All SAT II Math II Resources
Example Questions
Example Question #41 : 2 Dimensional Geometry
Note: figure NOT drawn to scale.
Refer to the triangle in the above diagram.
Evaluate
. Round to the nearest tenth, if applicable.
By the Law of Cosines,
Substitute
:
Example Question #2 : Finding Sides
In triangle
, and .Which of the following statements is true about the lengths of the sides of
?
In a triangle, the shortest side is opposite the angle of least measure; the longest side is opposite the angle of greatest measure. Therefore, if we order the angles, we can order their opposite sides similarly.
Since the measures of the three interior angles of a triangle must total
,
Since
,
we can order the lengths of their opposite sides the same way:
.
Example Question #1 : Finding Sides
Note: figure NOT drawn to scale.
Refer to the above diagram.
.
Which of the following expressions is equal to
?
By the Law of Sines,
.
Substitute
, , and :
Solve for
:
Example Question #41 : Geometry
Which of the following describes a triangle with sides of length 9 feet, 3 yards, and 90 inches?
The triangle is acute and equilateral.
The triangle is acute and scalene.
The triangle is obtuse and isosceles, but not equilateral.
The triangle is obtuse and scalene.
The triangle is acute and isosceles, but not equilateral.
The triangle is acute and isosceles, but not equilateral.
One yard is equal to three feet; One foot is equal to twelve inches. Therefore, 9 feet is equal to
inches, and 3 yards is equal to inches. The triangle has sides of measure 90, 108, 108.We compare the squares of the sides.
The sum of the squares of the two smaller sidelengths exceeds that of the third, so the triangle is acute.
The correct response is acute and isosceles.
Example Question #2 : Finding Sides
Note: figure NOT drawn to scale.
Refer to the above diagram.
.
Which of the following expressions is equal to
?
By the Law of Sines,
.
Substitute
, , and :
We can solve for
:
Example Question #6 : Finding Sides
Note: figure NOT drawn to scale.
Refer to the triangle in the above diagram.
.
Evaluate
.
By the Law of Sines,
Substitute
and solve for :
Example Question #2 : Finding Sides
The above figure is a regular decagon. Evaluate
to the nearest tenth.
Two sides of the triangle formed measure 6 each; the included angle is one angle of the regular decagon, which measures
.
Since we know two sides and the included angle of the triangle in the diagram, we can apply the Law of Cosines,
with
and :
Example Question #352 : Sat Subject Test In Math Ii
Regular Pentagon
has perimeter 35. has as its midpoint; segment is drawn. To the nearest tenth, give the length of .
The perimeter of the regular pentagon is 35, so each side measures one fifth of this, or 7. Also, since
is the midpoint of , .Also, each interior angle of a regular pentagon measures
.Below is the pentagon in question, with
indicated and constructed; all relevant measures are marked.A triangle
is formed with , , and included angle measure . The length of the remaining side can be calculated using the law of cosines:
where
and are the lengths of two sides, is the measure of their included angle, and is the length of the third side.Setting
, and , substitute and evaluate :
;
Taking the square root of both sides:
,
the correct choice.
Example Question #51 : Geometry
Regular Hexagon
has perimeter 360. and have and as midpoints, respectively; segment is drawn. To the nearest tenth, give the length of .
The perimeter of the regular hexagon is 360, so each side measures one sixth of this, or 60. Since
is the midpoint of , .Similarly,
.Also, each interior angle of a regular hexagon measures
.Below is the hexagon with the midpoints
and , and with constructed. Note that perpendiculars have been drawn to from and , with feet at points and respectively.is a rectangle, so .
.
This makes
and the short leg and hypotenuse of a 30-60-90 triangle; as a consequence,.
For the same reason,
Adding the segment lengths:
.
Example Question #4 : Finding Sides
Regular Pentagon
has perimeter 60.To the nearest tenth, give the length of diagonal
.
The perimeter of the regular pentagon is 60, so each side measures one fifth of this, or 12. Also, each interior angle of a regular pentagon measures
.The pentagon, along with diagonal
, is shown below:
A triangle
is formed with , and included angle measure . The length of the remaining side can be calculated using the Law of Cosines:
where
and are the lengths of two sides, the measure of their included angle, and the length of the side opposite that angle.Setting
, and , substitute and evaluate :
Taking the square root of both sides:
,
the correct choice.
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