\fdO'^ - Old Forge Coal Mines

GEOMETRY AND TRIGONOMETRY. 163
(351) Area of parallelogram equals
7 X 10 I (129 inches = 10?- ft. = ) 75^ sq. ft. Ans.
'
4 4
(352) (a) See Art. 778.
Area of the trapezoid = I^tV + ^Ifi x 7^ = 143.75 sq. ft.
4
Ans.
{d) In the equilateral triangle ABC, Fig. 18, the area,
143.75 square feet, is given to find a side.
Since the triangle is equilateral all the
angles are equal to — of 180° or 60°. In
o
the triangle AB D = A DC, we have A
D = ABxsin 60°. The area of any tri- B^
angle is equal to one-half the product
F'g- is-
of the base by the altitude, therefore, ^r = 143.75.
BC=AB and ^ Z> = ^ -5 x sin 60° ; then, the above becomes
ABxAB sin 60' = 143.75,
2
^^X. 86603
or ,,__
=143.75,
z
or AB' = ^l}*^f.
.86603
_ V287.50
Therefore, AB = y "
;"" = 18 ft.
.obOUo
2.64 in. Ans.
(353) (a) Side of square having an equivalent area =
/143775 = 11.99 feet. Ans.
/'^ = /l80277 = 13lfeet. Ans.
(^) Diameter of circle having an equivalent
{c)
area =
Perimeter of square = 4 X 11.99 = 47.96 ft.
Circumference or perimeter of circle = IS^- x 3.1416 = 42.41 ft.
5 feet 6.6 inches. Ans.
Difference of perimeter = 5.55 ft =