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kherson [118]
3 years ago
9

Given: MNOK is a trapezoid, MN=OK, m∠M=60°, NK ⊥ MN , MK=16cm Find: The midsegment of MNOK

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
4 0

Answer:

12 cm

Step-by-step explanation:

1. Consider right triangle MNK. In this triangle angle N is right and m∠M=60°, then m∠K=30°. Thus, this triangle is special 30°-60°-90° right triangle with legs MN and NK and hypotenuse MK=16 cm. The leg MN is opposite to the angle with measure of 30°, then this leg is half of the hypotenuse, MN=8 cm.

2. Consider right triangle MNH, where NH is the height of trapezoid drawn from the point N. In this triangle m∠M=60°, angle H is right, then m∠N=30°. Similarly, the leg MH is half of the hypotenuse MN, MH=4 cm.

3. Trapezoid MNOK is isosceles, because MN=OK=8 cm. This means that NO=MK-2MH=16-8=8 cm.

4. The midsegment of the trapezoid is

\dfrac{MK+NO}{2}=\dfrac{16+8}{2}=12\ cm.

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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 root5 Find: KM
Alla [95]

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.

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 KM is \sqrt{109} units.

5 0
3 years ago
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