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The hypotenuse of a right-angled triangle is 6 cm more than twice the shortest side.If the third side is 2 cm less than the hypotenuse,find the sides of the triangle.
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Solution
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Let the shortest side $$(AC)$$ be $$x \text{ cm}$$, so the hypotenuse $$(BC)$$ will
be $$(2x+6) \text{ cm}$$, as demonstrated in the figure drawn above.
$$\therefore AB=2x+6-2=2x+4 \text{ cm}$$
According to Pythagoras Theorem,
$$\because (BC)^2=(AC)^2+(AB)^2$$
$$\therefore (2x+6)^2=x^2+(2x+4)^2$$
On simplifying further,
$$4x^2+24x+36=x^2+4x^2+16x+16$$
[Using the identity of $$a^2+b^2+2ab=(a+b)^2$$]
$$x^2-8x-20=0$$..............(i)
On applying the formula,
$$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(-20)}}{2(1)}$$
$$x=\dfrac{8\pm \sqrt{64+80}}{2}$$
$$x=\dfrac{8 \pm \sqrt{144}}{2}$$
$$x=\dfrac{8 \pm 12}{2}$$
$$\therefore x=-2 \text{ or } 10$$
$$\therefore$$ The
shortest side $$(AC)$$ is $$x = 10 \text{ cm}$$ (Only positive values are to be taken), hypotenuse is $$(2x+6)$$ i.e., $$(BC)$$ is $$26 \text{ cm}$$ and the other side $$(AB)$$ is $$(2x+4)$$ i.e., $$24 \text{ cm}$$ .
According to Pythagoras Theorem,
$$\because (BC)^2=(AC)^2+(AB)^2$$
$$\therefore (2x+6)^2=x^2+(2x+4)^2$$
On simplifying further,
$$4x^2+24x+36=x^2+4x^2+16x+16$$
[Using the identity of $$a^2+b^2+2ab=(a+b)^2$$]
$$\therefore (2x+6)^2=x^2+(2x+4)^2$$
On simplifying further,
$$4x^2+24x+36=x^2+4x^2+16x+16$$
[Using the identity of $$a^2+b^2+2ab=(a+b)^2$$]
$$x^2-8x-20=0$$..............(i)
On applying the formula,
$$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(-20)}}{2(1)}$$
$$x=\dfrac{8\pm \sqrt{64+80}}{2}$$
$$x=\dfrac{8 \pm \sqrt{144}}{2}$$
$$x=\dfrac{8 \pm 12}{2}$$
$$\therefore x=-2 \text{ or } 10$$
$$\therefore$$ The
shortest side $$(AC)$$ is $$x = 10 \text{ cm}$$ (Only positive values are to be taken), hypotenuse is $$(2x+6)$$ i.e., $$(BC)$$ is $$26 \text{ cm}$$ and the other side $$(AB)$$ is $$(2x+4)$$ i.e., $$24 \text{ cm}$$ .
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