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3 years ago

12

Is the following shape a right triangle? How do you know?

2 answers:

Natalka [10]3 years ago

6 0

Answer:

B) Yes , two sides are perpendicular and side length fir the Pythagorean theorem.

Step-by-step explanation:

Given : A triangle on the graph.

To find : Is the following shape a right triangle? How do you know.

Solution : We have given that

A triangle ABC We need to explain is it right triangle or not.

Right triangle : A triangle which have two sides perpendicular or one of the angle is 90 degree.

We can see from the graph side AB and AC are perpendicular to each other.

Therefore, B) Yes , two sides are perpendicular and side length fir the Pythagorean theorem.

Marat540 [252]3 years ago

5 0

Answer:

Yes

Step-by-step explanation:

A has a right angle because A on the triangle is turned at a 90 degree angle.

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Rationalise the denominator of (12)/(\sqrt(10)+\sqrt(7)+\sqrt(3))

jasenka [17]

Answer:

$\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210} }{21 }$

Step-by-step explanation:

$\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }$

$=\frac{12\left( \sqrt{10} -\left( \sqrt{7} +\sqrt{3} \right) \right) }{(\sqrt{10}+(\sqrt{7} +\sqrt{3 } ))( \sqrt{10} -( \sqrt{7} +\sqrt{3}))}$

$=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{10^2} -(\sqrt{7} +\sqrt{3})^2}$

$\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{10-(10+2\sqrt{21} )} }$

$=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{-2\sqrt{21} }$

$=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{21} }$

$=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{\sqrt{21} } \times\frac{\sqrt{21} }{\sqrt{21} }$

$=\frac{-6\sqrt{21} \left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right) }{21 }$

$=\frac{-6\sqrt{210} +6\sqrt{147} +6\sqrt{63} }{21 }$

$=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210} }{21 }$

<u><em>Remark</em></u>:

You can simplify moreover if you want to.

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