Download Chapter 8 Review Answers

Geometry H
Chapter 8
Review Worksheet answers
1. A convex polygon has 17 sides.
(a) Find the sum of its interior angles.
(b) Find the sum of its exterior angles.
(a) Formula for sum of the interior angles :
S I  180(n  2)
S17 = 180(17-2) = 180(15) = 2700
(b) The sum of the exterior angle is always
360.
2. The sum of the interior angles of a
complex polygon is 3,600.
How many sides does it have?
S I  180(n  2)
3,600 = 180(n-2)
20 = n – 2
22 = n
The polygon has 22 sides.
3. A regular polygon has 12 sides.
(a) Find the measure of each interior angle.
(b) Find the measure of each exterior angle.
(a) Formula for int. angle of regular polygon :
180( n  2)
mi =
n
180(12  2) 1800

 150
mi =
12
12
(b) Formula for ext. angle of regular polygon :
360
me =
n
360
 30
me =
12
4. In a regular polygon, the measure of
each interior angle is 120.
How many sides does it have?
Method 1
180( n  2)
mi =
n
180( n  2)
120 =
n
120n = 180(n-2)
120n = 180n – 360
-60n = -360
n = 6
The polygon has 6 sides.
Method 2
If each interior angle has a measure of 120,
each exterior angle measures 60.
360
me =
n
360
60 =
n
60n = 360
n = 6
5. In a regular polygon, the measure of
each exterior angle is 15.
How many sides does it have?
360
n
360
15 =
n
15n = 360
n = 24
The polygon has 24 sides.
me =
The figure below is for questions #6-9.
Quad. ABCD is a parallelogram.
6. AB = 8x – 7 DC = 2x + 11
x = ___
Since opposite sides of a parallelogram are
congruent, AB = DC .
8x – 7 = 2x + 11
6x = 18
x = 3
Geometry H
Chapter 8
Review Worksheet answers
7. mB = 25 – y mD = 3y – 11 y = ___
Since opposite angles of a parallelogram are
congruent, mB = mD.
25 – y = 3y – 11
36 = 4y
9 = y
8. mA = 5z + 10 mD = 12z
z = ___
11. RS  UT and RU || ST
No. Theorem 8-9 requires that the sides that
are parallel also be the sides that are
congruent.
12. URS  STU and RST  TUR
Yes, by Theorem 8-8
Since consecutive angles of a parallelogram
are supplementary, mA + mD = 180.
5z + 10 + 12z = 180
17z + 10 = 180
17z = 170
z = 10
13. RT  SU
No. Theorem 8-10 requires that the
diagonals bisect each other, but not that they
be congruent to each other. A quad with
congruent diagonals might be an isosceles
trapezoid.
9. Draw the diagonals of ABCD. Call the
point of intersection Q.
AQ = w2 QC = 2w + 3 w = ____
Use this figure for questions #14-15
Since the diagonals of a parallelogram bisect
each other, AQ = QC.
w2 = 2w + 3
w2 - 2w – 3 = 0
(w – 3)(w + 1) = 0
w – 3 = 0 or w + 1 = 0
w=3
or w = -1
If w = 3, AQ = 9 and QC = 9
If w = -1, AQ = 1 and QC = 1
So, both 3 and -1 are possibilities for w.
Use this figure for questions #10-13.
With the given information, does quad. RSTU
have to be a parallelogram?
10. RS  UT and RU  ST
Yes, by Theorem 8-7
ABCD is a rectangle.
14. AC = 3x – 20 and BD = x + 30. x = ___
The diagonals of a rectangle are congruent.
So, 3x – 20 = x + 30
2x = 50
x = 25
15. BC = 8 and mBDC = 30. QC = ___
DCB is a right triangle.
BC is the leg opposite BDC.
BD is the hypotenuse.
BC
sinBDC =
BD
8
sin30 =
BD
8
 16
BD =
sin 30
Since BD = 16, AC = 16.
Since Q is the midpoint of AC , QC = 8.
Geometry H
Chapter 8
Review Worksheet answers
Use this figure for questions #16-18.
GHIJ is a rhombus
16. GH = 4y+120 and HI = 6y-20. y = ___
Consecutive sides of a rhombus are
congruent.
So, 4y + 120 = 6y - 20
140 = 2y
70 = y
17. GI = 24 and HJ = 10. GH = ___
The diagonals of a rhombus bisect each other.
So, GK = 12 and KH = 5.
The diagonals of a rhombus are .
Using the Pythagorean Theorem,
(GH)2 = (GK)2 + (KH)2
(GH)2 = 122 + 52
(GH)2 = 169
GH = 13
10
cos 25
10
40
Perimeter = 4∙GH = 4∙
=
cos 25
cos 25
= 44.13511676 = 44.1
Area
HK
In right GKH, tanHGK =
GK
HK
tan 25 =
10
10∙tan 25 = HK
Area of GHI = ½ ∙ GI ∙ HK
= ½ ∙ 20 ∙ 10∙tan 25
= 100∙tan 25
Since GJI  GHI,
Area of GJI = 100∙tan 25
Area GHIJ = Area of GHI + Area of GJI
= 100∙tan 25 + 100∙tan 25
= 200∙tan 25
= 93.26153163 = 93.3
GH =
Use this figure for questions #19-20
WXYZ is a trapezoid with WX || ZY .
18. mHGJ = 50. GI = 20.
Find the perimeter and area of GHIJ.
(Round to the nearest tenth.)
Since mHGJ = 50, mHGK = 25
(Diagonals of rhombi bisects opposite angles.)
Perimeter
19. mZWX = 110. mWZY = ___
Since they are same side interior angle of
parallel lines, the are spupplementary.
So, mWZY = 70.
Since GI = 20, GK = 10.
(Diagonals of rhombi bisects each other.)
GK
In right GKH, cosHGK =
GH
10
cos 25 =
GH
20. Is it possible for WY and XZ to bisect
each other? Explain.
No. That would make WXYZ a
parallelogram by Theorem 8-10.
Geometry H
Chapter 8
Review Worksheet answers
Use this figure for questions #21-24
Use this figure for questions #25-27
ABCD is a kite with AB = AD
21. BC = 2x + 10 and DC = 6x – 30
x = _____
Since BC = DC,
2x + 10 = 6x – 30
40 = 4x
10 = x
22. BE = 6 and AE = 8
AB = ___
Since the diagonals of a kite are perpendicular
AEB is a rt. . By Pythagorean Theorem,
(AB)2 = (AE)2 + (BE)2
(AB)2 = (8)2 + (6)2
(AB)2 = 100
AB = 10
23. mABC = 8y and m ADC = 10y – 30
y = ___
Since ABC and ADC are concruent,
mABC = mADC
8y = 10y – 30
-2y =
-30
y = 15
24. BD = 4z + 10 and BE = 3z – 8
z = ___
Since E is the midpoint of BD ,
BD = 2∙BE
4z + 10 = 2(3z – 8)
4z + 10 = 6z – 16
26 = 2z
13 = z
ABCD is a trapezoid with AB || DC
25. AB = 20 and DC = 56
JK = ___
Since JK is the midsegment of the trapezoid,
AB  DC
JK 
2
20  56 76
JK 

 38
2
2
26. AB = 12 and JK = 22
DC = ___
Since JK is the midsegment of the trapezoid,
AB  DC
JK 
2
12  DC
22 
2
44 = 12 + DC
32 = DC
27. A is 20 larger than AJK.
mA = ___
Since JK is the midsegment of the trapezoid,
AB || JK
Since A and AJK are same-side interior
angles of parallel lines,
mA + mAJK = 180
Since mA = mAJK + 20
mAJK + 20 + mAJK = 180
2mAJK + 20 = 180
mAJK = 80
So, mA = 80 + 20 = 100