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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

Excercises (Contains Trigonometry ~ Grade 9 standard)***Diagrams located below***

**1. **The lines PQ and SR are parallel and are 6 cm apart. T is the midpoint of QR. Find the area, in square centimeters of the shaded region.

**2. **PQR is an isosceles triangle with a base of 18cm and sides of 15cm and is inscribed in the circle as shown. Find the radius length of the circle.

**3. **Ribbon is run around a box so that it makes a complete loop with two parallel pieces of ribbon on the top (and on the bottom) of the box.

The ribbon crosses every face once, except the top and bottom, which it crosses twice. The ribbon rests tightly against the box all the way round because the angle at which it meets a corner is continued onto the next face.

I can cut the ribbon in advance of placing it around the box and I can slide the ribbon around a little to position it.

If the box is 60cm by 30cm by 15cm, how long will the ribbon be?

**4.** For the diagram given, show that

**5.** A square of side length 12cm is placed inside a circle so that each of the corners touches the circumference of the circle. Find the area of the shaded region to 2 decimal places.

**6.** In the rectangle shown, ST and UQ are perpendicular to PR. Given PQ = 18 and QR = 12, find:

(a) Length of UR

(b) Area of the Parallelogram TQUS.

**7.** The bearing of a buoy from the end of a jetty (Point A) is 320° and the bearing of the same buoy from the other end of the jetty (Point B) is 338°. The jetty is on a North-South line, that is A is due North of B. The distance between the points A and B is 150 metres. Calculate the distance of the buoy from point A correct to the nearest metre.

**8.** A ship leaves an island and sails on a bearing of 130° for 150km and then on a bearing of 250° for 320km. If the ship is to return to the island sailing on a direct course,

(a) How far does it have to travel?

(b) What bearing should it sail on?

**9.** A man is observing a yacht at sea from the top of a cliff. The angle of depression of the yacht from the top of the cliff is 44°. If the yacht, from its current position, sails a further 200 metres away from the cliff, the angle of depression is now 38°. How high is the cliff? Give your answer to the nearest metre.

*Last edited by Toast (2006-11-30 01:59:06)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

Nice Exercises ... need more like them!

What title would you prefer?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

*ANSWERS:*

*Diagrams Uploaded beneath again*

*Last edited by Toast (2006-10-27 21:11:00)*

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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

very cool exercises. could you please make more like problem nr 6?

and btw in nr 6 you said in the text that the sides are 9 and 6, but in the solution 18 and 12...

*Last edited by Kurre (2006-11-26 06:09:51)*

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**Mandy Mayday****Member**- Registered: 2006-11-29
- Posts: 4

Perfect exercise - tried it with my son and he understood it better than me!!!!!

Greetings

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