How many triangles exist with the given side lengths? 2mm,6mm,10mm

Mathematics

1 answer:

kondaur [170]3 years ago

4 0

Answer:

Zero

Step-by-step explanation:

2+6=8 which means it can't be. It has to be a length higher than 10

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The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!