12. Find the present value of a perpetual annuity of $540, beginning ten years hence, rate of interest 6%. $5025.55 present value. = 13. Find the present value of a certain annuity of $650, beginning at this time and continuing 15 years, rate 5%. $6746.78 P. V., C. I. = 14. Find the present value of an annuity of $150, to commence in 10 years and then continue 10 years, rate of interest 6%. Present value of annuity deferred Present value of annuity for 10 yr. $616.48, Pres. value. 15. What is the value of an annuity of $300, unpaid for 12 years, rate 6%? Ans. $5060.98. 16. A man pays $100 a year for wine and tobacco; what will that amount to at the end of 50 years, allowing 6% compound interest? Ans $29033.59. REM.-Strict attention must be given to the formulas of annuities, and the students may repeat the examples given until they become familiar with the various formulas. MENSURATION. DEFINITIONS. 1. Determining the areas of surfaces and the volume of solids is called Mensuration. 2. A Surface has two dimensions, as length and breadth; a Solid has three dimensions, viz., length, breadth and thickness. A level surface is called a plane. 3. A Line has length only. 4. Any position on a surface or a line is a Point. 5. The divergence of two straight lines from a point is an Angle. 6. If the two straight lines are perpendicular to each other, the angle is called a Right Angle. 7. An angle less than a right angle is an Acute Angle. 8. An angle greater than a right angle is an Obtuse Angle. 9. A plane figure having three angles is a Triangle. 10. A plane figure having four angles is a Quadrilateral. 11. If the angles of the quadrilateral are right angles it is a Rectangle, and if the sides of the rectangle are equal it is a Square. 12. If the opposite sides of the quadrilateral are parallel without the angles being right angles, it is a Parallelogram. 13. A plane figure of any number of sides has the general name of Polygon. MEASUREMENT OF SURFACES. PROBLEMS. 1. To find the area of a rectangle, multiply the base by the altitude. 2. To find the area of a parallelogram, multiply the base by the altitude (Geometry). 3. To find the area of a triangle. It is also proved by Geometry that a triangle is one-half a rectangle or parallelogram of the same base and altitude; therefore one-half the product of the base and altitude is the area of a triangle. EXAMPLES. 1. How many square feet in a board 12 ft. long and 1 ft. wide? Ans. 12 sq. ft. 2. How many square feet in a board 12 ft. long and 10 in. wide? 3. How many 9 in. wide? 12 x 10 sq. ft. square feet in a board 12 ft. long and 4. How miny square feet in a room 18 ft. long and 12 ft, wide? 18 × 12 = 216 sq. ft. 5. How many square yards in a room 18 ft. long and 12 ft. wide? 7. How many square feet in a wall 18 ft. 9 in. long and 9 ft. 6 in. high? 25 57 = 25x57 = 1425 = 1781 sq. ft., Ans. 2 8. How many square feet in a wall 16 ft. long and 9 ft. 6 in. high? Ans. 152 sq. ft. 9. How many square feet in the four walls of a room having the dimensions of 7th and 8th examples? Ans. 6601 sq. ft. 10. What number of square feet in the floor or ceiling? 11. How many square feet of plastering, in the walls and ceiling of this room? How many square yards? 1st Ans. 9601 sq. ft. 2d Ans. 106 sq. yd. 12. How many square feet in the roof of a house 50 ft. long and the rafters on each side of the roof 25 ft. long? 50 × 50 = 2500 sq. ft. 13. How many shingles, each shingle covering 6 in. by 4, will it take to roof the house? 6 2500 sq. ft. x 144 24 =15000 shingles. 14. How many square yards of plastering in a house 20 ft. front, 60 ft. deep and 36 ft. high; three stories in height? and how many shingles to roof it, each covering 8 in. by 6, and the entire length of the roof 64 ft., the width 20 ft.? 1st Ans. 1040 sq. yd. plastering. 2d Ans. 3840 shingles. 15. What is the area of a triangle whose base is 24 in. and the perpendicular distance from the vertical angle to the base 9 in. ? 12 22 108 inches. MEASUREMENT OF CIRCLES. DEF-A Circle is a plane figure bounded by a curved line called the circumference, every point of which is equally distant from a point within, called the centre. PROBLEMS. 1. To find the circumference of a circle, having given the diameter. Multiply the diameter by 3.1416 (Geometry). |