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=
=<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=
:.
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>
Answer:
They are congruent
Step-by-step explanation:
All of the sides and the angles are the same and it is just a rotation.
Let’s assume the following given:
Speed = 129 m/s
Time = 2,900,000 s
Find:
Distance = ?
Solution:
In order to get the distance, we multiply speed and time.
Distance = Speed * Time
Distance = 129 m/s * 2,900,000 s
Distance = 374,100,000 m
Next, we will convert the distance from meters to kilometers
by dividing the distance with 1,000 m.
374,100,000 m / 1,000 m = 374,100 km
<span>The last step would be converting the travelled distance in
kilometers to the nearest ten thousand.
Therefore, the travelled distance in kilometers to the nearest ten thousand is
370,000 km.</span>