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46913 Assignment
7.34
Area = ½ bh =96
In radius = Area/semiperimeter = 96/(12+16+20)/2=4
Circumradius =
7.35
KT = sqrt 53
7.36
a) Right triangle
b)
c) Area of XYZ = sq units.
d) XE/EM = 2/3 (centroid is E)
e) In radius = 1/3 of median = 1/3 (XM0 = units
f) Circumradius = 2/3 of median = units
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Angle CHD = 90-angle HCD (right angled triangle)
= 90-(FCB) = 90-(90-ABC) from right triangle CBF
= Angle ABC
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If altitudes are of same length then we have
Consider triangles ABE and ACF
BE=CF (given)
Angle F = Angle B (right angle)
Angle A = Angle A
Congruent triangles hence AB = AC
Similarly we prove that AC=BC
Hence equilateral triangle
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Use angle bisector theorem that if XD is angle bisector then
XY/XZ = YD/YZ
XY/8 = ¾ or xy = 6
7.40
Given that triangle ABC has area 48 sq units.
Triangle ADC will have same height but half base
So ADC area = 24
Triangle AGC will have same base but 1/3 height as that of altitude on AC
Hence area of AGC=1/3 (48) = 16
Triangle DEF will have all sides half of ABC
So area = ¼ (48) = 12
Triangle AEF also will have all sides half of ABC
So area of AEF = ¼ (48) =12
7.41
In triangles TUY and VUY,
AngleTUY = VUY (given since angle bisector)
UT = VT (given)
UY = UY (reflexive)
So congruent and hence TY = YV
a) Y is mid point of TV
b) By CPCT property angles TYU = VYU or UY is perpendicular to TV
c) TY = ½ TV = 18: UY =
d) In triangle TUM use angle bisector theorem....