Question

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

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {Let,\:base\:of\:triangle\:be\:x}$

$\mathsf {And, \:Let\:altitude\:of\:triangle\:be\:(x - 7)\:Cm}$

$\mathsf {It\:is\:given\:that\:hypotenuse\:of\:triangle\:is\:13\:Cm}$

$\underline \texttt {According\:to\:Pythagoras\:Theorem,}$

$\mathsf {13^2 = x^2 + (x - 7)^2 \: \:(a + b)^2 = a^2 + b^2 + 2ab}$

$\implies \mathsf {169 = x^2 + x^2 + 49 - 14x}$

$\implies \mathsf {169 = 2x^2 - 14x + 49}$

$\implies \mathsf {2x^2 - 14x - 120 = 0}$

$\underline \texttt {Dividing\:equation\:by\:2}$

$\implies \mathsf {x^2 - 7x - 60 = 0}$

$\implies \mathsf {x^2 - 12x + 5x - 60 = 0}$

$\implies \mathsf {X(x - 12) + 5(x -12) = 0}$

$\implies \mathsf {(x - 12) (x + 5}$

$\implies \mathsf {x = -5,\:12}$

$\mathsf {We\:discard\:x = -5\:because\:length\:of\:side}$

$\mathsf {of\:triangle\:cannot\:be\:negative.}$

$\therefore \mathsf \blue {Base\:of\:triangle = 12\:Cm}$

$\mathsf \blue {Altitude\:of\:triangle = (x - 7) = 12 - 7 = 5\:Cm}$

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