Download Solutions to Homework 4

Answers to Math 117 Homework #4:
Page 574
2. On a plane perpendicular to segment AB
4. a) construction
b) construction
c) a scalene right triangle
d) not possible
7. a) yes, SAS
b) yes, SSS
c) no
11. a) construction
b) construction
c) construction
12. a) construction
b) construction
14. a) one method is to start at anyh point on the circle and with a compass opened to the
length of the radius of the circle mark off equally spaced points; there will be six
equal arcs.
b) construction
c) congruent by SSS
d) The length of each side of the hexagon constructed in part (a) is equal to r, the
radius of the circle.
Consider one of the six congruent triangles in the hexagon; let A and B be the two
vertices of the triangle which are on the circle and the third vertex be O, at the
center of the circle.
OA = OB = r, implying the triangle is isosceles; the measure of the angle at O
equals 360°/6 = 60°
The base angles in an isosceles triangle are congruent and the sum of their
measures is 180° – 60° = 120°, implying each must measure 120°/2 = 60°.
Since all the angles of ∆AOB are congruent (60°) and the triangle is equilateral,
AB = r.
22. The center of the circle is the intersection of the perpendicular bisectors of any two
sides of the triangle. Because the triangle is obtuse, its center is exterior to the
triangle. The radius of the circle is the distance from the center to any of the vertices
of the triangle.
30. a) ∆ABC ≅ ADC by SSS. Hence ∠BAC ≅ ∠DAC and ∠BCM ≅∠DCM by CPCTC.
Therefore line AC bisects ∠A and ∠C
b) The angles formed are right angles. By part (a), ∠BAM ≅∠DAM. Hence
∆ABM≅∆ADM by SAS. ∠BMA ≅∠DMA by CPCTC. Since ∠ABM and
∠DMA are adjacent congruent angles, each must be a right angle. Since vertical
angles formed are congruent, all four angles formed by the diagonals are right
angles.
c) By part (b) segment BM ≅ segment MD; CPCTC.
Page 580
5. a) yes, ASA
b) yes, AAS
c) no, SSA does not assure congruence
d) no, AAA does not assure congruence
17. Either the arcs or the central angles must have the same measure (radii are the same
since the sectors are part of the same circle).
20. h = (a–b)/2
23. a) The lengths of one side of each square must be equal.
b) The lengths of the sides of two perpendicular sides of the rectangles must be equal.
c) Answers vary; one solution is that two adjacent sides must have equal lengths and
the included angle of one must be congruent to the other.
25. a) Use the definition of a parallelogram and ASA to prove that ∆ABD ≅ ∆CDB and
∆ADC ≅ ∆CBA.
b) Use a pair of taingles from (a)
c) hint: prove that ∆ABF ≅ ∆CDF
d) hint: extend segment AB and look for corresponding angles.
Page 591
4. a) The perpendicular bisectors of the sides of an acute triangle meet inside the
triangle.
b) The perpendicular bisectors of the sides of a right triangle meet at the midpoint of
the hypotenuse.
c) The perpendicular bisectors of the sides of an obtuse triangle meet outside the
triangle.
8. construction
10. construction
11. a) ray PQ is the perpendicular bisector of segment AB
b) Q is on the perpendicular bisector of segment AB because segment AQ ≅
segment QB; similarly, P is on the perpendicular bisector of seg AB. A unique
line contains two points so the perpendicular bisector contains ray PQ.
c) ray PQ is the angle bisector of ∠APB; ray QC is the angle bisector of ∠AQB.
d) Show that ∆APQ ≅ ∆BPQ by SSS; then ∠APQ ≅∠BPQ by CPCTC. Show that
∆AQC ≅∆BQC by SAS so that ∠AQC ≅∠BQC.
16. a) Construct an equilateral triangle and bisect one of its angles.
b) Bisect a 30° angle.
c) Add 30° and 15° angles or bisect a right angle.
d) Add 60° and 15° angles or 45° and 30° angles.
e) Add 60°, 30°, and 15° angles.
17. construction
20. a) The point is determined by the intersection of the angle bisector of ∠A and the
perpendicular bisector of segment BC. Because the point is on the angle bisector
of ∠BAD, it is equidistant from its sides. Because it is on the perpendicular
bisector of segment BC, it is equidistant form B and C.
b) the point is determined by the intersection of the angle bisector of ∠A and ∠B.
c) The point is determined by the intersection of the perpendicular bisectors of
segments AB and BC.
d) Not possible. If such a point existed it would have to be on the perpendicular
bisector of al the sides, but because the perpendicular bisectors of the four sides of
a quadrilateral do not necessarily intersect in a single point (this is the case for the
given quadrilateral), such a point does not exist.
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8. a) (i) ∆ABC ~ ∆DEF by AA
(ii) ∆ABC ~ ∆EDA by AA
(iii) ∆ACD ~ ∆ABE by AA
(iv) ∆ABE ~ ∆DBC by AA
b) (i) 2/3
(ii) ½
(iii) ¾
(iv) ¾
9. a) 7
b) 24/7
c) 3
d) 96/13
10. construction
13. 15 m
16. About 232.6 in. or 19.4 ft.
17. a) (i) Connect B with D and then apply Theorem 10-8 to triangles ABD and BCD.
Let P be the midpoint of segment MN.
(ii) Theorems 10-7 and 10-8 imply that MP = ½a and PN = ½b.
Thus, MN = MP + PN = ½a + ½b = ½(a+b)