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Solution. With extra help the other side of the wall will be painted more quickly than the front side. As one quantity (the number of painters) increases, the other quantity (time to paint the wall) decreases, and so this is an example of inverse proportion. One method of solving inverse proportion problems is the “unity method” as set out below:
♦ Five people paint the wall in 7 hours.
♦ One person paints the wall in 35 hours (Number of people x number of hours = 5 x 7).
♦ Eight people paint the wall in 4| hours (number of hours -h number of people = 35 -j- 8).
The eight friends paint the wall in 4| hours, which is 4 hours, 22 minutes, and 30 seconds.
Take care with some problems in which two quantities may appear to be in direct proportion, but on a closer inspection are not. For example, Luke weighs 20 kg at age 6 years. How much will he weigh at age 9 years? The two quantities weight and age are not in direct proportion, because people do not grow uniformly. This problem cannot be solved from the information given.
References: Cross Multiply, Enlargement, Hyperbola, Rate, Ratio, Similar Figures.
A protractor is a flat, transparent plastic semicircle with two scales on it that are used for measuring angles in degrees (see figure a). Each scale measures from 0° to 180°, and each marking on the scale is 1°. In figure a, the 10° markings are shown but not the 1° markings. The protractor will measure angles to an accuracy of 1°. The diameter of the semicircle is called the straight edge and the center of the semicircle is called the center. A protractor may also be fully circular.
Each angle has two arms, or rays, which intersect in a point, or vertex. In figure b the vertex of the angle is O and the arms are OA and OB. When we are using the protractor to measure the angle we place the center of the protractor on the vertex O of the angle and the straight edge on one of the arms of the angle as shown in the figure. We place the straight edge of the protractor on the arm OA (or we could use the arm OB).
Starting at the arm of the angle where the straight edge is placed, we examine the two scales on the protractor and notice that the inside scale begins at zero and the outside scale starts at 180°. Using the inside scale that begins at zero, we measure around the scale until we come to the other arm OB. The arm OB crosses the inside scale at 123°. The size of the angle AOB is 123°.
Always use the scale that begins at zero on the arm of the angle on which you place the straight edge. If the arms of the angle are not long enough to reach the scale on the protractor, they will have to be lengthened before using the protractor. It is a good idea to have an approximate estimate of the angle before you measure it in case you use the wrong scale.
Example. Use the protractor to measure the reflex angle shown in figure c.
Solution. Since a protractor can only measure up to 180°, you cannot measure a reflex angle directly unless you have a circular protractor. The method is to measure the acute angle and then subtract your answer from 360° to obtain the desired angle. You will recall there are 360° in a full turn. We place the protractor on top of the angle so that the straight edge lies on the arm OB of the angle, as shown. The scale on the protractor that begins at zero on the arm OB is the outside scale. The acute angle measured is 62°. We now subtract this angle from 360°: 360° — 62° = 298°. The required angle to be measured is 298°.
We use a similar method for drawing angles, and it is always necessary to draw one arm of the angle first and measure from that arm the angle you require.
References: Acute Angle, Degree, Ray, Reflex Angle, Vertex.
This is a solid figure (a polyhedron) that has a base (which can be any polygon) and edges drawn from each vertex of the base that meet at a single point V called the apex of the pyramid (see figure a). The base of a pyramid can be any polygon, but its other faces are always triangles. We will study the right pyramids, and in addition those that have regular polygons for their bases. The shape of its base categorizes a pyramid. The pyramid shown in figure a is a square-based pyramid.
Here V is the vertex, and point X is the center of the square base, which is shaded. VX is the altitude and the pyramid has four sloping triangular faces that meet at V. The triangles are either isosceles or equilateral. The volume of a pyramid is given by
Volume = | x area of base x altitude
The surface area of a pyramid is found using the descriptive formula
Surface area = area of base + sum of areas of triangular faces
The net of the square-based pyramid is drawn on the right-hand side of figure a. The four triangles are congruent, and may be either isosceles or equilateral, depending on length of the altitude.
Various kinds of pyramids are drawn in figure b.