Week 8 · Similarity of triangles: the AA, SSS and SAS cases, right triangles and real heights · 3 lessons
3
Similarity in right triangles, and heights from shadows and mirrors
Mathematics Prep 3 Term 1, Unit 3, Lesson 2
GoalUse the altitude to the hypotenuse to find lengths, and use similar triangles to find heights from shadows and mirrors.
New words
hypotenuse · the longest side of a right triangle, opposite the right anglealtitude · the perpendicular line from a vertex to the opposite sideshadow · the dark region behind an object; the sun’s rays are parallelreflection · a ray hits a mirror and leaves at the same angle it arrived
Explanation
Draw the altitude AD from the right angle A of the right triangle ABC to the hypotenuse BC. It cuts the triangle into two smaller right triangles. △DBA shares angle B with △ABC and both have a right angle, so △DBA ~ △ABC by AA. In the same way △DAC ~ △ABC, and so △DBA ~ △DAC. From the ratios we get three useful facts: AD² = BD × DC, AB² = BD × BC and AC² = CD × BC. Similar triangles also measure heights. At the same moment the sun makes equal angles with the ground, so an object and its shadow form a triangle similar to the triangle of any other object and its shadow. With a mirror on the ground the angle of arrival equals the angle of reflection, so the triangle of the person and the triangle of the tower are similar.
Worked examples
In right △ABC (∠A = 90°) with AB = 3, AC = 4, BC = 5, AD ⊥ BC. Find AD, BD and DC. △DBA ~ △ABC gives BD ÷ AB = AB ÷ BC, so BD = 3² ÷ 5 = 1.8. Similarly DC = 4² ÷ 5 = 3.2. Check BD + DC = 5 ✓. AD² = BD × DC = 1.8 × 3.2 = 5.76, so AD = 2.4.
A student 1.6 m tall casts a shadow 1.2 m long. At the same time a pole casts a shadow 7.5 m long. How tall is the pole? The two right triangles have equal angles (AA), so height ÷ shadow is the same: 1.6 ÷ 1.2 = h ÷ 7.5, so h = 7.5 × 1.6 ÷ 1.2 = 10 m.
Eyes of a boy are 1.5 m above the ground. He stands 2 m from a mirror lying on the ground and sees the top of a tower in it. The mirror is 16 m from the tower’s foot. Find the tower’s height. Both triangles are right-angled and the angles at the mirror are equal, so they are similar: h ÷ 16 = 1.5 ÷ 2, so h = 16 × 1.5 ÷ 2 = 12 m.
Class activity
Shadow measuring. On a sunny day measure your own height and your shadow, and then the shadow of a tree, a flagpole or a wall in the same minutes. Use similar triangles to estimate its height.
Practice
1
In right △ABC (∠A = 90°), AD ⊥ BC, BD = 9 and DC = 16. Find AD, AB and AC.
AD² = 9 × 16 = 144, AD = 12. BC = 25. AB² = 9 × 25 = 225, AB = 15. AC² = 16 × 25 = 400, AC = 20.
2
Why is △DBA ~ △ABC in a right triangle with the altitude AD?
Both have a right angle (∠ADB = ∠BAC = 90°) and they share ∠B, so AA applies.
3
A person 1.8 m tall has a shadow 2.4 m long. A building’s shadow is 20 m at the same time. Find its height.
1.8 ÷ 2.4 = h ÷ 20, so h = 20 × 1.8 ÷ 2.4 = 15 m.
4
Eyes 1.6 m high, the person stands 2.5 m from a mirror on the ground, and the mirror is 20 m from a tree. Find the tree’s height.
h ÷ 20 = 1.6 ÷ 2.5, so h = 20 × 1.6 ÷ 2.5 = 12.8 m.
5
In the 3-4-5 triangle, find the area two ways: with legs 3 and 4, and with base 5 and the altitude AD. Do you get the same area?
