Question

unit 4 test remediation
state if the two triangles are congruent. if they are, state by which congruence theorem.
1.
2.
3.
4.
state if the two triangles are similar. if they are, state by which similarity theorem.
5.
6.
7.
8.
find the missing length. the triangles in each pair are similar.
9.

1. \t$\triangle wvu sim \triangle wed$

Explanation:

Step1: Analyze Triangle 1 (Congruence)

The first triangle has two equal sides (marked with ticks) and a common perpendicular segment. So, by SAS (Side - Angle - Side) congruence theorem, the two triangles formed are congruent. The angle is the right angle (between the equal side and the perpendicular), and the two sides are equal. So, congruent by SAS.

Step2: Analyze Triangle 2 (Congruence)

Looking at the second figure, we have vertical angles (equal) and two sides with ticks (equal). So, by ASA (Angle - Side - Angle) or SAS? Wait, the angles at the vertices: the marked angles and the vertical angles. Wait, the figure has two triangles with a pair of equal sides (ticks) and a pair of equal angles (the right angle and the vertical angle? Wait, maybe AAS or ASA. Wait, the first triangle (problem 1) is an isosceles triangle with a perpendicular bisector, so the two triangles are congruent by SAS (the perpendicular is common, two sides equal, right angle included).

Step3: Analyze Triangle 3 (Congruence)

The third figure has two triangles with vertical angles (equal) and two sides with ticks (equal). So, by SAS (if the angle is included) or ASA. Wait, the angles at the non - included sides? Wait, the triangles are congruent by SAS (vertical angle, two equal sides).

Step4: Analyze Triangle 4 (Congruence)

The fourth figure has three angles equal (corresponding angles, vertical angles). So, by AAA (Angle - Angle - Angle) which for congruence is equivalent to ASA or AAS. So, the triangles are congruent by ASA (since two angles and the included side? Wait, the vertical angles and the marked angles, so ASA.

Step5: Analyze Triangle 5 (Similarity)

Triangle 5: We have angle D equal to angle L (marked), and vertical angles at K. So, by AA (Angle - Angle) similarity theorem, the two triangles are similar.

Step6: Analyze Triangle 6 (Similarity)

Check the ratios of sides. \( \frac{168}{42}=\frac{196}{37}? \) Wait, \( 168\div42 = 4 \), \( 196\div37\approx5.3 \). Wait, no, maybe I miscalculated. Wait, \( 42\times4 = 168 \), \( 37\times4 = 148 eq196 \). Wait, maybe the other way. Wait, \( 168 + 42=210 \), \( 196+37 = 233 \). No, maybe the triangles are similar by AA? Wait, the vertical angles are equal, and if there is another angle equal. Wait, maybe I made a mistake. Wait, the sides: \( \frac{168}{42}=4 \), \( \frac{196}{37}\approx5.3 \). No, maybe the triangles are not similar? Wait, no, maybe the labels are different. Wait, \( \triangle LKM \) and \( \triangle DKC \). So, \( \angle LKM=\angle DKC \) (vertical angles), and if \( \frac{LK}{DK}=\frac{MK}{CK} \)? Wait, \( LK = 168 + 42=210 \), \( DK = 42 \), \( MK = 196+37 = 233 \), \( CK = 37 \). \( 210\div42 = 5 \), \( 233\div37\approx6.29 \). No, maybe AA. Wait, if there are two angles equal. Maybe the triangles are similar by AA (vertical angle and another angle).

Step7: Analyze Triangle 7 (Similarity)

Check the ratios of sides. \( \frac{32}{56}=\frac{32}{56}=\frac{4}{7} \), \( \frac{56}{98}=\frac{4}{7} \), \( \frac{52}{39}=\frac{4}{3}? \) Wait, no, \( 32 + 56=88 \)? Wait, no, \( MD = 32 \), \( MK = 56 \), so \( \frac{MD}{MK}=\frac{32}{56}=\frac{4}{7} \), \( MC = 52 \), \( ML = 39 + 52=91 \)? Wait, no, \( MC = 52 \), \( CL = 39 \), so \( ML = 52 + 39 = 91 \)? Wait, \( \frac{56}{98}=\frac{4}{7} \), \( \frac{32}{56}=\frac{4}{7} \), \( \frac{52}{91}=\frac{4}{7} \) (since \( 52\div13 = 4 \), \( 91\div13 = 7 \)). So, by SSS (Side - Side - Side) similarity theorem, the triangles are similar (ratios of corresponding sides equal).

Step8: Analyze Triangle 8 (Similarity)

Since \( LM\parallel CB \), by the Basic Proportionality Theorem (Thales' theorem), the triangles are similar by AA (corresponding angles equal, since \( LM\parallel CB \), \( \angle ALM=\angle ACB \), \( \angle AML=\angle ABC \)). So, similar by AA.

Step9: Analyze Triangle 9 (Missing Length)

The triangles are similar. Let the missing length be \( x \). The ratio of sides: \( \frac{7}{12}=\frac{10}{15}=\frac{6}{9} \)? Wait, \( \frac{7}{12}\approx0.58 \), \( \frac{10}{15}=\frac{2}{3}\approx0.67 \), no, wait the big triangle has height 12, base 9, small triangle height 7, base 6. Wait, the ratio of similarity is \( \frac{6}{9}=\frac{2}{3} \), so the height should be \( 12\times\frac{2}{3}=8 \), but the small triangle has height 7? Wait, maybe I mixed up. Wait, the two triangles: let's denote the big triangle as \( \triangle SRJ \), small as \( \triangle BRJ \)? No, \( \triangle SRT \) and \( \triangle BRC \). Wait, \( \frac{BR}{SR}=\frac{7}{12} \), \( \frac{BC}{ST}=\frac{6}{9}=\frac{2}{3} \), \( \frac{BR}{SR}=\frac{7}{12}\approx0.58 \), \( \frac{BC}{ST}=\frac{2}{3}\approx0.67 \). Wait, no, the correct ratio: since the triangles are similar, \( \frac{7}{12}=\frac{10}{x} \)? Wait, no, \( \frac{7}{12}=\frac{6}{9} \)? No, \( 7\times9 = 63 \), \( 12\times6 = 72 \), not equal. Wait, maybe the ratio is \( \frac{10}{15}=\frac{6}{9}=\frac{2}{3} \), so the height of the big triangle is \( 7\div\frac{2}{3}=\frac{21}{2} = 10.5 \), but the given height is 12. Wait, I think I mislabeled. Let's take the two similar triangles: the small one with sides 7, 10, 6 and the big one with sides 12, 15, 9. Check \( \frac{7}{12}=\frac{10}{15}=\frac{6}{9} \). \( \frac{10}{15}=\frac{2}{3} \), \( \frac{6}{9}=\frac{2}{3} \), \( \frac{7}{12}\approx0.58 \), \( \frac{2}{3}\approx0.67 \). Wait, no, the correct similar triangles: the small triangle has sides 7, 10, 6 and the big triangle has sides \( 7 + y \), 15, 9. Wait, \( \frac{6}{9}=\frac{10}{15}=\frac{2}{3} \), so the ratio is \( \frac{2}{3} \). So, \( \frac{7}{h}=\frac{2}{3} \), \( h=\frac{21}{2}=10.5 \), but the given height is 12. Wait, maybe the missing length is in the base. Wait, the base of the small triangle is 6, big is 9, ratio \( \frac{2}{3} \). The other side of the small triangle is 10, big is 15 (ratio \( \frac{2}{3} \)), the height of the small is 7, big is \( 7\times\frac{3}{2}=10.5 \), but the given height is 12. I must have misread. Wait, the problem 9: Find the missing length. The triangles are similar. Let the missing length be \( x \). Let's assume the similar triangles have sides in ratio. For the first similar triangles (problem 9), the small triangle has sides 7, 10, 6 and the big triangle has sides 12, 15, 9. Wait, \( \frac{6}{9}=\frac{10}{15}=\frac{2}{3} \), so the ratio of small to big is \( \frac{2}{3} \). So, if the small side is 7, the big side is \( 7\times\frac{3}{2}=10.5 \), but that's not matching. Wait, maybe the missing length is 10.5? No, maybe I made a mistake. Wait, the second problem 10: \( \triangle WVU\sim\triangle WED \). So, the ratio of sides: \( \frac{WV}{WE}=\frac{WU}{WD}=\frac{VU}{ED} \). \( WV = 104 \), \( WU = 117 \), \( WD = 27 + x \), \( WE=104 +? \) Wait, no, \( \triangle WVU\sim\triangle WED \), so \( \frac{WV}{WE}=\frac{WU}{WD} \). \( WV = 104 \), \( WU = 117 \), \( WD = 27 + x \), \( WE = 104 +? \) Wait, no, the sides: \( WV = 104 \), \( WE \) is the base, \( WU = 117 \), \( WD = 27 + x \). Since they are similar, \( \frac{WV}{WE}=\frac{WU}{WD} \). Wait, maybe \( \frac{104}{104 + y}=\frac{117}{117 + 27} \)? No, \( 117+27 = 144 \), \( \frac{117}{144}=\frac{13}{16} \), \( 104\div13 = 8 \), \( 16\times8 = 128 \), \( 128 - 104 = 24 \). No, this is getting too complicated. Let's focus on problem 1:

Answer:

1. Congruent by SAS.
2. Congruent by ASA.
3. Congruent by SAS.
4. Congruent by ASA.
5. Similar by AA.
6. Similar by AA (if angles are equal) or SSS (if ratios match).
7. Similar by SSS (\( \frac{32}{56}=\frac{56}{98}=\frac{52}{91}=\frac{4}{7} \)).
8. Similar by AA (Thales' theorem).
9. Missing length: Let the ratio be \( \frac{6}{9}=\frac{2}{3} \), so if the small side is 7, big side is \( 7\times\frac{3}{2}=10.5 \) (but likely miscalculation, correct ratio \( \frac{10}{15}=\frac{2}{3} \), so missing length is \( 12\times\frac{2}{3}=8 \)? No, \( 15\times\frac{2}{3}=10 \), \( 9\times\frac{2}{3}=6 \), so the height of the big triangle is 12, so the height of the small is \( 12\times\frac{2}{3}=8 \), but the given small height is 7. I think the correct missing length for problem 9 is 8 (if the ratio is \( \frac{2}{3} \) and the big height is 12, small height is 8, but the given small height is 7. I must have misread the problem. For problem 10: \( \triangle WVU\sim\triangle WED \), so \( \frac{WV}{WE}=\frac{WU}{WD} \), \( \frac{104}{104 + x}=\frac{117}{117 + 27} \), \( \frac{117}{144}=\frac{13}{16} \), \( 104\times16 = 13\times(104 + x) \), \( 1664=1352+13x \), \( 13x = 312 \), \( x = 24 \). So, the missing length is 24.