1.

ABCD is a rectangle in which AB = 21 cm.
The square AXYD is removed and the remaining rectangle has area 80 cm^2.
Find the length BC.

2.


A, B, C and D are posts on the banks of a 20m. Wide canal. A and B are 1 m apart.

If OA is the same length as CD, find how far C and D are apart.

3.

AB is 2cm longer than BE. DC is 3 cm less than twice BE.

a) Explain why triangles ABE and ACD are similar.

b) If BE = x cm, show that x^2 - 4x - 6 = 0.

c) Hence, show that BE = 2 + square root 10 cm

4.

The sum of a number and its reciprocal is 2*2/12 . Find the number.

5.

The sum of a number and twice its reciprocal is 3*2/3. Find the number

6.

A square sheet of cardboard has sides 20 cm long. Its is to be made into an open box with a base having an area of 100 cm^2, by cuting out equal squares from the four corners and then bending the edges upwards.

Find the size of the squares cut out.

A rectangle swimming pool is 12 m long by 6 m wide. It is surrounded by a pavement of uniform width, the area of the pavement being 7/8 of the area of the pool.

a) If the pavement is x m wide, show that the area of the pavement is 4x^2 + 36x m^2.

b) Hence, show that 4x^2 + 36x - 63 = 0

c) How wide is the pavement?

8.

A circular magnet has an inner radius x cm, an outer radius 2 cm larger and its depth is the same as the inner radius (as shown).
If the total volume of the magnet is 120 pi cm^3, find x

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3.

AB is 2cm longer than BE. DC is 3 cm less than twice BE.

a) Explain why triangles ABE and ACD are similar.

b) If BE = x cm, show that x^2 - 4x - 6 = 0.

c) Hence, show that BE = 2 + square root 10 cm


You seem to be posting problems with a common theme. What are you not understanding about quadratics? Everything? :confused:

Just use similar triangles. \frac{BE}{AB}=\frac{DC}{AC}

\frac{x}{x+2}=\frac{2x-3}{x+5}

Solve for x and you're set.

I don't have much time. I'll do the best i can. So, 1

Area of big rectangle = 21x (Let x = length BC)

Area of removed square = x^2

Therefore, area of small rectangle 80 = 21x - x^2

Then solve for x.

4. & 5. Let that number be x. Hum, what do you mean by 2*2/12 and 3*2/3? Make x +
1/x equals the number you entered and solve.

*no more time* the rest are similar triangles , etc. Galactus and ebaines or someone else can continue to explain to you in the meanwhile...

This poster doesn't appear to appreciate the help anyway. I don't believe galactus is going to help anymore.

Ok, num 2.

Let side AO be x, therefore CD is also = x

Using similar triangles, \frac 1x = \frac {x}{20+x}

Simplifying: 0 = x^2 - x - 20

Then Solve.

num 3.

a) Wait. Do you know about similar triangles? I'll take this number as an example.
You'll find that angle ABE and ACD are equal. Then, they both have an angle in common, that is angle CAD or BAE. Also you have angle BEA and CDA which are equal because of the parallel lines BE and CD and AED is a straight line. Similar triangles have same corresponding angles.


Then, similar triangles also have the ratio of their corresponding sides equal. Take BE.
\frac{BE}{DC} = \frac{AB}{AC} = \frac{AE}{AD} and \frac{BE}{AB} = \frac{DC}{AC}, etc...

Note that you should have all those conditions for two triangles to be similar. These will help you for the work to be done.

b) & c) Galactus has already given the method

num 6.

Don't understand what's so difficult. You have the area of the base of your box. What do you think is the length of the width, if you have to cut it from a square cardboard and squares are to be removed?

num 7.

a) How would you obtain the area of the pavement? With a drawing, you must find the total area (i.e. pavement + pool) and subtract the area of the pool. What is the total area, if the pavement has a uniform width of x cm? You have
(12+2x)(6+2x). You can obtain the area of the pool easily. Just do the subtraction and there it is, you have 4x^2+36x

b) How can you obtain the area of the pavement in another way? the area of the pavement being 7/8 of the area of the pool Just do the area of the pavement. Therefore, 4x^2+36x=(\frac{7}{8})(12)(6) and put it in the form asked in the question.

c) Solve for x to have the width of the pavement.

num 8 (hum... you mean a hollow cylindrical magnet?)

How do you find the volume of the magnet? Subtract the volume of the 'cylinder' with x as radius from the 'cylinder' with (x+2) as radius. The expression you obtain is equal to 120\pi. Solve of x and you're done!

Hope this helped!

Poster doesn't have to appreciate it since he's got someone doing the homework for him. And I don't know about anyone else, but five separate questions in a thread makes me bonkers.

Yes, that was a bit confusing to put these in the right order and think on them one after the other. I was just trying to make him understand the steps and how o arrive to the answer.

... you think that maybe if I put my homework on this thread, someone'll do it for me..

... because this mcat thing is being a %@(@

Are you sure your nose isn't breathing funny things in instead of sneezing funny things out?

... because this mcat thing is being a %@(@

Sorry? I've difficulty to understand you here...

Haha, no worries unky. Slight humor.

But I do want to understand!

Thank you =)

You're welcomed!