The hypotenuse of a right angled triangle is 1m less than twice the shortest side. if the third side is 1m more than the shortest side, find the sides of the triangle
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Hi...☺
Here is your answer...✌
Let the shortest side of the right triangle be x m
Then,
Hypotenuse = (2x - 1) m
And Third side = (x + 1) m
Using Pythagoras Theorem
Hypotenuse² = (Side 1)² + (Side2)²
(2x - 1)² = x² + (x + 1)²
4x² - 4x + 1 = x² + x² + 2x + 1
4x² - 4x + 1 = 2x² + 2x + 1
⇒ 4x² - 2x² - 4x - 2x + 1 - 1 = 0
⇒ 2x² - 6x = 0
⇒ 2x (x - 3) = 0
⇒ 2x = 0 or x - 3 = 0
⇒ x = 0 or x = 3
⇒ x = 3
[ By rejecting x = 0 Since , Side cannot be 0 ]
Thus, x = 3
⇒ 2x - 1 = 2(3) - 1 = 5 { Hypotenuse }
⇒ x + 1 = 3 + 1 = 4 { Third side }
Hence ,
The sides of given right angled triangle are 3m , 4m and 5m
Here is your answer...✌
Let the shortest side of the right triangle be x m
Then,
Hypotenuse = (2x - 1) m
And Third side = (x + 1) m
Using Pythagoras Theorem
Hypotenuse² = (Side 1)² + (Side2)²
(2x - 1)² = x² + (x + 1)²
4x² - 4x + 1 = x² + x² + 2x + 1
4x² - 4x + 1 = 2x² + 2x + 1
⇒ 4x² - 2x² - 4x - 2x + 1 - 1 = 0
⇒ 2x² - 6x = 0
⇒ 2x (x - 3) = 0
⇒ 2x = 0 or x - 3 = 0
⇒ x = 0 or x = 3
⇒ x = 3
[ By rejecting x = 0 Since , Side cannot be 0 ]
Thus, x = 3
⇒ 2x - 1 = 2(3) - 1 = 5 { Hypotenuse }
⇒ x + 1 = 3 + 1 = 4 { Third side }
Hence ,
The sides of given right angled triangle are 3m , 4m and 5m
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55
Given solution is in the pic
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