15.0 Review
Objects having the same shape but different sizes are called similar figures. Maps of Nepal of different sizes and photographs of the same person taken in different sizes show similarity because their shapes remain the same though their sizes change.
Geometrical figures of different sizes but having the same shape are called similar figures.
From discussion:
All circles have the same shape.
All squares have the same shape.
All rhombuses do not have the same shape.
All rectangles do not have the same shape.
Two triangles are similar if:
Their corresponding sides are proportional.
Or, their corresponding angles are equal.
15.1 Similarity of Triangles (Clear Concept)
Triangles having the same shape but not necessarily the same size are called similar triangles.
Key idea: In similar triangles, angles decide the shape and sides decide the size.
Conditions for similarity of triangles (with explanation):
1.AA Similarity (Angle–Angle):
If two angles of one triangle are respectively equal to two angles of another triangle, then the triangles are similar.
Reason: The third angle also becomes equal automatically.
2. SSS Similarity (Side–Side–Side):
If the ratios of all three pairs of corresponding sides of two triangles are equal, then the triangles are similar.
3.SAS Similarity (Side–Angle–Side):
If two sides of one triangle are proportional to two sides of another triangle and the included angle is equal, then the triangles are similar.
15.2 Similar Polygons
Polygons having the same shape but different sizes are called similar polygons.
Conditions for similarity of polygons:
The number of sides must be equal.
Corresponding angles must be equal.
Ratios of corresponding sides and perimeters must be equal.
Important Examples
Example 1: Shadow Problem
Given:
Height of lamp post = 3.9 m
Shadow of lamp post = 6.5 m
Height of girl = 1.2 m
Solution:
Let DE be the shadow of the girl.
Or, triangles formed are similar.
Or, corresponding sides are proportional.
Or, DE / 6.5 = 1.2 / 3.9
Or, 3.9 DE = 6.5 × 1.2
Or, DE = 2 m
Example 2: Similar Quadrilaterals
Given:
ABCD ∼ EFGH
Solution:
Corresponding sides are proportional.
Or, a / 16 = 5 / 12
Or, a = 3 cm
Or, b / 15 = 5 / 12
Or, b = 4 cm
Or, c / 16.5 = 5 / 12
Or, c = 6.88 cm
Example 3: Similar Pentagons
Given:
ABCDE ∼ KLMNO
Solution:
Five pairs of similar triangles are formed.
Or, ratio of perimeters equals ratio of sides.
Or, MN / 10 = 12.5 / 25
Or, MN = 5 cm
Example 4: Right-angled Triangles
Given:
Two right-angled triangles.
Solution:
Angles are equal.
Or, triangles are similar by AAA.
Or, BP × PD = EP × PC
IMPORTANT QUESTIONS WITH SOLUTIONS (Clear Step-wise)
Question 1: Map Scale Problem
Given:
Length of room = 20 ft
Breadth of room = 18 ft
Length on map = 5 cm
To find:
Breadth on map = ?
Solution:
Step 1: Length and breadth of room are similar to length and breadth on map.
Step 2: Therefore, ratios of corresponding sides are equal.
Breadth on map / 5 = 18 / 20
Breadth on map = (18/20) × 5
Breadth on map = 4.5 cm
Answer: Breadth of the map = 4.5 cm
Question 2: Shadow of Pole Problem
Given:
Height of man = 6 ft
Shadow of man = 3 ft
Shadow of pole coincides with shadow of man
To find:
Height of pole = ?
Solution:
Step 1: The man and the pole form right-angled triangles with the ground.
Step 2: Angles of elevation of the sun are equal.
Step 3: Therefore, the triangles are similar by AA similarity.
Height of pole / Height of man = Shadow of pole / Shadow of man
Height of pole / 6 = Shadow of pole / 3
Hence, height of pole = (6 × Shadow of pole) / 3
Question 3: Triangle with Parallel Line
Given:
EF ∥ BC
EF = 3 cm, AC = 12 cm, AE = 4.5 cm, BC = 6 cm
To find:
AF and AB
Solution:
Step 1: Since EF ∥ BC, corresponding angles are equal.
Step 2: Therefore, ΔAEF ∼ ΔABC (AA similarity).
AE / AC = EF / BC
4.5 / 12 = 3 / 6
AF = AC − AE = 12 − 4.5 = 7.5 cm
AB = (AC × AE) / AF
Question 4: Prove Similarity (Proof)
Given:
MN ∥ QR
To prove:
ΔPMN ∼ ΔPQR
Proof:
Step 1: ∠PMN = ∠PQR (alternate interior angles)
Step 2: ∠PNM = ∠PRQ (alternate interior angles)
Step 3: Two angles of one triangle are equal to two angles of another triangle.
Therefore, ΔPMN ∼ ΔPQR (AA similarity).
Question 5: Rectangle Frame Problem
Given:
Outer frame breadth = 2 × inner frame breadth
Inner frame length = 3 cm
Solution:
Step 1: Rectangular frames are similar.
Step 2: Ratio of corresponding sides is equal.
Outer length / Inner length = Outer breadth / Inner breadth
Outer length / 3 = 2 / 1
Outer length = 6 cm
Exam Tip:
Always write:
Given → To find / To prove → Construction (if needed) → Steps → Conclusion