perimeter= 70 units

x+y+hypotenuse = 70
=} x+y+29 = 70

=} x+y= 41

=} y = 41-x

by using pythagoras theorem
(hypotenuse)² = x²+y²

29² = x² + (41-x)²
=}841 = x² + x² -82x+ 1681

=} 2x² -82x + 840= 0

=} x²-41x+420=0

=} x²-21x-20x+420=0

=} x(x-21)-20(x-21)= 0

=} (x-20)(x-21)=0

=} x=20 or x=21

=} y=21 or x=20

∴ the sides are 20 units and 21 units

That's all🙂

Answer

How to find the length of the other sides of a right angled triangle using quadratic equations?

Given Information:

Perimeter of the triangle = 70 units

Hypotenuse of the triangle = 29 units

Solution:

Let the other two sides of the triangle be 'a' and 'b'

Using the Pythagorean theorem, we can write:
a² + b² = (29)² = 841

As the perimeter of the triangle is 70 units, we can write:
a + b + 29 = 70

Simplifying this equation, we get:
a + b = 41

Solving these two equations simultaneously, we get:
a² + b² = 841

a + b = 41

Multiplying the second equation by 2, we get:
2a + 2b = 82

Squaring both sides of the second equation, we get:
(a + b)² = 1681

a² + 2ab + b² = 1681

Substituting a² + b² = 841 from the first equation, we get:
841 + 2ab = 1681

2ab = 840

ab = 420

Now, we can write one of the sides in terms of the other as:
a = 41 - b

Substituting this in the equation ab = 420, we get:
(41 - b)b = 420

b² - 41b + 420 = 0

This is a quadratic equation in 'b', which can be solved using the quadratic formula:
b = (41 ± √(41² - 4(1)(420))) / (2(1))

b = (41 ± 9) / 2

b = 25 or 16

Substituting these values of 'b' in the equation a = 41 - b, we get:
a = 16 or 25

Answer:

The other two sides of the triangle are 16 units and 25 units.

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