Question 1 :
If the measure of angle A of triangle ABC is 3x, the measure of angle B is 5x and the measure of angle C is 4x, what is the value of x?
a) 12 b) 15 c) 20 d) 30 e) 45
Solution :
In a triangle, sum of interior of a triangle is 180.
<A + <B + <C = 180
3x+4x+5x = 180
12x = 180
x = 180/12
x = 15
Question 2 :
Which of the following number sentences is NOT true?
a) angle 1 + angle 2 = angle 5
b) angle 4 + angle 1 = angle 3 + angle 6
c) angle 3 + angle 2 = angle 4
d) angle 1 + angle 2 + angle 3 = 180
e) angle 4 + angle 5 + angle 6 = 360
Solution :
Option (a) is incorrect.
Problem 3 :
If the circle inscribed in square ABCD has a radius of r, what is the size of the shaded area in terms of r?
a) r^{2} – πr^{2} b) 2r – πr^{2} c) 2r – π r^{2}/4
d) r^{2} – (πr^{2}/4) e) (r^{2} – πr^{2})/4
Solution :
Required area
= (Area of half of the square - area of semicircle)/2
= [a^{2}/2 - (πr^{2}/2)]/2
Here a = 2r
= [(2r)^{2}/2 - (πr^{2}/2)]/2
= [2r^{2} - (πr^{2}/2)]/2
= r^{2} - (πr^{2}/4)
Problem 4 :
The figure is an equilateral inscribed in a circle with radius 10.
What is the measure of ∠AOC?
Solution :
<AOB = <AOC = <COB
<AOC = (1/3) ⋅ 360
<AOC = 120
Problem 5 :
What is the area of the isosceles trapezoid above?
A) 238 B) 252 C) 276 D) 308
Solution :
Since BE is perpendicular to AD, triangle ABE is a right triangle.
AB^{2} = AE^{2} + EB^{2}
13^{2} = AE^{2} + 12^{2}
169- 144 = AE^{2}
AE = √25
AE = 5
FD = 5
AD = 5+18+5
AD = 28
Area of trapezoid = (1/2)⋅h (a+b)
= (1/2)⋅12 (18+28)
= 6(18+28)
= 276
Problem 6 :
In the figure above, EO is the midsegment of trapezoid TRAP and RP intersect EO at point Z. If 15 = RA and 18 =EO, what is the length of EZ
Solution :
The length of the midsegment of a trapezoid is average of lengths of bases.
EO = (1/2)(RA + TP)
18 = (1/2)(15 + TP)
36 = 15 + TP
TP = 36 - 15
TP = 21
EZ = (1/2)(21)
EZ = 10.5
Problem 7 :
In the figure above, ABCD is a rectangle and BCFE is a square. If AB = 40, BC = 16 and <AGD = 45, what is the area of the shaded region?
Solution :
Triangle ADG is special 45-45-90 right triangle.
FC = 16
DF = DC - FC
DF = 40-16
DF = 24 = AE
BC = AD = 16 = DG
GF = DF - DG
GF = 24 - 16
GF = 8
Area of the shaded region is in the form of trapezoid.
Area of trapezoid = (1/2)⋅h (a+b)
= (1/2) 16 (24+8)
= 8 (32)
= 256
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