Question

the altitude of a right triangle is 7 cm less than its base . if hypotenuse is 13 cm find other two sides​

Answer 1

Step-by-step explanation:

let base be x cm

perpendicular = x -7 cm

hypotenuse = 13 cm

h² = p²+ b²

h² = (x-7)² + x²

h² = x² -49+x²

13² = 2x² -49

169 + 49 = 2x²

218 = 2x²

x² = 109

x = √109

Answer 2

Question :-

The altitude of a right triangle is 7 cm less than its base. If hypotenuse is 13 cm, find other two sides.

$\large\underline{\sf{Solution-}}$

Given that,

The altitude of a right triangle is 7 cm less than its base.

So, Let assume that

Base of right angle triangle is x cm.

So,

Altitude of right angle triangle is x - 7 cm.

Thus, we have In right angle triangle

Base = x cm

Altitude = x - 7 cm

Hypotenuse = 13 cm

Now, By using Pythagoras Theorem, we have

$\rm \: {Hypotenuse}^{2} = {Base}^{2} + {Altitude}^{2}$

So, on substituting the values, we get

$\rm \: {13}^{2} = {x}^{2} + {(x - 7)}^{2}$

$\rm \: 169 = {x}^{2} + {x}^{2} + 49 - 14x$

$\rm \: 169 = 2{x}^{2} + 49 - 14x$

$\rm \: 2{x}^{2} + 49 - 14x - 169 = 0$

$\rm \: 2{x}^{2} - 14x - 120 = 0$

$\rm \: 2({x}^{2} - 7x - 60) = 0$

$\rm \: {x}^{2} - 7x - 60= 0$

$\rm \: {x}^{2} - 12x + 5x - 60= 0$

$\rm \: x(x - 12) + 5(x - 12) = 0$

$\rm \: (x - 12)(x + 5) = 0$

$\rm\implies \:x = 12 \: \: or \: \: x = - 5 \: \{rejected \}$

So,

Base of right angle triangle = 12 cm

Altitude of right angle triangle = 12 - 7 = 5 cm

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac