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wolverine [178]
3 years ago
12

Show Workings.Question is in attached image.​

Mathematics
2 answers:
Anna35 [415]3 years ago
6 0

Answer:

<h3>Question 1</h3>

a)

d = 40 cm ⇒ r = 20 cm

Let the perpendicular distance is x.

Connecting the center with  the chord we obtain a right triangle with hypotenuse of r and leg x with adjacent angle of 70/2 = 35°.

<u>From the given we get:</u>

  • x/20 = cos 35°
  • x = 20 cos 35°
  • x = 16.383 cm (rounded)

b)

The minor arc is 70° and r = 20

<u>The length of the arc is:</u>

  • s = 2πr*70/360° = 2*3.142*20*7/36 = 24.437 cm (rounded)
<h3 /><h3>Question 2</h3>

Since XZ is diameter, the opposite angle is the right angle, so the triangle XYZ is a right triangle.

  • r = 15/2 cm ⇒ XZ = d = 2r = 2*15/2 = 15 cm

<u>Find the missing side, using Pythagorean:</u>

  • YZ = \sqrt{XZ^2 - XY^2} = \sqrt{15^2-12^2} = \sqrt{81} = 9

<u>The area of the triangle:</u>

  • A = 1/2*XY*YZ = 1/2*12*9 = 54 cm²

Mashutka [201]3 years ago
3 0

Answer:

A.]A chord of a circle of diameter 40 cm subtends an angle of 70° at the centre of the circle.

Solution given;

diameter [d]=40cm

centre angle [C]=70°

(a) Find the perpendicular distance be tween the chord and the centre of the circle.

Answer:

we have

the perpendicular distance be tween the chord and the centre of the circle=[P]let

we have

P=d Sin (C/2)

=40*sin (70/2)

=22.9cm

<u>the</u><u> </u><u>perpendicular distance be tween the chord and the centre of the </u><u>circle</u><u> </u><u>is</u><u> </u><u>2</u><u>2</u><u>.</u><u>9</u><u>c</u><u>m</u><u>.</u>

(b) Using = 3.142, find the length of the minor arc.

Solution given;

minor arc=\frac{70}{360}*πd=\frac{7}{36}*3.142*40

=<u>2</u><u>4</u><u>.</u><u>4</u><u>4</u><u>c</u><u>m</u>

<u>the length of the minor arc.</u><u> </u><u>is</u><u> </u><u>2</u><u>4</u><u>.</u><u>4</u><u>4</u><u>cm</u><u>.</u>

B.]In the diagram, XZ is a diameter of the cir cle XYZW, with centre O and radius 15/2 cm.

If XY = 12 cm, find the area of triangle XYZ.

Solution given:

XY=12cm

XO=15/2cm

XZ=2*15/2=15cm

Now

In right angled triangle XOY [inscribed angle on a diameter is 90°]

By using Pythagoras law

h²=p²+b²

XZ²=XY²+YZ²

15²=12²+YZ²

YZ²=15²-12²

YZ=\sqrt{81}=9cm

:.

base=9cm

perpendicular=12cm

Now

Area of triangle XYZ=½*perpendicular*base

=½*12*9=54cm²

the area of triangle XYZ is <u>5</u><u>4</u><u>c</u><u>m</u><u>²</u><u>.</u>

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