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Mathematics 17 Online
OpenStudy (nerdyotter):

Check my answers please: 3. For triangle TOE the following facts are given: mTOS = mSOE TE = 6 cm OT = 2 cm OE = 5.8 cm OG = 4.35 cm AU = 0.45 cm Use this information to answer the following: a) Why is triangle OBG similar to triangle OTE? Now find the following missing lengths. Show all work or reasoning. Round non-integral lengths to the nearest hundredth. b) GE c) TS d) OA e) BT (Use the Side-Splitting Theorem.) f) SE g) OU Answers will be in a comment below! Thank you!

OpenStudy (nerdyotter):

A) First note that both triangles OBG and OTE have the same common angle at vertex O. Therefore, the measure of angle BOG = the measure of angle TOE. Because OS bisects both angles BOG and TOE, it divides TS and SE into the same ratio as OT and OE. Furthermore, because BG is parallel to TE, OS also divides BG = BN + NG into the same ratio. That implies that the sides of the angle BOG are in the same ratio as the sides of angle TOE. Since OB in triangle OBG corresponds to OT in triangle OTE, and OG in triangle OBG corresponds to OE in triangle OTE, then triangles OBG and OTE are similar by equal ratios of corresponding parts. B) GE = OE - OG GE = 5.8 cm - 4.35 cm GE = 1.45 cm C) TS/SE = OT/OE TS = SE (OT/OE) Let TS = x and SE = 6 - x. (Recall that TE = 6.) x = (6 - x)(2/5.8) x = (6 - x)(0.344) x = 6 (0.344) - 0.344x x = 2.064 - 0.344x x + 0.344x = 2.064 1.344x = 2.064 x = 2.064/1.344 x = 1.54 to the nearest hundredth Since TS = x, then TS = 1.54 cm. D) OA/OU = OG/OE OA = OU (OG/OE) OA = (OA + AU)(OG/OE) OA = (OA + 0.45)(4.35/5.8) OA = (OA + 0.45)(0.75) OA = (0.75)OA + 0.3375 OA - (0.75) OA = 0.3375 (0.25)OA = 0.3375 OA = 0.3375/0.25 OA = 1.35 cm E) BT = OT - OB BT = 2 - OB OB/OT = OA/OU OB = OT(OA/OU) OB = 2 [1.35/(1.35 + 0.45)] OB = 2 (0.75) OB = 1.50 Now plug OB = 1.50 into the equation BT = 2 - OB. BT = 2 - OB BT = 2 - 1.50 BT = 0.50 cm F) SE = TE - TS SE = 6 - 1.54 SE = 4.46 cm G) OU = OA + AU OU = 1.35 + 0.45 OU = 1.80 cm

OpenStudy (nerdyotter):

Attached diagram:

jimthompson5910 (jim_thompson5910):

The way I'd do part A) is this BG || TE (given) angle BOG = angle TOE (common angle) angle OBG = angle OTE (corresponding angles) triangle OBG ~ triangle OTE (AA similarity theorem)

jimthompson5910 (jim_thompson5910):

that's the basic outline. You can either do a 2 column proof or paragraph proof of that

OpenStudy (nerdyotter):

@jim_thompson5910 Thank you for your help. :)

jimthompson5910 (jim_thompson5910):

Everything else looks great. Nice work

OpenStudy (nerdyotter):

@jim_thompson5910 Thank you! I appreciate the help is such a timely fashion. Could I have help with one more proof question? It's essentially the same as question A in the last problem set.

jimthompson5910 (jim_thompson5910):

sure, go ahead

OpenStudy (nerdyotter):

For the attached image, would this proof (based off the proof you gave me earlier) be correct? AN || RI (Given.) angle ATN = angle RTI (Common angle.) angle TAN = angle TRI (Corresponding angle.) triangle TAN ~ triangle TRI (AA Similarity theorem.)

jimthompson5910 (jim_thompson5910):

everything looks good

jimthompson5910 (jim_thompson5910):

To be more formal, I would place it in a 2 column proof. If that isn't possible, then convert it into paragraph form

OpenStudy (nerdyotter):

@jim_thompson5910 Thank you for all the help. I appreciate it. :)

jimthompson5910 (jim_thompson5910):

you're welcome

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