QQuestion
In △ABC, ∠ABC = 90°, AB = 20 cm and AC = 25 cm. DE is perpendicular to AC such that ∠DEA = 90° and DE = 3 cm, as shown.
(a) Prove that △ABC ∼ △AED.
(b) Find BC, AD and AE.
(c) If BCED represents land on a map whose actual area is 576 m², find the scale factor of the map.
✓Answer
(a) In △ABC and △AED:
∠ABC = ∠AED = 90°
∠BAC = ∠DAE, as it is the common angle at A.
Therefore, △ABC ∼ △AED by the AA criterion.
(b) In right-angled △ABC:
AC² = AB² + BC²
⇒ 25² = 20² + BC²
⇒ 625 = 400 + BC²
⇒ BC² = 225
⇒ BC = 15 cm
Corresponding sides of similar triangles are proportional:
\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AD}} \frac{20}{\mathrm{AE}}=\frac{15}{3}⇒ \frac{20}{\mathrm{AE}}=5
⇒ AE = 4 cm
\frac{15}{3}=\frac{25}{\mathrm{AD}}⇒ 5=\frac{25}{\mathrm{AD}}
⇒ AD = 5 cm
(c) Area of △ABC = \frac{1}{2} × 20 × 15 = 150 cm²
Area of △AED = \frac{1}{2} × 4 × 3 = 6 cm²
Area of BCED on the map = 150 − 6 = 144 cm²
Actual area = 576 m² = 5,760,000 cm²
If the linear scale factor is k:
k2 = \frac{144}{5760000}=\frac{1}{40000}
⇒ k = \frac{1}{200}
Therefore, the scale factor is 1 : 200.
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