PRACTICE QUESTIONS OF GEOMETRY FOR SAT

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) r2 – πr2   b)  2r – πr2   c) 2r – π r2/4

d) r2 – (πr2/4)   e) (r2 – πr2)/4

Solution :

Required area

=  (Area of half of the square - area of semicircle)/2

=  [a2/2 - (πr2/2)]/2

Here a  =  2r

=  [(2r)2/2 - (πr2/2)]/2

=  [2r2 - (πr2/2)]/2

=  r2 - (πr2/4)

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.

AB2  =  AE2 + EB2

132  =  AE2 + 122

169- 144  =  AE2

AE  =  √25

AE  =  5

FD  =  5

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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