9 min readPathway sheet3-mark and 5-mark questions; shadow and height problems often come as the 4-mark case-based question

Using similarity to solve problems

Find two similar triangles, name them in matching order, then write equal ratios of corresponding sides to find a length or prove a result.

MethodMark angles → pick two triangles → prove them similar → write ratios → solve

Shadow problemsAt the same time the sun's rays make equal angles, so height/shadow is the same for every object

Perimeters of similar trianglesRatio of perimeters = ratio of corresponding sides (every side is k times as long, so their sum is k times too)

Product formFrom a/b = c/d, write a × d = b × c

The four-step method

1
Draw the figure and mark equal angles, right angles and parallel lines.
2
Pick the two triangles that contain the lengths you need.
3
Prove them similar with a criterion (AA is the most common), naming them in matching order.
4
Write the ratios of corresponding sides, put in the values and solve. Give the unit.

Example 1: height from a shadow

Problem

At 4 pm, a boy 1.5 m tall casts a shadow 2 m long. At the same time, a temple gopuram casts a shadow 48 m long. Find the height of the gopuram.

Working

The boy and the gopuram stand upright, so each makes a 90° angle with the ground.
At the same time, the sun's rays make the same angle with the ground.
So the two right triangles are similar (AA).
Height/shadow is the same for both: h/48 = 1.5/2
h = (1.5 × 48)/2 = 72/2 = 36 m

Answer

The gopuram is 36 m tall.

Example 2: a shadow from a street lamp

Problem

A lamp is fixed 4.8 m high on a pole. A girl 1.2 m tall stands 4.5 m from the foot of the pole. How long is her shadow?

Working

Let AB be the pole (4.8 m) and CD the girl (1.2 m). Let her shadow be DE = x m.
BD = 4.5 m, so BE = 4.5 + x.
In ΔEDC and ΔEBA: ∠E is common and ∠D = ∠B = 90°. So ΔEDC ~ ΔEBA (AA).
CD/AB = DE/BE, so 1.2/4.8 = x/(4.5 + x)
1/4 = x/(4.5 + x), so 4.5 + x = 4x
3x = 4.5, so x = 1.5 m

Answer

Her shadow is 1.5 m long.

Example 3: proving a product

Problem

In ΔABC, D is on AB and E is on AC such that ∠ADE = ∠ACB. Prove that AD × AB = AE × AC.

Working

In ΔADE and ΔACB: ∠DAE = ∠CAB (common angle).
∠ADE = ∠ACB (given).
So ΔADE ~ ΔACB (AA). Note the order: D matches C and E matches B.
Corresponding sides: AD/AC = AE/AB
Cross-multiply: AD × AB = AE × AC. Hence proved.

Answer

AD × AB = AE × AC

How to draw shadow and lamp-post figures

Draw the ground as a horizontal line.
Draw each pole, tower or person as a vertical line on it, and mark the right angle at the foot.
Draw the sun's ray or the lamp's light as a slanting line. It runs from the top of the tall object (or the lamp) to the tip of the shadow.
Label the heights and the shadow lengths, and write x for the unknown.
Mark the common angle (or the equal angle of the sun) in both triangles.

In real life

Long ago, people found the height of temple gopurams and palm trees by comparing shadows with a stick. You can try it at school with a metre scale on a sunny day.

Exam angle

3 marks: “A vertical stick 1.8 m long casts a shadow 1.2 m long. At the same time a tower casts a shadow 30 m long. Find the height of the tower.” The triangles are similar (AA). h/30 = 1.8/1.2, so h = 30 × 1.5 = 45 m.

5 marks: “In ΔABC, DE ∥ BC with D on AB and E on AC. AD = 3 cm, DB = 2 cm and BC = 7.5 cm. Prove ΔADE ~ ΔABC, find DE, and find the ratio of their perimeters.” ∠ADE = ∠ABC (corresponding angles) and ∠A is common, so AA. AB = 5 cm. DE/7.5 = 3/5, so DE = 4.5 cm. Perimeters are in the ratio 3 : 5.

Remember it

Mark, pick, prove, ratio. And keep the letters in matching order.