Question

(a) In the given figure, $△ A BC$ is right angled triangle with $∠ A = 90°$ . 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-angled $△ A BC$ with $∠ A = 90°$ and $A D ⊥ BC$ : (i) Prove $△ D B A \sim △ D A C$ by AA similarity (both have a right angle and share $∠ B A D = ∠ A CB$ ). (ii) From similarity, $D A / D C = D B / D A$ so $D A^{2} = D C \cdot D B$ . (iii) Taking $D C = 16$ cm and $D B = 9$ cm, $D A^{2} = 16 \times 9 = 144$ , so $D A = 12$ cm. Then the area of $△ A BC = \frac{1}{2} \times BC \times D A = \frac{1}{2} \times 25 \times 12 = 150$ cm².

Step-by-step working

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

similar trianglesright angled triangleperpendicularprovebasic proportionality theoremarea of triangle

Approach

Step-by-step working

Concepts used

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