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egoroff_w [7]
3 years ago
8

URGENT HELP NEEDED WITH RIGHT TRIANGLE TRIG!

Mathematics
2 answers:
WINSTONCH [101]3 years ago
7 0

Since this is a right triangle, we could use the pythagorean theorem: a²+b²=c² where this problem already gave us a=10 and b=√60, and we want to solve for c.


10²+(√60)²=c² ← 10² is 100, and the square cancels out the square root on the 60

100 + 60 = c²

160 = c² ← To isolate c, we take the square root of both sides

√160 = c ← √160 is the same as 4√10


Answer: 4√10


Aleksandr [31]3 years ago
6 0
We can notice that , it is a right angled triangle.

Let the unknown side be ' c ' units.

Now by using Pythagoras we have a relation for all sides of the right angled triangle.

i.e
{ a}^{2}  +  {b}^{2}  =  {c}^{2}
Now by substituting the values of a and b ;

{ c}^{2}  =  {10}^{2}  +  { \sqrt{60} }^{2}  \\  \\  {c}^{2}  = 100 + 60 = 160 \\  \\ c =  \sqrt{160}  = 4 \sqrt{10}
Hence the third option has the correct answer.
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Answer:

The length of KM is \sqrt{109} units.

Step-by-step explanation:

Given information:  KLMN is a trapezoid, ∠N= ∠KML, FD=8, \frac{LM}{KN}=\frac{3}{5}, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3\sqrt{5}.

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is 3\sqrt{5}.

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since \frac{LM}{KN}=\frac{3}{5}, therefore LM is 3x and KN is 5x.

\frac{3x+5x}{2}=8

\frac{8x}{2}=8

x=2

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

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10=6+2(EN)                (KA=EN, isosceles trapezoid)

EN=2

KE=KN-EN=10-2=8

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

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(KM)^2=(KE)^2+(ME)^2

(KM)^2=(8)^2+(3\sqrt{5})^2

KM^2=64+9(5)

KM=\sqrt{109}

Therefore the length of KM is \sqrt{109} units.

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