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Geometry · Similar Triangles

Similar Triangles
Mastery Set

From basic ratios to real-world shadow and mirror problems — 20 carefully crafted questions with full explanations.

20Problems
4Levels
∞Attempts
⚡ Quick Memory
AA Similarity → 2 angles equal Corresponding sides are proportional Scale factor = small ÷ big Cross-multiply to solve Shadow = similar triangles Mirror = equal angles

Level 1 · Foundations

Problems 1–5 · Basic ratios & scale factor

📐 Core Concept — Similar Figures

Two figures are similar (~) if they have the same shape but different size. Their angles are equal and their sides are proportional — meaning ratios between matching sides are all the same.

If △ABC ~ △DEF, then   AB/DE = BC/EF = AC/DF = k   (scale factor)
01 Easy Scale Factor

△ABC ~ △DEF. If AB = 6 and DE = 9, what is the scale factor of △ABC to △DEF?

A B C D E F 6 9 ~
💡 KEY: Scale Factor = first triangle ÷ second triangle Write as a reduced fraction: (side of ABC) ÷ (matching side of DEF).
02 Easy Finding a Side

△PQR ~ △XYZ. PQ = 4, XY = 10, QR = 6. Find YZ.

💡 KEY: Set up proportion, cross-multiply PQ/XY = QR/YZ  →  4/10 = 6/YZ  →  cross-multiply.
03 Easy AA Similarity

In △MNP and △STU, ∠M = ∠S = 52° and ∠N = ∠T = 73°. Are the triangles similar? Which theorem applies?

💡 KEY: AA (Angle-Angle) Similarity Theorem If 2 angles of one △ = 2 angles of another △, the triangles are similar. (The 3rd angle is automatically equal too, since angles sum to 180°.)
04 Easy Missing Side

△ABC ~ △DEF with scale factor 3 : 5. If BC = 12, find EF.

💡 KEY: Scale factor ratio = side ratio BC/EF = 3/5. Solve for EF: EF = BC × (5/3).
05 Easy Perimeter Ratio

Two similar triangles have sides in ratio 2 : 7. The smaller triangle has perimeter 18. What is the perimeter of the larger triangle?

💡 KEY: Perimeter ratio = side ratio If sides are in ratio k : 1, then perimeters are ALSO in ratio k : 1.

Level 2 · Parallel Lines

Problems 6–10 · Parallel lines create similar triangles

📐 Core Concept — Parallel Lines & AA

When a line is drawn parallel to one side of a triangle, it creates a smaller similar triangle inside. This is the Triangle Proportionality Theorem. The two triangles share an angle at the vertex, and parallel lines create equal corresponding angles — giving us AA Similarity.

If DE ∥ AB in △ABC, then △CDE ~ △CAB   (AA)  →  CD/CA = CE/CB = DE/AB
06 Easy Parallel Line

In △ABC, DE ∥ AB with D on CA and E on CB. If CD = 4, DA = 8 and CE = 5, find EB.

C A B D E CD=4 DA=8 CE=5 ∥ AB
💡 KEY: Triangle Proportionality → CD/DA = CE/EB 4/8 = 5/EB  →  cross-multiply.
07 Medium DE ∥ AB · find DC

In the diagram, DE ∥ AB. Given CB = 12.6, CE = 9, and AC = 8.4, find the length of DC.

C A B E D AC=8.4 CB=12.6 CE=9
💡 KEY: △CDE ~ △CAB because DE ∥ AB (AA) CE/CB = DC/AC  →  9/12.6 = DC/8.4. Cross-multiply.
08 Medium IJ ∥ FG · find HJ

In the diagram below, IJ ∥ FG. If HG = 13.2, IH = 9.6, and FH = 17.6, find the length of HJ.

H F G I J HI=9.6 FH=17.6 HG=13.2 ∥ FG
💡 KEY: △HIJ ~ △HFG (AA — parallel lines) HI/HF = HJ/HG  →  9.6/17.6 = HJ/13.2
09 Medium Full Side Length

In △ABC, DE ∥ BC with D on AB and E on AC. AD = 5, DB = 10, AE = 7. Find AC (the full length, not just EC).

💡 KEY: AD/AB = AE/AC (not AD/DB!) Common mistake: using segment ratios instead of full-side ratios. AB = AD + DB = 15.   5/15 = 7/AC.
10 Medium Find the Missing x

In △RST, UV ∥ ST with U on RS and V on RT. RU = x, US = 9, RV = 8, VT = 12. Find x.

💡 KEY: RU/US = RV/VT (segment ratio form) x/9 = 8/12  →  simplify 8/12 first, then solve.

Level 3 · Shared Angle Problems

Problems 11–15 · Overlapping triangles with ∠ congruence

📐 Core Concept — Overlapping Triangles

When one angle is shared between two triangles (or two angles are given as congruent), use AA Similarity. The key is identifying the correct correspondence — which vertex in the small triangle matches which vertex in the large triangle. Always write the similarity statement first!

Shared angle at S + ∠QRS ≅ ∠TUS → △QRS ~ △TUS  →  QS/TS = RS/US = QR/TU
11 Medium ∠ congruence · find SQ

In △QSR, point T is on RS and point U is on QS so that ∠QRS ≅ ∠TUS. If ST = 16, RT = 14, and US = 12, find the length of SQ.

Q R S T U RT=14 ST=16 US=12
💡 KEY: ∠S is shared + ∠QRS ≅ ∠TUS → △QRS ~ △TUS (AA) Write the proportion: QS/TS = RS/US. RS = RT + TS = 14 + 16 = 30.   QS/16 = 30/12.
12 Medium Shared Vertex

In △ABC, point D is on AB and point E is on AC such that ∠ADE ≅ ∠ABC. If AD = 8, DB = 4, and AE = 10, find EC.

💡 KEY: ∠A shared + ∠ADE ≅ ∠ABC → △ADE ~ △ABC AD/AB = AE/AC. AB = AD + DB = 12. 8/12 = 10/AC → AC = 15. EC = AC − AE = 15 − 10 = 5.
13 Medium Vertical Angles

Two triangles △AEB and △CED share the vertex E where the diagonals of a quadrilateral cross. ∠A ≅ ∠C (alternate interior angles). If AE = 6, BE = 9, CE = 4, find DE.

💡 KEY: ∠AEB ≅ ∠CED (vertical angles) + ∠A ≅ ∠C → AA Similarity AE/CE = BE/DE → 6/4 = 9/DE.
14 Hard Find x in segment

In △PQR, S is on PQ and T is on PR so that ∠PST ≅ ∠PQR. If PS = x, SQ = 20, PT = 9, and TR = 16, find x.

💡 KEY: △PST ~ △PQR (∠P shared, ∠PST ≅ ∠PQR) PS/PQ = PT/PR PQ = x + 20, PR = 9 + 16 = 25. x/(x+20) = 9/25 → cross-multiply → 25x = 9x + 180 → 16x = 180.
15 Hard Solve for x (quadratic)

In △ABC, D on AB and E on AC, DE ∥ BC. AD = x + 2, DB = 6, AE = x, EC = 4. Find x.

💡 KEY: AD/DB = AE/EC (Triangle Proportionality) (x+2)/6 = x/4 → cross-multiply → 4(x+2) = 6x → 4x + 8 = 6x → 2x = 8.

Level 4 · Real World

Problems 16–20 · Shadows, mirrors & indirect measurement

📐 Core Concept — Indirect Measurement

Similar triangles let us measure things we can't reach directly — like the height of a building. Two common setups: Shadow Problems (person and tree cast shadows at the same angle of sunlight) and Mirror Problems (angle of incidence = angle of reflection creates two similar triangles).

Shadow:   height₁ / shadow₁ = height₂ / shadow₂ Mirror:   object height / distance to mirror = eye height / distance from mirror
16 Easy Shadow Problem

A 5 ft tall person casts a 8 ft shadow. At the same time, a nearby flagpole casts a 32 ft shadow. How tall is the flagpole?

5 ft 8 ft ? 32 ft
💡 KEY: height/shadow = height/shadow (same sun angle) 5/8 = h/32. Solve for h.
17 Medium Shadow (overlap)

Travis is 1.55 m tall. At 10 a.m., he measures the length of a tree's shadow to be 21.15 m. He stands 16.8 m away from the tree so that the tip of his shadow meets the tip of the tree's shadow. Find the height of the tree to the nearest hundredth of a meter.

21.15 m (tree's shadow) 16.8 m 1.55m ?
💡 KEY: Travis's shadow = 21.15 − 16.8 = 4.35 m Both triangles have the same shadow tip → same angle. 1.55/4.35 = tree height/21.15.
18 Medium Mirror Problem

Morgan places a mirror on the ground 13.85 m from a goalpost. She walks 4.25 m past the mirror until she can see the top of the goalpost in the mirror. Her eyes are 1.15 m above the ground. How tall is the goalpost? Round to the nearest hundredth.

✕ 13.85 m 4.25 m 1.15m ?
💡 KEY: Mirror creates two similar right triangles goalpost height / distance(goalpost to mirror) = eye height / distance(mirror to Morgan) h / 13.85 = 1.15 / 4.25
19 Hard Mirror — find person height

A building is 24 m tall. A mirror is placed on the ground 36 m from the building. A person stands 3 m from the mirror (on the opposite side from the building) and can just see the top of the building. What is the eye height of the person?

💡 KEY: Same mirror ratio building / dist(building→mirror) = eye height / dist(person→mirror) 24 / 36 = eye height / 3
20 Hard Shadow — two unknowns

At the same time of day, a 1.8 m tall student and a tree both cast shadows. The student's shadow is 2.4 m. The tree's shadow extends 18 m from the base of the tree, and the student stands 9.6 m from the tree (shadow tips meet). Find the height of the tree.

💡 KEY: Student's shadow = 18 − 9.6 = 8.4 m ✗ Wait — the student's shadow is 2.4 m (given directly). Tree total shadow = 18 m. 1.8/2.4 = tree height/18. (Same sun angle ratio.)