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In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE.

In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

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Answer

Join A and B as shown in the figure below:

In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

From figure,

Since, ∠CDE = 90° so, ∠ADB = 90° (∵ vertically opposite angles are equal.)

In right angle triangle △ADB, by pythagoras theorem,

AB2=AD2+BD252=AD2+4225=AD2+16AD2=2516AD2=9AD=3 cm.\Rightarrow AB^2 = AD^2 + BD^2 \\[1em] \Rightarrow 5^2 = AD^2 + 4^2 \\[1em] \Rightarrow 25 = AD^2 + 16 \\[1em] \Rightarrow AD^2 = 25 - 16 \\[1em] \Rightarrow AD^2 = 9 \\[1em] \Rightarrow AD = 3 \text{ cm}.

Chords AE and CB intersect each other at D.

In △ADB and △CDE,

∠BAD = ∠DCE (∵ angles in same segment are equal.)

∠ADB = ∠CDE (∵ vertically opposite angles are equal.)

△ADB ~ △CDE. (By AA axiom)

Since △ADB ~ △CDE, Hence, the ratio of corresponding sides are equal.

ADCD=BDDE\therefore \dfrac{AD}{CD} = \dfrac{BD}{DE}

∴ AD × DE = CD × BD

⇒ 3 × DE = 9 × 4

⇒ DE = 363\dfrac{36}{3}

⇒ DE = 12 cm.

Hence, the length of DE = 12 cm.

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