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Alex777 [14]
3 years ago
10

Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM

Mathematics
2 answers:
denis23 [38]3 years ago
8 0

Answer:

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

Step-by-step explanation:

Given information:  KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=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 LM:KN=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.

KN=KA+AE+EN

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.

Hypotenuse^2=base^2+perpendicular^2

(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 side KM is \sqrt{109} units.

olchik [2.2K]3 years ago
7 0

The answer is \sqrt{70}

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