Question

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Answer 1

Answer:

Question 1

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°.

From the given we get:

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

b)

The minor arc is 70° and r = 20

The length of the arc is:

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

Question 2

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

Find the missing side, using Pythagorean:

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

The area of the triangle:

• A = 1/2*XY*YZ = 1/2*12*9 = 54 cm²

Answer 2

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

the perpendicular distance be tween the chord and the centre of the circle is 22.9cm.

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

Solution given;

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

=24.44cm

the length of the minor arc. is 24.44cm.

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 54cm².