Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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D is the mid point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that c2 = p2-ax+a24 - Mathematics

Sum

D is the mid point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that c2 = `"p"^2 - "a"x + "a"^2/4`

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Solution

From the figure, D is the midpoint of BC.

We have ∠AED = 90°

∴ ∠ADE < 90° and ∠ADC > 90°

i.e. ∠ADE is acute and ∠ADC is obtuse,

In ∆ABD, ∠ADE is an acute angle.

AB2 = AD2 + BD2 – 2BD . DE

⇒ AB2 = AD2 + (12BC)2 – 2 × 12 BC . DE

⇒ AB2 = AD2 + 14 BC2 – BC . DE

⇒ AB2 = AD2 – BC . DE + 14 BC2

⇒ c2 = p2 – ax + 14 a2

Hence proved.

Concept: Similarity
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APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 4 Geometry
Unit Exercise – 4 | Q 6. (ii) | Page 201
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