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3 years ago

Which of the following is the correct name for LM in ABC below

Mathematics

2 answers:

Lena [83]3 years ago

6 0

FM is midsegment of triangle ABC

Answer

C. Midsegment

Irina-Kira [14]3 years ago

5 0

Answer:

C) midsegment

Step-by-step explanation:

An angle bisector cuts an angle in half. LM does not do this.

An altitude extends from a vertex perpendicular to the opposite side. LM does not do this.

A midsegment goes from the midpoint of one side to the midpoint of another side. LM does this.

A perpendicular bisector splits a side in half, forming a right angle. LM does not do this.

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3 years ago

Read 2 more answers

1)a chord of length 18cm midway the radius of a circle. calculate the radius of the circle correct to 1d.p. 2)if two parallel ch

kykrilka [37]

Part 1:

Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.

Thus, half the angle formed by the chord at the centre of the circle is given by:

$\cos\theta=\frac{\left( \frac{1}{2} r\right)}{r}= \frac{1}{2} \\ \\ \Rightarrow\theta=\cos^{-1}\left( \frac{1}{2} \right)=60^o$

Now,

$\sin60^o= \frac{9}{r} \\ \\ \Rightarrow r= \frac{9}{\sin60^o} =10.392$

Therefore, the radius of the circle is 10.4 cm to 1 d.p.

Part 2I:

Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

$\sin\theta= \frac{4}{10} = \frac{2}{5} \\ \\ \theta=\sin^{-1}\left( \frac{2}{5} \right)=23.6^o$

Now,

$\cos23.6^o= \frac{x}{10} \\ \\ \Rightarrow x=10\cos23.6^o=9.165\approx9.2cm$

Part 2II:

Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

$\sin40^o= \frac{y}{10} \\ \\ \\ \Rightarrow y=10\sin40^o=6.428$

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately 12.9 cm.

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