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chapter2.1r
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à 2.1ïArea of a Triangle
ä Please find the area of the following triangles.
â
êêêêïAreaè=è1/2∙b∙c∙sin A
êêêêêï=è(1/2)∙(18)∙(12)∙sin 47°
êêêêêï≈è78.99 square cm
@fig2101.bmp,45,90
éSêêëThe area of a triangle is known to be one
@fig2102.bmp,15,118
êêêïhalf of the base times the altitude of the tri-
êêêïangle.êè Aè=è1/2∙b∙h
êêêïIn the figure, using triangle ABD, the altitude
êêêïis seen to be h = c∙sin A. Thus, the formula for
êêêïarea is:êïAreaè=è1/2∙b∙c∙sin A
êêêëWe could choose a different altitude and it's
êêêïbase to find the formula for area. The other two
êêêïaltitudes generate the following formulas.
êêêëAreaè=è1/2∙a∙c∙sin B
êêêëAreaè=è1/2∙a∙b∙sin C
êêêëThus, there are three formulas for the area
êêêïof a triangle.ïThe pattern that is evident
êêêïin ç three formulas allows us to make a
êêêïgeneral statement about area.ïThe area of
êêêïany triangle equals one half the
product of two sides times the sine of the included angle.ïIf two sides
and the included angle are not given, it may be necessary to find addi-
tional parts before finding the area.
è It is possible to develop a formula for area when you know two an-
gles and one side.ïSuppose you know angle A, angle B, and side b.ïFrom
the Law of Sines, a/sin A = b/sin B, so a = (b∙sin A)/(sin B).ïAlso,
sin C = sin (180° - (A + B)) = sin (A + B).ïThese two substitutions
generate the following formula.
Areaï=ï1/2∙a∙b∙sin Cï=ï1/2∙(b∙sin A)/(sin B)∙b∙sin(A + B)ï=
#êêêëbì∙sin A∙sin(A + B)
#êêê =è───────────────────
êêêêè 2∙sin B
èAnother formula can be derived for the case when you know three sides.
#êêêêêêê ┌─────────
#êê Areaï=ï1/2∙b∙c∙sin Aï=ï1/2∙b∙c∙á1 - cosìA
In the above formula, if you replace "cos A" using the Law of Cosines
and simplify the result, the following form involving only the lengths
of the sides can be derived.
#êêë┌───────────────────────────────────────────────
#ë Areaï=ï1/4∙á(a + b + c)∙(a + b - c)∙(a + c - b)∙(b + c - a)
è Since it is possible to break up a polygon into adjoining triangles,
you can find the area of an irregular polygon by adding up the area of
the triangle components.ïFor example, the area of a piece of land with
boundry composed of line segments can be found by breaking up a drawing
of the land into triangles and adding the area of the triangles.
1êë Find the area of triangle ABC if angle A is 35°,
êêê side b is 26 m, and side c is 12 m.
#êêêèA)ï93.68 mìêêïB)ï87.61 mì
#êêêèC)ï89.48 mìêêïD)ïå of ç
@fig2102.bmp,25,118
ü
#ëAreaï=ï1/2∙b∙c∙sin Aï=ï1/2∙26∙12∙sin 35°ï≈ï89.48 mì
Ç C
2êë Find the area of triangle ABC if angle C is 107°,
êêê side a is 22 m, and side b is 28 m.
#êêêèA)ï294.54 mìêê B)ï186.4 mì
#êêêèC)ï197.38 mìêê D)ïå of ç
@fig2102.bmp,25,118
ü
#ëAreaï=ï1/2∙a∙b∙sin Cï=ï1/2∙22∙28∙sin 107°ï≈ï294.54 mì
Ç A
3êë Find the area of triangle ABC if angle B is 73.5°,
êêê side a is 1.45 in, and side c is 2.83 in.
#êêêèA)ï2.01 inìêêïB)ï1.97 inì
#êêêèC)ï2.56 inìêêïD)ïå of ç
@fig2102.bmp,25,118
ü
#èAreaï=ï1/2∙a∙c∙sin Bï=ï1/2∙(1.45)∙(2.83)∙sin 73.5°ï≈ï1.97 inì
Ç B
4êë Find the area of triangle ABC if angle A is 27.3°,
êêê angle B is 42.1°, and side b is 38 yd.
#êêêèA)ï328.2 ydìêê B)ï462.35 ydì
#êêêèC)ï298.46 ydìêêD)ïå of ç
@fig2102.bmp,25,118
ü
#êëbì∙sin A∙sin(A + B)ë(38)ì∙sin 27.3°∙sin 69.4°
#èAreaï=ï───────────────────ï=ï─────────────────────────ï=
êêè 2∙sin Bêêë2∙sin 42.1°
#êêêê =è462.35 ydì
Ç B
5êë Find the area of triangle ABC if angle A is 65°,
êêê angle C is 35°, and side "a" is 30 in.
#êêêèA)ï280.47 inìêêB)ï291.68 inì
#êêêèC)ï220.3 inìêê D)ïå of ç
@fig2102.bmp,25,118
ü
#êëaì∙sin C∙sin(A + C)ë(30)ì∙sin 35°∙sin 100°
#èAreaï=ï───────────────────ï=ï─────────────────────────ï=
êêè 2∙sin Aêêë2∙sin 65°
#êêêê =è280.47 inì
Ç A
6êë Find the area of triangle ABC if side "a" is 14,
êêê side b is 18, and side c is 6.
êêêèA)ï51.62êêëB)ï42.38
êêêèC)ï35.14êêëD)ïå of ç
@fig2102.bmp,25,118
ü
#êêï┌────────────────────────────────────────────────
#èAreaï=ï1/4∙á(a + b + c)∙(a + b - c)∙(a + c - b)∙(b + c - a )
#êêêê┌──────────────────
#êêê=ï1/4∙á(38)∙(26)∙(2)∙(10)
êêêê =è35.14
Ç C