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9.3 Area of a Triangle
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The area of a triangle equals one-half the product of the length of a side and the length of the altitude to that side. (A proof of this theorem is given in 16.)
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9.4 Area of a triangle Find the area of the triangle in Fig. 9-9.
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9.5 Formulas for the area of an equilateral triangle Derive the formula for the area of an equilateral triangle (a) whose side has length s; (b) whose altitude has length h.
CHAPTER 9 Areas
Fig. 9-10
Solutions
See Fig. 9-10. (a) Here A Then A (b) Here A Then A
1 2 bh, 1 2 bh 1 2 bh, 1 2 bh
where b
s and h2
s2 (1s)2 or h 2
1 2 s 23.
1 1 2 s(2 s 23)
1 2 4 s 23. 1 2 s 23
where b
1 2 sh
s and h
1 2h . 2 ( 23 ) h
or s
2h 23
9.6 Area of an equilateral triangle In Fig. 9-11, find the area of (a) an equilateral triangle whose perimeter is 24; (b) a rhombus in which the shorter diagonal has length 12 and an angle measures 60 ; (c) a regular hexagon with a side of length 6.
Fig. 9-11
Solutions
(a) Since p 3s 24, s 8. Then A
1 2 4 s 23 1 4 (64) 23
16 23.
(b) Since m/A 60 , ^ADB is equilateral and s d 12. The area of the rhombus is twice the area of 2 A 1 B (144) 23 72 23. ^ABD. Hence A 2 A 1s2 23 B 4 4 (c) A side s of the inscribed hexagon subtends a central angle of measure 1(360 ) 6 60 . Then, since OA OB radius R of the circumscribed circle, m/OAB m/OBA 60 . Thus ^ AOB is equilateral. 6 A 1 B (36 23) 54 23. Area of hexagon 6(area of ^AOB) 6 A 1s2 23 B 4 4
9.4 Area of a Trapezoid
The area of a trapezoid equals one-half the product of the length of its altitude and the sum of the lengths of its bases. (A proof of this theorem is given in 16.) Thus if h 20, b 27, and b 23 in Fig. 9-12, then A 1(20)(27 23) 500. 2
CHAPTER 9 Areas
The area of a trapezoid equals the product of the lengths of its altitude and median. Since 1 br) and m 1 (b b ), A hm. 2 h(b 2
Fig. 9-12
SOLVED PROBLEMS
9.7 Area of a trapezoid (a) Find the area of a trapezoid if the bases have lengths 7.3 and 2.7, and the altitude has length 3.8. (b) Find the area of an isosceles trapezoid if the bases have lengths 22 and 10, and the legs have length 10. (c) Find the bases of an isosceles trapezoid if the area is 52 23, the altitude has length 4 23, and each leg has length 8.
Solutions
See Fig. 9-13.
Fig. 9-13
(a) Here b (b) Here b
7.3, b 22, b
2.7, h 10, AB 10
3.8. Then A 10. Also EF 64 so h 264 48
1 2h
1 2 h(b
(3.8
2.7) 10)
1 2 (8)(22
19. 6. 10) 128. FD) b 8. Then 17. Then
10 and AE b) AE
1 2 (22
In ^BEA, h (c) AE A b
1 2 h(b
8. Then A
2(AB)2 b) 17 8 b 8
h2 9.
4. Also FD
4, and b
b (AE
1 2 h(2b
8) or 52 23
1 2 (4 23)(2b
8), from which 26
2b 8 or b
9.5 Area of a Rhombus
The area of a rhombus equals one-half the product of the lengths of its diagonals. Since each diagonal is the perpendicular bisector of the other, the area of triangle I in Fig. 9-14 is 1(1d)(1d ) 2 2 2 4(1dd ) or 1dd 8 2
1 8 dd
. Thus the rhombus, which consists of four triangles congruent to ^I, has an area of
CHAPTER 9 Areas
Fig. 9-14
SOLVED PROBLEMS
Area of a rhombus (a) Find the area of a rhombus if one diagonal has length 30 and a side has length 17. (b) Find the length of a diagonal of a rhombus if the other diagonal has length 8 and the area of the rhombus is 52.
Solutions
See Fig. 9-15.
Fig. 9-15
(a) In right ^AEB, s2 (1d)2 (1d )2 or 172 2 2 1 240. A 1dd 2 2 (16)(30) (b) We have d 8 and A 52. Then A
1 2 dd
(1d)2 2 or 52
152. Then 1d 2
1 2 (d)(8)
8 and d 13.
16. Now
and d
9.6 Polygons of the Same Size or Shape
Figure 9-16 shows what we mean when we say that two polygons are of equal area, or are similar, or are congruent.
Fig. 9-16
CHAPTER 9 Areas
Parallelograms have equal areas if they have congruent bases and congruent altitudes.
PRINCIPLE 1:
Thus, the two parallelograms shown in Fig. 9-17 are equal.
Fig. 9-17
PRINCIPLE 2:
Fig. 9-18
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