Name: Geometry Review: Packet #4: Def GHF X | PDF | Trigonometric ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: If △DEF ~ △GHF, find the value of x.

We are told that triangle DEF is similar to triangle GHF. That means their corresponding sides are proportional.

Looking at the diagram (even though we don’t describe it, we use the labels and numbers given):

- Side DF corresponds to side HF → DF = 9x - 13, HF = 27
- Side EF corresponds to side GF → EF = 90, GF = 15

So set up the proportion:

> DF / HF = EF / GF
> (9x - 13) / 27 = 90 / 15

Simplify right side: 90 ÷ 15 = 6

So:

> (9x - 13) / 27 = 6

Multiply both sides by 27:

> 9x - 13 = 162

Add 13 to both sides:

> 9x = 175

Wait — 162 + 13 = 175? Let me check: 162 + 10 = 172, +3 = 175 → yes.

Then divide by 9:

> x = 175 / 9 ≈ 19.44... Hmm, that’s not a nice number. Did I match the sides correctly?

Wait — maybe I matched wrong. In similar triangles, order matters: △DEF ~ △GHF

That means:

D ↔ G
E ↔ H
F ↔ F (common vertex?)

Actually, looking at the naming: △DEF ~ △GHF

So:

D → G
E → H
F → F

Therefore:

Side DE ↔ GH
EF ↔ HF
FD ↔ FG

But in the diagram, we have lengths on segments from F:

From F to E is 90, from F to G is 15 → so EF and GF are parts of the same line? Actually, since F is common, and triangles share angle at F, likely vertical angles or something.

Actually, better approach: Since they are similar with correspondence D→G, E→H, F→F, then:

Side opposite D is EF, opposite G is HF → so EF corresponds to HF? Wait no.

Better: Use the sides adjacent to the common angle at F.

In △DEF and △GHF, angle at F is common (or vertically opposite? But labeled as same point).

Actually, standard way: when two triangles share a vertex and are similar with that vertex corresponding, the sides around that vertex correspond.

So for △DEF ~ △GHF, with F corresponding to F, then:

DF corresponds to GF
EF corresponds to HF

Because:

In △DEF: sides from F are FD and FE
In △GHF: sides from F are FG and FH

And since D↔G, E↔H, then:

FD ↔ FG
FE ↔ FH

So:

FD / FG = FE / FH

Given:

FD = 9x - 13
FG = 15
FE = 90
FH = 27

So:

(9x - 13) / 15 = 90 / 27

Simplify 90/27: divide numerator and denominator by 9 → 10/3

So:

(9x - 13)/15 = 10/3

Cross-multiply:

3*(9x - 13) = 10*15
27x - 39 = 150

Add 39 to both sides:

27x = 189

Divide by 27:

x = 189 / 27 = 7

Check: 27 * 7 = 189 → yes.

✔ So x = 7

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Problem 2: If △PQR ~ △TSU, find the value of x.

Correspondence: P→T, Q→S, R→U

So sides:

PQ ↔ TS
QR ↔ SU
RP ↔ UT

Given:

PQ = 7x + 2
TS = ? Wait, look at diagram values:

Triangle PQR: PQ = 7x+2, PR = 35
Triangle TSU: TU = 14, SU = x+8

Since R↔U, and P↔T, then side PR corresponds to TU? Let's see:

Vertices: P-Q-R and T-S-U

So:

PQ ↔ TS
QR ↔ SU
RP ↔ UT

Yes.

So RP corresponds to UT → RP = 35, UT = 14

QR corresponds to SU → QR = ? Not given directly. Wait, in triangle PQR, we have PQ and PR, but not QR. In triangle TSU, we have TU and SU.

Actually, perhaps better to use the sides we know.

We know:

PR = 35 (side from P to R)
UT = 14 (side from U to T) → which is same as TU, length 14

Since R↔U and P↔T, then side PR corresponds to side TU.

Similarly, side PQ corresponds to side TS — but we don’t have TS.

Wait, another pair: QR corresponds to SU.

But we don’t know QR.

Wait — perhaps we can use PR and TU, and also PQ and... what?

Hold on — maybe the diagram shows:

In △PQR: side PR = 35, side PQ = 7x+2
In △TSU: side TU = 14, side SU = x+8

If R↔U and P↔T, then PR ↔ TU → ratio = 35/14 = 5/2

Also, if Q↔S, then PQ ↔ TS — but we don’t have TS.

But SU is side from S to U, which corresponds to QR (since Q↔S, R↔U)

But we don’t have QR.

Wait — perhaps the side given as “x+8” is SU, which corresponds to QR, but we don’t have QR.

Alternatively, maybe the side labeled “x+8” is actually TS? Let me re-read.

The problem says: △PQR ~ △TSU

Diagram description (from memory of typical problems): usually, in such diagrams, the sides are labeled near them.

Assume:

In △PQR:
- Side PQ = 7x + 2
- Side PR = 35

In △TSU:
- Side TU = 14
- Side SU = x + 8

Now, since correspondence is P→T, Q→S, R→U

Then:

Side PQ (between P and Q) corresponds to side TS (between T and S)
Side QR (Q to R) corresponds to side SU (S to U)
Side RP (R to P) corresponds to side UT (U to T)

So RP corresponds to UT → RP = 35, UT = 14 → ratio = 35/14 = 5/2

Also, QR corresponds to SU → but we don’t know QR.

PQ corresponds to TS — but we don’t know TS.

Wait — unless “x+8” is actually TS? The problem says “x+8” is next to SU? Or is it next to TS?

Looking back at user input: “x + 8” is written near SU? Actually, in the text: “x + 8” is listed under triangle TSU, and “14” is under TU.

Perhaps “x+8” is the length of SU, which corresponds to QR.

But we don’t have QR.

Alternative idea: Maybe the side labeled “x+8” is actually TS? Because sometimes diagrams label the side between T and S as x+8.

Let me assume that “x+8” is the length of TS, which corresponds to PQ.

Because:

PQ corresponds to TS → PQ = 7x+2, TS = x+8

And PR corresponds to TU → PR = 35, TU = 14

So ratios should be equal:

PQ / TS = PR / TU

So:

(7x + 2) / (x + 8) = 35 / 14

Simplify 35/14 = 5/2

So:

(7x + 2)/(x + 8) = 5/2

Cross-multiply:

2*(7x + 2) = 5*(x + 8)
14x + 4 = 5x + 40

Subtract 5x from both sides:

9x + 4 = 40

Subtract 4:

9x = 36

x = 4

Check: Left side: 7*4 + 2 = 30, x+8=12, 30/12=2.5=5/2 → matches 35/14=2.5 → correct.

✔ So x = 4

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Problem 3: If △JKL ~ △NML, find JL.

Correspondence: J→N, K→M, L→L (common vertex)

So sides:

JK ↔ NM
KL ↔ ML
LJ ↔ LN

Given:

In △JKL: KL = 4x - 12, JL = 2x + 9
In △NML: ML = 24, NL = 21

Since K→M, L→L, then KL corresponds to ML → KL = 4x-12, ML=24

Also, J→N, L→L, so JL corresponds to NL → JL = 2x+9, NL=21

So set up proportion using these two pairs:

KL / ML = JL / NL

So:

(4x - 12) / 24 = (2x + 9) / 21

Cross-multiply:

21*(4x - 12) = 24*(2x + 9)

Calculate both sides:

Left: 21*4x = 84x, 21*(-12) = -252 → 84x - 252
Right: 24*2x = 48x, 24*9 = 216 → 48x + 216

So:

84x - 252 = 48x + 216

Subtract 48x from both sides:

36x - 252 = 216

Add 252 to both sides:

36x = 468

Divide by 36:

x = 468 / 36

Simplify: divide numerator and denominator by 12: 468÷12=39, 36÷12=3 → 39/3=13

Or 36*13=468 → yes, x=13

Now, JL = 2x + 9 = 2*13 + 9 = 26 + 9 = 35

Check with other side: KL = 4*13 - 12 = 52 - 12 = 40
ML = 24, so ratio KL/ML = 40/24 = 5/3
JL/NL = 35/21 = 5/3 → same → correct.

✔ So JL = 35

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Topic #2: Proving Triangles are Similar

Question: What are the methods to prove triangles are similar?

Answer: There are three main methods:

1. AA (Angle-Angle): If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. (You don’t need side lengths.)

2. SAS (Side-Angle-Side) Similarity: If two sides of one triangle are proportional to two sides of another triangle, and the included angles are congruent, then the triangles are similar.

3. SSS (Side-Side-Side) Similarity: If all three sides of one triangle are proportional to all three sides of another triangle, then the triangles are similar.

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Problem 4: Determine if the triangles are similar. If yes, state which method.

Triangles: △ABC and △DEF

Given:

△ABC: AB = 21, BC = 11.2, AC = 7
△DEF: DE = 15, EF = 8, DF = 5

Check if sides are proportional.

List sides in order. Assume correspondence A→D, B→E, C→F

Then:

AB/DE = 21/15 = 7/5 = 1.4
BC/EF = 11.2 / 8 = 1.4
AC/DF = 7 / 5 = 1.4

All ratios equal → 1.4

So SSS similarity applies.

✔ Yes, similar by SSS.

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Problem 5: Determine if the triangles are similar. If yes, state which method.

Diagram: Triangle VWZ with points X on VZ and Y on WZ, and XY drawn. Arrows on VX and WY? Wait, description: "arrows" probably indicate parallel lines.

Typical setup: If XY is parallel to VW, then triangles XYZ and VWZ are similar by AA (corresponding angles).

But here, the triangle is VWZ, with X on VZ, Y on WZ, and XY connecting them. Arrows on VX and WY? That might mean VX and WY are marked, but probably the arrows are on XY and VW indicating they are parallel.

Standard interpretation: If a line segment connects two sides of a triangle and is parallel to the third side, then it creates a smaller triangle similar to the original.

Here, if XY || VW, then △XYZ ~ △VWZ by AA similarity (because corresponding angles are equal due to parallel lines).

The arrows likely indicate that XY is parallel to VW.

So yes, similar by AA.

Even without assuming, if arrows mean parallel, then angle at Z is common, and angle ZXY = angle ZVW (corresponding angles), so AA.

✔ Yes, similar by AA.

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Final Answers:

1. x = 7
2. x = 4
3. JL = 35
4. Yes, SSS
5. Yes, AA

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Final Answer:
1. 7
2. 4
3. 35
4. Yes, SSS
5. Yes, AA