Solved Examples

How to Use This Section

This is your practice powerhouse for Triangles. Below are 30+ fully worked problems spanning the whole chapter — similar figures, the BPT, similarity criteria, area ratios, and Pythagoras — arranged roughly easy to hard.

How to study: Try each problem with the solution covered, then check the steps. In geometry, marks are awarded for stating the correct theorem/criterion and for clear step-by-step reasoning.

Keep these handy:

• BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ when $DE \parallel BC$.
• Similarity: AA, SSS, SAS.
• Area ratio of similar triangles $= (\text{side ratio})^2$.
• Pythagoras: $\text{hyp}^2 = \text{leg}_1^2 + \text{leg}_2^2$.

Solved Examples

Example 1: BPT length

In $\triangle ABC$, $DE \parallel BC$, $AD = 3$, $DB = 4$, $AE = 6$. Find $EC$.

Solution:

1. $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{3}{4} = \dfrac{6}{EC}$.
2. $EC = \dfrac{6 \times 4}{3} = 8$.

Final Answer: $EC = 8$.

Takeaway: Apply BPT and cross-multiply.

Example 2: Prove similar (AA)

In $\triangle ABC$, $DE \parallel BC$. Prove $\triangle ADE \sim \triangle ABC$.

Solution:

1. $\angle A$ is common.
2. $\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$).
3. By AA, $\triangle ADE \sim \triangle ABC$.

Final Answer: $\triangle ADE \sim \triangle ABC$ (AA).

Takeaway: Parallel line ⇒ equal corresponding angles ⇒ AA similarity.

Example 3: SSS similarity

Are triangles with sides $(4, 6, 8)$ and $(6, 9, 12)$ similar?

Solution:

1. $\dfrac{4}{6} = \dfrac{6}{9} = \dfrac{8}{12} = \dfrac{2}{3}$.
2. All ratios equal ⇒ SSS similarity.

Final Answer: Yes, similar (SSS).

Takeaway: Equal side ratios ⇒ SSS.

Example 4: Find a side by similarity

$\triangle ABC \sim \triangle PQR$, $AB = 5$, $PQ = 10$, $AC = 6$. Find $PR$.

Solution:

1. $\dfrac{AB}{PQ} = \dfrac{AC}{PR} \Rightarrow \dfrac{5}{10} = \dfrac{6}{PR}$.
2. $PR = 12$.

Final Answer: $PR = 12$.

Takeaway: Equal corresponding ratios give the unknown side.

Example 5: Area ratio

Two similar triangles have sides in ratio $5 : 6$. Find the ratio of their areas.

Solution:

1. Area ratio $= 5^2 : 6^2 = 25 : 36$.

Final Answer: $25 : 36$.

Takeaway: Square the side ratio.

Example 6: Pythagoras find side

A right triangle has legs 9 cm and 12 cm. Find the hypotenuse.

Solution:

1. Hypotenuse $= \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ cm.

Final Answer: 15 cm.

Takeaway: $(9, 12, 15) = 3 \times (3, 4, 5)$.

Example 7: Converse of Pythagoras

Is a triangle with sides 9, 40, 41 right-angled?

Solution:

1. $9^2 + 40^2 = 81 + 1600 = 1681 = 41^2$.
2. Equal, so right-angled (right angle opposite 41).

Final Answer: Yes, right-angled.

Takeaway: $(9, 40, 41)$ is a Pythagorean triple.

Example 8: Converse of BPT

In $\triangle ABC$, $D$ on $AB$, $E$ on $AC$, $AD = 4$, $DB = 6$, $AE = 6$, $EC = 9$. Is $DE \parallel BC$?

Solution:

1. $\dfrac{AD}{DB} = \dfrac{4}{6} = \dfrac{2}{3}$; $\dfrac{AE}{EC} = \dfrac{6}{9} = \dfrac{2}{3}$.
2. Equal ratios ⇒ by the converse, $DE \parallel BC$.

Final Answer: Yes, $DE \parallel BC$.

Takeaway: Equal ratios prove parallelism.

Example 9: Shadow problem

A vertical pole 6 m high casts a shadow 4 m long. At the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

1. Triangles are similar (equal sun angle, AA).
2. $\dfrac{6}{4} = \dfrac{h}{28} \Rightarrow h = \dfrac{6 \times 28}{4} = 42$ m.

Final Answer: 42 m.

Takeaway: Height : shadow is constant for objects at the same time.

Example 10: Area to side

The areas of two similar triangles are 81 cm² and 144 cm². If a side of the smaller is 9 cm, find the corresponding side of the larger.

Solution:

1. Side ratio $= \sqrt{\dfrac{81}{144}} = \dfrac{9}{12} = \dfrac{3}{4}$.
2. Corresponding side $= 9 \times \dfrac{4}{3} = 12$ cm.

Final Answer: 12 cm.

Takeaway: Square-root the area ratio for the side ratio.

Example 11: Diagonal of a square

Find the diagonal of a square of side 10 cm.

Solution:

1. Diagonal $= \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$ cm.

Final Answer: $10\sqrt{2}$ cm.

Takeaway: A square's diagonal is $\sqrt2 \times$ side.

Example 12: BPT with algebra

In $\triangle ABC$, $DE \parallel BC$, $AD = x$, $DB = x - 2$, $AE = x + 2$, $EC = x - 1$. Find $x$.

Solution:

1. $\dfrac{x}{x-2} = \dfrac{x+2}{x-1} \Rightarrow x(x-1) = (x+2)(x-2)$.
2. $x^2 - x = x^2 - 4 \Rightarrow x = 4$.

Final Answer: $x = 4$.

Takeaway: Cross-multiply and simplify.

Example 13: Similar triangles, perimeter

$\triangle ABC \sim \triangle DEF$ with $\dfrac{AB}{DE} = \dfrac{2}{5}$. If the perimeter of $\triangle ABC$ is 16 cm, find the perimeter of $\triangle DEF$.

Solution:

1. Perimeter ratio = side ratio = $\dfrac{2}{5}$.
2. $\dfrac{16}{P_{DEF}} = \dfrac{2}{5} \Rightarrow P_{DEF} = 40$ cm.

Final Answer: 40 cm.

Takeaway: Perimeters share the side ratio.

Example 14: Altitude on hypotenuse

In right $\triangle ABC$ (right angle at $B$), $BD \perp AC$. If $AB = 6$, $BC = 8$, find $BD$.

Solution:

1. $AC = \sqrt{36 + 64} = 10$.
2. Area two ways: $\dfrac{1}{2}(AB)(BC) = \dfrac{1}{2}(AC)(BD) \Rightarrow 6 \times 8 = 10 \times BD$.
3. $BD = \dfrac{48}{10} = 4.8$.

Final Answer: $BD = 4.8$.

Takeaway: Equate the two area expressions to find the altitude.

Example 15: Midpoint theorem

In $\triangle ABC$, $D$ and $E$ are midpoints of $AB$ and $AC$. If $BC = 10$ cm, find $DE$.

Solution:

1. By the midpoint theorem, $DE \parallel BC$ and $DE = \dfrac{1}{2}BC$.
2. $DE = \dfrac{1}{2}(10) = 5$ cm.

Final Answer: $DE = 5$ cm.

Takeaway: The segment joining midpoints is half the third side.

Example 16: SAS similarity proof

In $\triangle ABC$ and $\triangle ADE$, $\angle A$ is common, $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{1}{2}$. Show $\triangle ADE \sim \triangle ABC$.

Solution:

1. $\angle A = \angle A$ (common, included angle).
2. $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (given, sides including the angle).
3. By SAS, $\triangle ADE \sim \triangle ABC$.

Final Answer: $\triangle ADE \sim \triangle ABC$ (SAS).

Takeaway: Common angle + proportional including sides ⇒ SAS.

Example 17: Area ratio in a figure

In $\triangle ABC$, $DE \parallel BC$ with $AD : DB = 2 : 3$. Find the ratio of areas of $\triangle ADE$ and trapezium $DBCE$.

Solution:

1. $\dfrac{AD}{AB} = \dfrac{2}{5}$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(ABC)} = \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}$.
2. ar(trapezium) $=$ ar($ABC$) $-$ ar($ADE$) $= 25 - 4 = 21$ parts.
3. Ratio ar($ADE$) : ar(trapezium) $= 4 : 21$.

Final Answer: $4 : 21$.

Takeaway: Subtract the small triangle's area from the whole for the trapezium.

Example 18: Pythagoras word problem

The foot of a 10 m ladder is 6 m from a wall. How high up the wall does it reach?

Solution:

1. Height $= \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ m.

Final Answer: 8 m.

Takeaway: $(6, 8, 10)$ is a scaled $(3, 4, 5)$ triple.

Example 19: Find both segments (BPT)

In $\triangle ABC$, $DE \parallel BC$, $\dfrac{AD}{DB} = \dfrac{2}{3}$, $AB = 10$ cm. Find $AD$ and $DB$.

Solution:

1. $AD : DB = 2 : 3$, total 5 parts $= AB = 10$.
2. $AD = \dfrac{2}{5}(10) = 4$, $DB = \dfrac{3}{5}(10) = 6$.

Final Answer: $AD = 4$ cm, $DB = 6$ cm.

Takeaway: Divide the whole side in the given ratio.

Example 20: Diagonals of trapezium

In trapezium $ABCD$ ($AB \parallel DC$), diagonals meet at $O$. If $AB = 4$ cm, $DC = 8$ cm, and $AO = 3$ cm, find $OC$.

Solution:

1. $\triangle AOB \sim \triangle COD$ (AA), so $\dfrac{AO}{OC} = \dfrac{AB}{DC} = \dfrac{4}{8} = \dfrac{1}{2}$.
2. $\dfrac{3}{OC} = \dfrac{1}{2} \Rightarrow OC = 6$ cm.

Final Answer: $OC = 6$ cm.

Takeaway: In such a trapezium, $\dfrac{AO}{OC} = \dfrac{AB}{DC}$.

Example 21: Find the third side type

The sides of a triangle are 6, 8, 11. Is the angle opposite the 11-side acute, right, or obtuse?

Solution:

1. Compare $11^2 = 121$ with $6^2 + 8^2 = 36 + 64 = 100$.
2. Since $121 > 100$, the angle opposite 11 is obtuse.

Final Answer: Obtuse.

Takeaway: If $c^2 > a^2 + b^2$, the angle opposite $c$ is obtuse; if $<$, acute; if $=$, right.

Example 22: Similar triangles in a figure

Two triangles $\triangle ABC$ and $\triangle AMP$ are right-angled at $B$ and $M$ respectively, sharing angle $A$. Prove they are similar and find $MP$ if $BC = 6$, $AB = 8$, $AM = 4$.

Solution:

1. $\angle A$ common, $\angle B = \angle M = 90°$ ⇒ $\triangle ABC \sim \triangle AMP$ (AA).
2. $\dfrac{BC}{MP} = \dfrac{AB}{AM} \Rightarrow \dfrac{6}{MP} = \dfrac{8}{4} = 2 \Rightarrow MP = 3$.

Final Answer: Similar (AA); $MP = 3$.

Takeaway: Common angle + right angles ⇒ AA, then use proportional sides.

Example 23: Area ratio of medians

The ratio of corresponding medians of two similar triangles is $4 : 7$. Find the ratio of their areas.

Solution:

1. Median ratio = side ratio = $4 : 7$.
2. Area ratio $= 4^2 : 7^2 = 16 : 49$.

Final Answer: $16 : 49$.

Takeaway: Medians scale like sides; areas like the square.

Example 24: Diagonal of cuboid-base (2D)

A rectangular field is 40 m long and 30 m wide. Find the length of the diagonal path across it.

Solution:

1. Diagonal $= \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$ m.

Final Answer: 50 m.

Takeaway: $(30, 40, 50) = 10 \times (3, 4, 5)$.

Example 25: Prove a relation (Pythagoras)

In an equilateral triangle $ABC$ with side $a$, prove that the square of the altitude is $\dfrac{3}{4}a^2$.

Solution:

1. The altitude splits the base into $\dfrac{a}{2}$ each, forming a right triangle with hypotenuse $a$.
2. $h^2 = a^2 - \left(\dfrac{a}{2}\right)^2 = a^2 - \dfrac{a^2}{4} = \dfrac{3a^2}{4}$.

Final Answer: $h^2 = \dfrac{3}{4}a^2$.

Takeaway: Use Pythagoras with half the base.

Note: so $h = \dfrac{\sqrt3}{2}a$.

Example 26: BPT to prove parallel and find length

In $\triangle ABC$, $D$ on $AB$ with $AD = 2.4$, $AB = 6$; $E$ on $AC$ with $AE = 3.2$, $AC = 8$. Show $DE \parallel BC$.

Solution:

1. $\dfrac{AD}{AB} = \dfrac{2.4}{6} = 0.4$; $\dfrac{AE}{AC} = \dfrac{3.2}{8} = 0.4$.
2. Equal ratios (part-to-whole) ⇒ by the converse of BPT, $DE \parallel BC$.

Final Answer: $DE \parallel BC$.

Takeaway: The part-to-whole form also tests parallelism.

Example 27: Areas with a common height

$\triangle ABC \sim \triangle PQR$, and the ratio of their areas is $9 : 25$. If $PR = 20$ cm, find $AC$.

Solution:

1. Side ratio $= \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$.
2. $\dfrac{AC}{PR} = \dfrac{3}{5} \Rightarrow \dfrac{AC}{20} = \dfrac{3}{5} \Rightarrow AC = 12$ cm.

Final Answer: $AC = 12$ cm.

Takeaway: Convert area ratio to side ratio, then find the length.

Example 28: Right triangle inside

In $\triangle ABC$ right-angled at $C$, $AC = 5$, $BC = 12$. A point $D$ on $AB$ satisfies $CD \perp AB$. Find $AB$ and the area of $\triangle ABC$.

Solution:

1. $AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$.
2. Area $= \dfrac{1}{2}(5)(12) = 30$ sq units.

Final Answer: $AB = 13$; area $= 30$.

Takeaway: $(5, 12, 13)$ is a Pythagorean triple; area uses the two legs.

Example 29: Prove triangles similar (vertical angles)

Two line segments $AC$ and $BD$ intersect at $O$ with $\dfrac{OA}{OC} = \dfrac{OB}{OD}$. Show $\triangle AOB \sim \triangle COD$.

Solution:

1. $\angle AOB = \angle COD$ (vertically opposite angles).
2. The sides including these angles satisfy $\dfrac{OA}{OC} = \dfrac{OB}{OD}$ (given).
3. By SAS, $\triangle AOB \sim \triangle COD$.

Final Answer: $\triangle AOB \sim \triangle COD$ (SAS).

Takeaway: Vertically opposite angle + proportional including sides ⇒ SAS.

Example 30: Combine BPT and similarity

In $\triangle ABC$, $DE \parallel BC$ meets $AB$ at $D$ and $AC$ at $E$. If $AD = 4$ cm, $DB = 8$ cm, and $DE = 5$ cm, find $BC$.

Solution:

1. $\triangle ADE \sim \triangle ABC$ with ratio $\dfrac{AD}{AB} = \dfrac{4}{12} = \dfrac{1}{3}$.
2. $\dfrac{DE}{BC} = \dfrac{1}{3} \Rightarrow \dfrac{5}{BC} = \dfrac{1}{3} \Rightarrow BC = 15$ cm.

Final Answer: $BC = 15$ cm.

Takeaway: The similarity ratio applies to $DE$ and $BC$ too.

Example 31: Mixed — tower and stick

A 1.5 m tall stick casts a 3 m shadow. A nearby tower casts a 60 m shadow at the same time. Find the tower's height.

Solution:

1. AA similarity (equal sun angle): $\dfrac{1.5}{3} = \dfrac{h}{60}$.
2. $h = \dfrac{1.5 \times 60}{3} = 30$ m.

Final Answer: 30 m.

Takeaway: Height : shadow is constant at a given time of day.