Math, asked by spanshsingh, 4 months ago

find the lengths of the medians of the triangle with vertices a(-1 1) b(5 -3) c(3 5)find the lengths of the medians of the triangle with vertices a(-1 1) b(5 -3) c(3 5)

Answers

Answered by Anonymous
1

Answer:

\star\:\:\bf\large\underline\blue{Question:-}

Question:−

Simplify the problem.

\star\:\:\bf\large\underline\blue{Solution:-}

Solution:−

\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }

Now, let

y = \sqrt{ab + (a + b)x + {x}^{2} }y=

Therefore,

\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }

= \sqrt{a + b + 2x + 2 + 2 \sqrt{y} }=

Now, we will factorise y.

\sqrt{ab + (a + b)x + {x}^{2} }

= \sqrt{ab + ax + bx + {x}^{2} }=

= \sqrt{a(b + x) +x(b+ x)}=

= \sqrt{(x + a)(x + b)}=

Now,

= \sqrt{a + b + 2x + 2 + 2 \sqrt{y} }=

= \small\sqrt{(x + a) + (x + b) + 2 \sqrt{(x + a)(x + b)} }=

= \sqrt{ {( \sqrt{x + a}) }^{2} + {( \sqrt{(x + b} )}^{2} + 2 \times \sqrt{(x + a} \times \sqrt{(x + b)} }=

= \sqrt{ {( \sqrt{x + a} + \sqrt{x + b} )}^{2} }=

= \sqrt{x + a} + \sqrt{x + b}=

\star\:\:\bf\large\underline\blue{Answer:-}

\sqrt{x + a } + \sqrt{x + b}

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