Question

The corresponding sides of $△ A BC$ and $△ PQR$ are in the ratio $3 : 5$ . $A D ⊥ BC$ and $PS ⊥ QR$ as shown in the following figures:

(i) Prove that $△ A D C \sim △ PSR$
(ii) If $A D = 4$ cm, find the length of PS.
(iii) Using (ii) find ar $(△ A BC)$ : ar $(△ PQR)$

Approach

$△ A BC \sim △ PQR$ with ratio $3 : 5$ . AD and PS are altitudes. (i) Prove $△ A D C \sim △ PSR$ by AA similarity (both have right angles and $∠ C = ∠ R$ ). (ii) From similarity, $\frac{A D}{PS} = \frac{3}{5}$ . Given AD = 4 cm, PS = $\frac{20}{3}$ cm. (iii) Ratio of areas = $(\frac{3}{5})^{2} = \frac{9}{25}$ .

Step-by-step working

1
Given $\frac{A B}{PQ} = \frac{BC}{QR} = \frac{A C}{PR} = \frac{3}{5} \Rightarrow △ A BC \sim △ PQR \Rightarrow ∠ C = ∠ R$ .
0.5 marks
2
(i) In $△ A D C$ and $△ PSR$ : $∠ A D C = ∠ PSR = 90°$ and $∠ C = ∠ R \Rightarrow △ A D C \sim △ PSR$ by AA.
1 mark
3
(ii) $\frac{A D}{PS} = \frac{A C}{PR} = \frac{3}{5} \Rightarrow \frac{4}{PS} = \frac{3}{5} \Rightarrow PS = \frac{20}{3}$ cm.
1.5 marks
4
(iii) $\frac{ar ( △ A BC )}{ar ( △ PQR )} = \frac{\frac{1}{2} \times BC \times A D}{\frac{1}{2} \times QR \times PS} = \frac{BC}{QR} \times \frac{A D}{PS} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ .
2 marks

Concepts used

similar trianglesproveratio of areasbasic proportionality theoremparallel lines

Approach

Step-by-step working

Concepts used

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