Answer:
In the given triangle, 9² + 12² = (x + y)² [Using the Pythagorean theorem]
⇒ 81 + 144 = (x + y)²
⇒ 225 = (x + y)²
⇒ = x + y
⇒ 15 = x + y --------------- 1
∴ x + y = 15 [from 1]
Now, z² + x² = 9²
⇒ z² + x² = 81
⇒ x² = 81 - z² ---------------2
Also, z² + y² = 12²
⇒ z² + y² = 144
But, y = 15 - x [from 1]
So, substituting y:
⇒ z² + (15 - x)² = 144
⇒ z² + 225 + x² - 30x = 144
⇒ x² = 144 - z² - 225 + 30x ------------------3
Now, from eq.2 and eq.3 (Since both are equal to x²) :
⇒ 81 - z² = 144 - z² -225 + 30x
⇒ 81 - z² = -81 - z² + 30x {solved -225 + 144}
⇒ 81 = -81 + 30x {Cancelled -z² from both sides}
⇒ 81 + 81 = 30x
⇒ 162 = 30x
⇒ 162/30 = x
⇒ x = 5.4
But, x² + z² = 81
⇒ 5.4² + z² = 81
⇒ 29.16 + z² = 81
⇒ z² = 51.84
⇒ z = 7.2
we already know from eq.1 that x + y = 15
So, x + y + z = 15 + 7.2
So, x + y + z = 22.2
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Answer:
In the given triangle, 9² + 12² = (x + y)² [Using the Pythagorean theorem]
⇒ 81 + 144 = (x + y)²
⇒ 225 = (x + y)²
⇒
= x + y
⇒ 15 = x + y --------------- 1
∴ x + y = 15 [from 1]
Now, z² + x² = 9²
⇒ z² + x² = 81
⇒ x² = 81 - z² ---------------2
Also, z² + y² = 12²
⇒ z² + y² = 144
But, y = 15 - x [from 1]
So, substituting y:
⇒ z² + (15 - x)² = 144
⇒ z² + 225 + x² - 30x = 144
⇒ x² = 144 - z² - 225 + 30x ------------------3
Now, from eq.2 and eq.3 (Since both are equal to x²) :
⇒ 81 - z² = 144 - z² -225 + 30x
⇒ 81 - z² = -81 - z² + 30x {solved -225 + 144}
⇒ 81 = -81 + 30x {Cancelled -z² from both sides}
⇒ 81 + 81 = 30x
⇒ 162 = 30x
⇒ 162/30 = x
⇒ x = 5.4
But, x² + z² = 81
⇒ 5.4² + z² = 81
⇒ 29.16 + z² = 81
⇒ z² = 51.84
⇒ z = 7.2
we already know from eq.1 that x + y = 15
So, x + y + z = 15 + 7.2
So, x + y + z = 22.2
Hope it helps! Please upvote :)