XV. RECTILINEAL FIGURE. 165. To find the area of any rectilineal figure. RULE. Divide the figure into convenient parts, find the area of every part, and the sum will be the area of the figure. In general the parts into which the rectilineal figure can be most conveniently divided will all be triangles; but in some cases we may have a square, a parallelogram, or a trapezoid, as one of the parts. The area of the triangle ABC= × 10′4 × 4·8=24'96; the area of the triangle ACD=× 10·4 × 6·5=33·8; the area of the triangle AED=}× 8·7 ×3·2=13·92. Thus the area of the rectilineal figure in square feet =24.96+33.8+13.92=72.68. (2) ABCDEF is a six-sided figure: BK, CL, EM, FN are perpendiculars on AD. The following lengths are in feet: BK=3, CL=4, EM=47, FN=5·1; also AK=34, KL=3·2, LD=4·1, AN=3′3, NM=5·3. It follows from these lengths that AD=10'7, and that AM=86; hence MD=107—8·6=2·1. The area of the triangle AKB x34x3 5'1, the area of the trapezoid BKLC=} × 7×3·2=11·2, =×41×4=8'2, the area of the trapezoid FNME=} × 9·8 × 5·3 = 25′97, the area of the triangle EMD =×2·1×4·7=4'935. Thus the area of the rectilineal figure in square feet =51+11·2+8′2+8·415+25′97+4·935 = 63.82. 167. We will now solve some exercises. (1) The side of a regular hexagon is one foot: find the area. By the aid of the figure in Art. 99, we see that a regular hexagon can be divided into six equilateral triangles; this can be done by drawing straight lines from 0 to A, B, C, D, E, and F. Now, by Art. 154, the area of each equilateral triangle in square feet is of the square root of 3; therefore the area of the hexagon in square feet is of the square root of 3, that is, of the square root of 3. (2) A regular polygon of twelve sides is inscribed in a circle of which the radius is one foot: find the area of the polygon. In Art. 99 it is shewn that AM is the side of a regular polygon of twelve sides inscribed in the circle; so that the area of the polygon is twelve times the area of the triangle OAM. The area of the triangle OAM=x OM × AL; now OM=1 foot, AL of AF a foot. Thus the area of the triangle OAM=‡ of a square foot. Therefore the area of the polygon 12 square feet=3 square feet. EXAMPLES. XV. 1. ABCDE is a five-sided figure; the following lengths are in feet: AC=16, AD=12, the perpendiculars from B and D on AC are 84 and 46 respectively, and the perpendicular from E on AD is 5 feet: find the area. 2. ABCDE is a five-sided figure; BK, CL, EM are perpendiculars on AD; the following lengths are in feet: AD=153, BK = 7·6, CL = 5′5, EM = 4·3, AK = 2·7, DL 39: find the area. 3. ABCDEF is a six-sided figure; BK, CL, EM, FN are perpendiculars on AD; the following lengths are in feet: AD=18'4, BK=5, CL=7, EM=6, FN=4, AK=47, AN=4·1, DL=5·3, DM=49: find the area. 4. ABCDEF is a figure having six equal sides; AB=578 feet, BF=644 feet, and the portion BCEF forms a rectangle: find the area. 5. ABCDE is a five-sided figure, having the angle at E a right angle; the following lengths are in chains: AB=14, BC=7, CD=10, DE=12, EA=5, AC=17. Find the area. 6. Find the area of a regular hexagon each side of which is 20 feet. 7. Find the area of a regular hexagon which is inscribed in a circle, the diameter of which is 100 feet. 8. The length of the side of a field, which is in the form of a regular hexagon, is 10 chains: find the area. 9. The radius of a circle is one foot: find the area of a regular polygon of eight sides inscribed in the circle. 10. Find the area of a regular polygon of 24 sides inscribed in a circle, the radius of which is one foot. if greater accuracy is required, multiply the square of the radius by 3'1416. 7 7 the area of the circle is about 784 square feet. (2) The radius of a circle is 3 miles. The square of 3 is 9; and 9 × 31416=28.2744. Thus the area of the circle is about 28-2744 square miles. 170. Both the rules in Art. 168 make the area of the circle a little greater than it ought to be; but the second rule is sufficiently accurate for most practical purposes. If a more accurate result is required we must take as many decimal places of the number 3.1415926... as may be ne cessary. 171. The area of a circle being given, to find the radius. 22 RULE. Divide the area by and extract the square 7 root of the quotient; or, if greater accuracy is required, divide the area by 3·1416, and extract the square root of the quotient. 172. Examples: (1) The area of a circle is 100 square feet. 100 ÷ 22 7 700 7 = 100 × = of this is 5'64... =318181...; the square root Thus the radius of the circle is 5'64 feet. (2) The area of a circle is an acre. An acre is 4840 square yards; dividing 4840 by 3.1416 we have the quotient 1540 61...; the square root of this is 39 25... Thus the radius of the circle is about 39-25 yards. 173. To find the area of a circular ring, that is of the space between the circumferences of two concentric circles. RULE. Find the area of each circle, and subtract the area of the inner circle from the area of the outer circle. Or, Multiply the sum of the radii by their difference 22 7 and the product by or if greater accuracy is required by 3.1416. 174. Examples: (1) The radii of the two circles are 10 feet and 12 feet respectively. The area of the inner circle in square feet = 10 × 10 × 3·1416=314·16; the area of the outer circle in square feet = 12 × 12 × 3·1416=452.3904, 452 3904-314 16=138*2304. Thus the area of the ring is 138.2304 square feet. Or thus, 12+10=22, 12-10=2, 22 × 2 × 3·1416=138°2304. |