Question

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

Answer 1

Answer:

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Let's take the base as X

Given, Altitude is seven less than the base

Then the base is (x-7)

By the Pythagorus theorum

(Hypotenuse^2) = (Base^2) + (Perpendicular^2)

(13m^2) = (x^2) + (x-7^2)

169sq.m = 2xsq.m-14x+49

2x^-14x+49 =169

Now interchange the side

Take the constant at one side

2x^2-14x-169 =0

2x^2-14x-120 =0

x^2-7-60 =0

x-12x+5x-60 =0

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

x=12,x=−5

Since the side of the triangle cannot be negative, so the base of the triangle is 12cm and the altitude of the triangle will be 12−7=5cm.

Answer 2

Given,

• Altitude of right triangle is 7 cm less than its base.
• Hypotenuse is 13 cm.

To find,

• The other two sides.

Solution,

• Let x be the base of the triangle
• Then altitude will be (x-7)

We know that,

$\sf{Base^2+Altitude^2=Hypotenuse^2}$

So, by pythagoras theorem,

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

So, x = 12 or x = -5

Since,the side of a triangle cannot be negative,so the base of the triangle is 12 cm.

And the altitude will be (12-7) = 5 cm