Word Problems on Similar Triangles
Word problems on similar triangles apply the concept of proportional sides and equal angles to practical situations — measuring heights using shadows, mirrors, scale models, and indirect measurement techniques.
These problems use the fact that in similar triangles, corresponding sides are proportional. This allows us to find unknown lengths when direct measurement is not possible.
The chapter on Triangles in Class 10 NCERT frequently includes these application-based problems in board examinations.
What is Word Problems on Similar Triangles?
Definition: Two triangles are similar if:
- Their corresponding angles are equal (AA criterion), OR
- Their corresponding sides are in the same ratio (SSS criterion), OR
- Two sides are in proportion and the included angle is equal (SAS criterion).
If △ABC ~ △DEF, then:
- ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
- AB/DE = BC/EF = CA/FD
Word Problems on Similar Triangles Formula
Proportionality of corresponding sides:
AB/DE = BC/EF = CA/FD
Ratio of areas of similar triangles:
Area(△ABC)/Area(△DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)²
Types and Properties
Common types of similarity word problems:
- Shadow problems: A person/tree and their shadow form a triangle similar to another object and its shadow (sun's rays are parallel).
- Mirror problems: Using the law of reflection, the angle of incidence equals the angle of reflection, creating similar triangles.
- Height measurement: Using a stick of known height to find the height of a building or tree.
- Scale model problems: A model and the actual object are similar figures.
- River width problems: Using similar triangles to measure a distance that cannot be directly measured.
- Photograph/map problems: Proportional enlargement or reduction.
Solved Examples
Example 1: Shadow Problem
Problem: A 6 m tall tree casts a shadow of 4 m. At the same time, a building casts a shadow of 24 m. Find the height of the building.
Solution:
Sun's rays are parallel, so the angles of elevation are equal. The tree-shadow triangle and building-shadow triangle are similar (AA).
Proportionality:
- Height of tree / Shadow of tree = Height of building / Shadow of building
- 6/4 = h/24
- h = 6 × 24/4 = 36 m
Answer: The height of the building is 36 m.
Example 2: Mirror Problem
Problem: A girl of height 1.5 m stands at a distance of 3 m from a mirror placed on the ground. She can see the top of a building in the mirror. If the mirror is 12 m from the building, find the height of the building.
Solution:
By the law of reflection, the angle of incidence = angle of reflection. The two right triangles formed (girl-mirror and building-mirror) are similar (AA).
Proportionality:
- Girl's height / Girl's distance from mirror = Building height / Building's distance from mirror
- 1.5/3 = h/12
- h = 1.5 × 12/3 = 6 m
Answer: The height of the building is 6 m.
Example 3: Height Using a Stick
Problem: A vertical stick 20 cm long casts a shadow of 15 cm. A tower beside it casts a shadow of 45 m at the same time. Find the height of the tower.
Solution:
The triangles are similar (sun's rays parallel, AA criterion).
- Height of stick / Shadow of stick = Height of tower / Shadow of tower
- 20/15 = h/45
- h = 20 × 45/15 = 900/15 = 60 m
Note: units of the stick (cm) cancel out in the ratio; the tower's units follow the shadow's units (m).
Answer: The height of the tower is 60 m.
Example 4: Scale Model Problem
Problem: A model of a building is made to a scale of 1:50. If the model is 30 cm tall, find the actual height. Also, if a room in the model has a floor area of 60 cm², find the actual floor area.
Solution:
Height:
- Scale = 1:50 → Actual height = 30 × 50 = 1500 cm = 15 m
Area:
- Areas are in the ratio (1:50)² = 1:2500
- Actual area = 60 × 2500 = 1,50,000 cm² = 15 m²
Answer: Actual height = 15 m, Actual floor area = 15 m².
Example 5: River Width Problem
Problem: A person standing at point A on one bank of a river looks at a tree T on the opposite bank. He walks 100 m along the bank to point B, then walks 40 m perpendicular to the bank to point C. From C, the tree T is in line with a point D on the bank such that BD = 20 m. Find the width of the river.
Solution:
Triangles TAB and DCB are similar (AA: right angles at A and D, and vertically opposite angles at B).
Wait — let us re-set up: △TAB and △CDB where TA is the river width.
- TA/CD = AB/BD
- TA/40 = 100/20
- TA = 40 × 5 = 200 m
Answer: The width of the river is 200 m.
Example 6: Lamp Post and Shadow
Problem: A person 1.8 m tall walks away from a 6 m lamp post at 1 m/s. Find the rate at which the shadow lengthens. (Also: When the person is 5 m from the post, how long is the shadow?)
Solution (static part):
Let the person be x m from the post, and the shadow be s m long.
By similarity of triangles (lamp-tip-shadow and person-tip-shadow):
- 6/(x + s) = 1.8/s
- 6s = 1.8(x + s)
- 6s = 1.8x + 1.8s
- 4.2s = 1.8x
- s = 1.8x/4.2 = 3x/7
When x = 5: s = 15/7 ≈ 2.14 m.
Since s = 3x/7, the shadow grows at 3/7 of the walking speed = 3/7 m/s.
Answer: Shadow length when 5 m away = 15/7 m. Shadow lengthens at 3/7 m/s.
Example 7: Areas of Similar Triangles
Problem: Two similar triangles have sides in the ratio 3:5. If the area of the smaller triangle is 54 cm², find the area of the larger triangle.
Solution:
- Ratio of areas = (Ratio of sides)² = (3/5)² = 9/25
- 54/Area₂ = 9/25
- Area₂ = 54 × 25/9 = 150 cm²
Answer: The area of the larger triangle is 150 cm².
Example 8: Map Distance Problem
Problem: On a map, 1 cm represents 5 km. Two cities are 3.5 cm apart on the map. If a park in the first city is shown as a triangle with sides 0.4 cm, 0.5 cm, and 0.6 cm, find the actual distances and the actual perimeter of the park.
Solution:
Distance between cities: 3.5 × 5 = 17.5 km
Park sides:
- 0.4 × 5 = 2 km
- 0.5 × 5 = 2.5 km
- 0.6 × 5 = 3 km
Actual perimeter: 2 + 2.5 + 3 = 7.5 km
Answer: Cities are 17.5 km apart. Park perimeter is 7.5 km.
Real-World Applications
Similar triangle word problems are used in:
- Surveying: Measuring heights of mountains, buildings, and towers indirectly.
- Cartography: Maps are scale models — all distances are proportional.
- Engineering: Scale models of bridges, buildings, and machines use similarity.
- Photography: Enlargement and reduction of images maintain similarity.
- Navigation: Triangulation methods use similar triangles to find positions.
Key Points to Remember
- In shadow problems, the sun's rays are parallel, ensuring AA similarity.
- In mirror problems, the angle of incidence equals the angle of reflection (AA similarity).
- Corresponding sides of similar triangles are proportional.
- Ratio of areas = square of the ratio of corresponding sides.
- Scale models maintain similarity — multiply by the scale factor for actual dimensions.
- Always identify the correct correspondence of vertices before writing proportions.
- Units must be consistent within each ratio (they cancel out).
- Draw the figure clearly and label all known and unknown quantities.
- These problems commonly appear as 3-4 mark questions in CBSE board exams.
- A proportion equation has three knowns and one unknown — always solvable.
Practice Problems
- A pole 8 m high casts a shadow of 6 m. A nearby tower casts a shadow of 30 m. Find the height of the tower.
- A boy 1.5 m tall can see the top of a 30 m building in a mirror placed 2 m from his feet. How far is the mirror from the building?
- Two similar triangles have areas 64 cm² and 121 cm². If a side of the first is 8 cm, find the corresponding side of the second.
- A model airplane has a wingspan of 40 cm. If the scale is 1:200, find the actual wingspan in metres.
- A tree breaks at a point 6 m from the ground and falls so that its top touches the ground 8 m from the base. Find the original height of the tree.
- In a triangle ABC, DE is drawn parallel to BC such that AD = 3 cm and DB = 5 cm. If BC = 16 cm, find DE.
Frequently Asked Questions
Q1. Why do shadows create similar triangles?
The sun is extremely far away, so its rays are effectively parallel. This means the angle of elevation of the sun is the same for all objects at the same time and place. The right angle at the base of each object completes the AA similarity.
Q2. How does a mirror create similar triangles?
By the law of reflection, the angle of incidence equals the angle of reflection. Combined with the right angles at the ground, we get two right triangles with equal acute angles, establishing AA similarity.
Q3. What is the relationship between similarity ratio and area ratio?
If corresponding sides are in the ratio k:1, then the areas are in the ratio k²:1. For example, if sides are in ratio 2:3, areas are in ratio 4:9.
Q4. Can similar triangles have different orientations?
Yes. Similar triangles need not be in the same orientation. One may be rotated, reflected, or flipped. What matters is that corresponding angles are equal and corresponding sides are proportional.
Q5. How do you identify corresponding sides?
Corresponding sides are opposite to equal angles. If ∠A = ∠D, then the sides opposite to them (BC and EF) are corresponding. Write the similarity statement carefully: △ABC ~ △DEF means A↔D, B↔E, C↔F.
Q6. How accurate are indirect measurements using similarity?
The accuracy depends on precise angle measurement and correct identification of similar triangles. In practice, small measurement errors get amplified by the scale factor, so careful measurement is essential.
Related Topics
- Similar Triangles
- Criteria for Similarity of Triangles
- Basic Proportionality Theorem (BPT)
- Areas of Similar Triangles
- Angle Sum Property of Triangle
- Exterior Angle Property of Triangle
- Properties of Isosceles Triangle
- Properties of Equilateral Triangle
- Triangle Inequality Property
- Medians and Altitudes of Triangle
- Right-Angled Triangle Property
- Congruent Triangles - Proofs
- Inequalities in Triangles
- Converse of Basic Proportionality Theorem
