Area of right-angled triangle. By drawing a diagonal of a rectangle we divide the rectangle into two equal right-angled Hence the area of a right-angled triangle may be found by regarding it as half a certain rectangle. triangles. Ex. 889. Find the number of squares contained by a triangle whose corners are (i) (0, 0), (0, 2), (6, 0). (Complete the rectangle.) Ex. 890. Find the areas of right-angled triangles in which the sides containing the right angle are (i) 2′′, 3′′, (ii) 6·5 cm., 4.4 cm., (iii) 4·32′′, 3·71′′, (iv) 112 mm., 45 mm. (in sq. mm. and also in sq. cm.). Area of any rectilinear figure (on squared paper). With the aid of rectangles and right-angled triangles we can find the area of any figure contained by straight lines (i.e. any rectilinear figure). This way is especially convenient when one side of the figure runs along a line of the squared paper. Fig. 163 shows how a 4-sided figure may be divided up into rectangles and right-angled triangles; the number inside each rectangle and triangle indicates the number of squares it contains; and the complete area is 1991 or 199.5 squares. Ex. 891. Measure the size of the small squares in fig. 163; hence find the area of the 4-sided figure in sq. inches. Ex. 892. Find the area (in squares of your paper) of each of the following figures by dividing up the figures into rectangles and right-angled triangles: (vi) (1, 4), (6, 3), (1, −3). (vii) (-4, −3), (−3, 3), (5, 6), (10, −3). (viii) (3, 5), (−3, 2), (−5, −3), (3, −7). (x) (2, 5), (5, 2), (5, -2), (2,-5), (-2, -5), (-5, -2), (5, 2), (-2, 5). Are all the sides of this figure equal? (xi) (3, 4), (4, 3), (4, −3), (3, −4), (−3, −4), (−4, −3), (−4, 3), (−3, 4). (xii) (5, 0), (4, 3), (3, 4), (0, 5), (-3, 4), (-4, 3), (-5, 0), (-4, -3), (−3, −4), (0, −5), (3, −4), (4, −3). Ex. 893. Draw the three following figures on the same axes; find the area and perimeter of each. (This exercise shows that two figures may have the same perimeter and different areas.) Ex. 894. Draw the two following figures on the same axes; find the area and perimeter of each. (i) (0, 0), (7, 0), (9, 5), (2, 5). (ii) (0, 0), (7, 0), (3, 5), (−4, 5). (This exercise shows that two figures may have the same area and different perimeters.) (i) ` (1, 0), (1, 8), (4, 14), (2, 14), (0, 10), (−2, 14), (−4, 14), (−1, 8), (−1, 0). (ii) (5, 7), (−4, 7), (−5, 5), (1, 5), (−5, −7), (5, −7), (6, −5), (−1, −5). If there is no side of the figure which coincides with a line of the paper (ABCD in fig. 164), it is generally convenient to draw lines outside the figure, parallel to the axes, thus making up a rectangle (PQRS); the area required can then be found by subtracting a certain number of right-angled triangles from the rectangle. Ex. 896. Find the areas of the following figures:— (i) (1, 1), (16, 5), (9, 14). (ii) (6,3), (12, 9), (3, 11). (iii) (10, -20), (20, −24), (12, 4). (v) (1, 0), (6, 1), (5, 6), (0, 5). Area of a curvilinear figure. This cannot be found exactly by the method of counting squares: the approximate value however is easily calculated as follows. C B fig. 165. To find the area of the fig. ACBA, notice that the curved boundary ACB cuts through various squares; in counting squares we have to decide what is to be done with these broken squares. The following rule gives a useful approach to the true value :— If the broken square is more than half a complete square, count 1; if less than half a square, count 0. Counting up the squares in ACBA on this system, we find that the area is 72 squares. As each of the above squares is 100 sq. inch, the area is 72 sq. inches. Ex. 897. On inch paper draw a circle of radius 1 inch; find its area as above, and reduce to square inches. (The counting can be shortened in various ways; e.g. by dividing the circle into 4 quarters by radii.) Calculate, Ex. 898. Find the area of circles of radii 2, and 3 inches. to 2 places, how many times each of these circles contains the 1-inch circle of Ex. 897. DEF. Any side of a parallelogram may be taken as the base. The perpendicular distance between the base and the opposite (parallel) side is called the height, or altitude. Thus in fig. 166 if BC be taken as base, MN (which may be drawn from any point of the base) is the height (or altitude). If AB be taken as base, GH is the height. A M B N fig. 166. Ex. 900. In fig. 166 what is the height if CD be taken as base? if AD be taken? Ex. 901. Prove that the altitudes of a rhombus are equal. PA Q D Area of parallelogram. Take a sheet of paper (a rectangle) and call the corners P, B, C, Q; BC being one of the longer sides (fig. 167). Mark a point A on the side PQ. Join BA, and cut (or tear) off the right-angled triangle PBA. You now have two pieces of paper; you will find that you can fit them together to make a parallelogram (ABCD in fig. 167). fig. 167. Notice (i) that the rectangle you had at first and the parallelogram you have now made, are composed of the same paper, and therefore have the same area. (ii) that the rectangle and the parallelogram are on the same base BC, and both lie between the same pair of parallel lines BC and PAQD. Or, we may say that they have the same height. Ex. 902. Make a paper parallelogram with sides of 6 and 4 ins. and an angle of 60°. Cut the parallelogram into two pieces which you can fit together to make up a rectangle. Find its area. Ex. 903. Repeat Ex. 902 with sides of 12 and 6 cm. and angle of 60°. |