Question

(a) In the given figure, $△ A BC$ is right angled triangle with $∠ A = 90º$ . AD is perpendicular to BC.
Prove that:
(i) $△ D B A \sim △ D A C$
(ii) $D A^{2} = D B \times D C$
(iii) Find the area of $△ A BC$ when $D B = 9$ cm and $D C = 16$ cm.

(b) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Approach

In right triangle ABC with right angle at A, AD perpendicular to BC. (i) Prove $△ D B A \sim △ D A C$ using AA similarity. (ii) From similarity, $D A^{2} = D C \times D B$ . (iii) With DC = 16 cm and DB = 9 cm, $D A = 12$ cm. Area of ABC is $\frac{1}{2} \times 25 \times 12 = 150$ cm².

Step-by-step working

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(i) In $△ D B A$ and $△ D A C$ : $∠ B D A = ∠ C D A = 90°$ , $∠ B A D = 90° - B = ∠ A CB$ . Hence $△ D B A \sim △ D A C$ (AA).
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(ii) From similarity: $\frac{D A}{D C} = \frac{D B}{D A} \Rightarrow D A^{2} = D C \times D B$
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(iii) $D A^{2} = 16 \times 9 = 144 \Rightarrow D A = 12$ cm
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Area of $△ A BC = \frac{1}{2} \times BC \times A D = \frac{1}{2} \times 25 \times 12 = 150$ cm²
1 mark

Concepts used

similar trianglesright triangleperpendicularprovearea of trianglebasic proportionality theorem

Approach

Step-by-step working

Concepts used

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