Mensuration of Surfaces.

5. What would be the width if the rafters on each

side were 14 feet?

Ans. 19,7989 feet.

PROBLEM V.

12. When the hypothenuse and one side of a right angled triangle are known to find the other side.

RULE.

Square the hypothenuse and also the other given side, and take their difference: extract the square root of this difference, and the result will be the required side.

EXAMPLES.

1. In the right angled triangle ABC, there are given

AC 50 ft., and AB=40 ft.

required the side BC.

We first square the hypothenuse and the other side, after which we take the dif

B

Operation.

502-2500

402=1600

ference, and then extract the

square root, which gives

Diff.-900

BC= √900 30 feet.

=

2. The height of a precipice on the brink of a river

QUEST.-12. How do you find one side of a right angled triangle, when the hypothenuse and the other are known?

Mensuration of Surfaces.

is 103 feet, and a line of 320 feet in length will just reach from the top of it to the opposite bank: required the breadth of the river.

Ans. 302,9703 feet.

3. The hypothenuse of a triangle is 53 yards, and the perpendicular 45 yards: what is the base?

Ans. 28 yards.

4. A ladder 60 feet in length, will reach to a window 40 feet from the ground on one side of the street, and by turning it over to the other side, it will reach a window 50 feet from the ground: required the breadth of the

street.

Ans. 77,8875 feet.

PROBLEM VI.

13. To find the area of a trapezoid.

RULE.

Multiply the sum of the parallel sides by the perpendicular distance between them, and then divide the product by two:—the quotient will be the area.

AB=321,51 ft., DC=214,24 ft., and CE=171,16 ft.

QUEST.-13. How do you find the area of a trapezoid;

2. What is the area of a trapezoid, the parallel sides of which, are 12,41 and 8,22 chains, and the perpendicular distance between them 5,15 chains?

Ans. 5 A. 1 R. 9,956 P..

3. Required the area of a trapezoid whose parallel sides are 25 feet 6 inches, and 18 feet 9 inches, and the perpendicular distance between them 10 feet and 5 inches?

Ans. 230 sq. ft. 5' 7".

4. Required the area of a trapezoid whose parallel sides are 20,5 and 12,25, and the perpendicular distance between them 10,75 yards.

Ans. 176,03125 sq. yds.

5. What is the area of a trapezoid whose parallel sides are 7,50 chains, and 12,25 chains, and the perpendicular height 15,40 chains?

Ans. 15 A. 0 R. 33,2 P.

Mensuration of Surfaces.

6. What is the content when the parallel sides are 20 and 32 chains, and the perpendicular distance between them 26 chains?

Ans. 67 A. 2 R. 16 P.

PROBLEM VII.

14. To find the area of a quadrilateral.

RULE.

Measure the four sides of the quadrilateral, and also one of the diagonals: the quadrilateral will thus be divided into two triangles, in both of which all the sides will be known. Then, find the areas of the triangles separately, and their sum will be the area of the quadrilateral.

QUEST.-14. How do you find the area of a quadrilateral ?

Mensuration of Surfaces.

REMARK. Instead of measuring the four sides of the quadrilateral, we may let fall the perpendiculars Bb, Dg, on the diagonal AC. The area of the triangles may then be determined by measuring these perpendiculars and the diagonal AC. The perpendiculars are Dg= 18,95 ch, and Bb=17,92 ch.

2. Required the area of a quadrilateral whose diagonal is 80,5 and two perpendiculars 24,5 and 30,1 feet ? Ans. 2197,65 sq. ft.

3. What is the area of a quadrilateral whose diagonal is 108 feet 6 inches, and the perpendiculars 56 feet 3 inches, and 60 feet 9 inches?

Ans. 6347 sq. ft. 3'.

4. How many square yards of paving in a quadrilateral whose diagonal is 65 feet, and the two perpendiculars 28 and 331 feet?

Ans. 222 sq. yds.

5. Required the area of a quadrilateral whose diagonal is 42 feet, and the two perpendiculars 18 and 16 feet. Ans. 714 sq. ft.

6. What is the area of a quadrilateral in which the diagonal is 320,75 chains, and the two perpendiculars 69,73 chains, and 130,27 chains?

Ans. 3207 A. 2 R.