Posted at 16:49 in Age Range: 11-16, Focus: Numbers & Calculation, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Posted at 17:07 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query: How do I estimate the height of this mast (see diagram)?
Response: The man's height has to be estimated first. Next, you calculate how many men of the same height (i.e. standing on each other's heads) it would take to reach the top of the mast.
Click on the graphic below to see how you would show this.
Posted at 10:13 in Age Range: 11-16, Focus: Numbers & Calculation, Focus: Shape & Space, Phase: GCSE Foundation, Phase: Key Stage 3 (UK) | Permalink | Comments (0) | TrackBack (0)
Posted at 13:43 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: Key Stage 3 (UK) | Permalink | Comments (0) | TrackBack (0)
Posted at 10:00 in Age Range: 11-16, Focus: Numbers & Calculation, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Posted at 10:05 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: Key Stage 3 (UK) | Permalink | Comments (0) | TrackBack (0)
Posted at 18:00 in Age Range: 11-16, Focus: Numbers & Calculation, Focus: Shape & Space, Numeracy, Phase: GCSE Foundation, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Posted at 06:00 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [From LAL]: A 20cm square sheet of metal has equal quadrants cut from the corners, as shown in the diagram. The area left is 246 sq.cm and we need to find the radius of the circle from which the quadrants are taken. [use pi as 22/7]
Response:
The answer is 7 cm.
Posted at 09:28 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from Aristotle]: There is a missing shape from this series...
The correct answer is below - can you explain why this is?
Response: We sense we're being tested here - so here goes....!
Take a look at the rows and columns of the original three by three shapes (minus the missing one, that is) and try to spot the following clues...
Posted at 16:43 in Age Range: 11-16, Focus: Logic & Puzzles, Focus: Shape & Space | Permalink | Comments (0) | TrackBack (0)
Query [from Ron Gale]: I run a ranch in Canada, where we train horses. I would like to know how many cubic yards of sand I will need for my arena measuring 100yds. X 300yds. (with rounded ends). Also, how do I calculate the cubic yards per inch of depth?
Response: Initially, you should disregard the effect of the rounded corners, as it will probably only actually represent a very small percentage of the top surface area.
A depth of 1 yd. of sand will require a volume of 100x300x1 cu.yds.
So, a 1" depth requires 1/36th of this, i.e. (100x300x1) ÷ 36 = 833.3 cu.yds.
If you wanted an 8" depth, for example, then simply multiply this figure by x8 (to get an approximate total of 6666 cu.yds., to the nearest cu.yd.).
At this stage, you should make a decision about the likely effects of the rounded corners. So, if the radius of these corners is less than 1 yd., you could just ignore it.
However, if the corners are very rounded then apply a 2.5% or a 5% (or even higher) reduction to the overall volume you have calculated, i.e. multiply your total by x0.975 or x0.95.
COMMENT [from DH]: You wouldn't believe how nice it is to work in imperial units once more!
Posted at 16:45 in Age Range: 11-16, Focus: Numbers & Calculation, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
The answer is 9 cm. These are the steps you need to take....
Posted at 12:01 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher, Problem Challenge | Permalink | Comments (0) | TrackBack (0)
A hemispherical bowl of internal diameter 12 cm. is full of water and all the water is emptied into a cylindrical can of base diameter 8 cm.
How deep will the water level in the cylinder become?
Please e-mail us your solution - you can attach a scan of your handwritten version, if it's more convenient. The solution will be posted at 12 noon [GMT] on Friday, 14/09/07.
Again, no prizes on offer - just the pleasure of being first to solve it! But can you solve it?
Posted at 07:00 in Age Range: 11-16, Focus: Algebra & Graphs, Focus: Shape & Space, Phase: GCSE Higher, Problem Challenge | Permalink | Comments (0) | TrackBack (0)
Query [from KP]: I've been given two triangles - one with sides of 6, 8 and 12 cm and the other with sides 9, 10 and 15 cm. I'm meant to prove they are not similar?
Response: Although the triangles in the diagram certainly look similar, they are not drawn to scale. You will need to compare corresponding sides to test for similarity, by seeing if one triangle is a true enlargement of the other.
In their lowest terms: Smallest (2:3), Middle (4:5) & Longest (4:5).
Since these ratios are not identical, it follows that the triangles are not similar.
Posted at 07:00 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from CarolK]: I've been given a triangle with angles of 30, 60 and 90 degrees. The hypotenuse length is 3 point 5 and I'm meant to find the length opposite the 60 degree angle in SURD form? I could do it using a calculator easily enough!
Response: All 30-60-90 triangles are enlargements of the one shown in the diagram [this should really be memorised].
In this case, comparing the lengths of both hypotenuses gives an enlargement ratio of 2:3.5, i.e. a scale factor of 1.75.
This means that the side opposite the 60° becomes 1.75 x √3.
You should now completely rationalise this surd term by removing the decimals.
Posted at 08:00 in Age Range: 16+, Focus: Pure Maths, Focus: Shape & Space | Permalink | Comments (1) | TrackBack (0)
Most students find the whole topic of similarity very challenging.
At its most simple, two figures (2 or 3 dimensions) are said to be similar if one is an enlargement of the other. This means that the angles associated with both shapes are identical and the lengths are in a fixed ratio, sometimes called the scale factor.
Query [from TerryL]: I've been given a diagram with two regular pentagons and told that the area of the inner pentagon is 5cm2. How do I find the shaded area?
Response: As both pentagons are regular we know immediately that they must be similar to each other.
The length scale factor of enlargement is x 2, worked out by comparing the lengths of y and 2y (since one is 2 'times' the other).
This means that the area scale factor becomes x 4, which was arrived at by taking the square of the length scale factor.
In this case, therefore the outer area must be 20cm2 , which leaves the shaded area as 15cm2.
Posted at 10:05 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
A prism is defined as a three-dimensional object which has an uniform cross-sectional area running along one axis. So, if the prism was cut perpendicular to this axis the new faces revealed would be exactly the same size and shape as the ends of the original object. Confusing, isn't it?
Query: You're given a pentagonal prism with a length of 2.5 cm. and a volume of 1.875 cubic cm. and you're meant to find the area of cross-section?
Response: The shaded area here refers to the uniform cross-section, which all prisms must have.
The Volume of any prism = 'Cross-Sectional Area (shaded)' x 'Length'
So the Cross-Sectional Area = 1.875 ÷ 2.5 = 0.75 cm2.
Posted at 07:20 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
This is a type of problem that we've had to explain many times in our teaching careers!
Query [from AL]: I was asked to find the perimeter of a semi-circular shape of diameter 20 cm. I worked out the circumference of the circle and then divided it by two. I got this question wrong, why?
Response: It is true that the circular part of the perimeter of the semicircle is exactly half the circumference of the whole circle.
However, a journey around any perimeter involves following the outline of the shape until you end up back where you started!
You forgot to add on the diameter, which joins the opposite ends of the semicircle, when calculating the whole perimeter.
Perimeter = ½ x (2 x π x 10) + 20 ≈ 51.4 cm.
Posted at 08:09 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: Key Stage 3 (UK) | Permalink | Comments (3) | TrackBack (0)
Drawing a line of best fit for data usually causes real problems for students.
At GCSE (11-16) level, however, you just have to draw your best 'guesstimate', as there is no elaborate theory of 'linear regression' to apply - this is only studied for Advanced, Higher or University courses! 
Query [from GOG]: In this scatter graph, I'm given the mean point and asked to draw a line of best fit?
Response: A line of best fit should pass through the coordinates of the mean points of the two 'graphed' variables and it is then drawn 'by eye' (that is, trying to get about as many points below as above!)
Here, we suggest you plot the given mean point and then using your ruler sideways (thin side down) you
NOTE: If the mean isn't given and the question is of low value (only worth 1 or 2 marks), then just draw it by 'eye', without calculating the means.
Posted at 12:38 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: Key Stage 3 (UK) | Permalink | Comments (2) | TrackBack (0)
Query [from PaulG]: I've been asked to find the distance from the point A(2,3) to another point B(5,-4) and I get the answer to be the square root of 10 when I apply the distance formula? What have I done wrong?
Response: The distance formula is actually an application of the Theorem of Pythagoras. If you draw a quick sketch of these points onto an X-Y coordinate grid, you'll get the picture that we've created here...
The XAB shift is +3 units and the YAB shift is -7 units (not -1)
This means that AB2 = 32 + (-7)2 = 58, giving AB = √58.
Posted at 10:43 in Age Range: 16+, Focus: Geometry & Trig., Focus: Pure Maths, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from DD007]: What's an annulus and how do you calculate the area?
Response: This diagram shows a shaded-in picture of an annulus, marked with an internal radius of 3 and an external radius 5.
NOTE: The two circles which make up the annulus are concentric, which means that they share the same centre.
Posted at 11:43 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from KL007]: I've been told that if you add up all the exterior angles of a regular polygon they add up to 360 degrees. How does this help you find the interior angles of a decagon?
Response: Calculating the exterior angles first is a very useful technique to quickly calculate the interior angles of any regular polygon.
Posted at 17:27 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from PL]: I've been given the lengths of the sides of a right-angled triangle as three algebra terms, 3x, x+1 & 2x+5 (longest). What do I do next?
Response: This is a disguised question which requires the application of the Theorem of Pythagoras - the clue is in the fact that it is a right-angled triangle and only side lengths are mentioned.
Posted at 18:31 in Age Range: 11-16, Focus: Shape & Space, Phase: GCSE Foundation, Phase: GCSE Higher | Permalink | Comments (0) | TrackBack (0)
Query [from SCJ]: I know the lengths of the sides of an isosceles triangle and my teacher has shown me how to work out the area of the triangle by splitting it in two and using trigonometry. I wondered how to work out the area for a triangle that isn’t isosceles?
Response: You have to use a formula named after the Greek philosopher Heron (usually referred to as Hero's Formula, in Maths text books - some texts even attribute the formula to Archimedes - so who knows?)
This formula is also very useful for modelling calculations involving maximum area for a fixed perimeter of a triangle whilst using a spreadsheet.
Posted at 18:45 in Age Range: 11-16, Focus: Algebra & Graphs, Focus: Shape & Space | Permalink | Comments (0) | TrackBack (0)
Posted at 16:32 in Age Range: 11-16, Focus: Algebra & Graphs, Focus: Shape & Space, Phase: GCSE Foundation, Phase: GCSE Higher | Permalink | Comments (0)