Question

the altitude of
a right triangle is 7cm less then it's base if the hypotenusevis 13 CM find the other two side​

Answer 1

Answer:

Hypotenuse = 13cm

Let one of the sides be x, then the other will be x - 7.

Using the Pythagoras Theorem.

$x^{2} + (x-7)^{2} = 13^{2} \\x^{2} + (x)^{2} + 7^{2} - 14x = 169\\2x^{2} - 14x + 49 - 169 = 0\\2x^{2} - 14x - 120 = 0\\x^{2} - 7x - 60 = 0\\x^{2} + 5x - 12x - 60 = 0\\x(x + 5) - 12(x + 5) = 0\\(x - 12)(x + 5) = 0\\\\Case 1\\x - 12 = 0\\x = 12\\\\Case 2\\x + 5 = 0\\x = -5$

As length cannot be negative, rejecting Case 2.

Hence, x = 12 and x - 7 = 5.

So the side lengths are 12cm and 5cm.

Answer 2

Given:-

• The altitude of a right triangle is 7cm less then it's base the hypotenus is 13 CM

To find:-

• find the other two side.

Solutions:-

• Let the base of the right triangle be x cm.
• it's altitude = (x - 7) cm

From Pythagoras theorem;

Base² + Altitude ² = Hypothesis ²

Therefore,

=> x² + (x - 7)² = 13²

=> x² + x² + 49 - 14x = 169

=> 2x² - 14x - 169 - 49 = 0

=> 2x² - 14x - 120 = 0

=> 2(x² - 7x - 60) = 0

=> x² - 7x - 60 = 0

=> x² - 12x + 5x - 60 = 0

=> x(x - 12) + 5(x - 12) = 0

=> (x - 12) (x + 5) = 0

=> x - 12 = 0 or x + 5 = 0

=> x = 12 0r x = -5

Since, sides are positive x can only be 12.

Therefore, the base of the triangle is 12cm and altitude of the triangle be (12 - 7)cm = 5 cm.

Hence, the others two sides is 12cm and 5cm.