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Pythagoras Theorem

In figure, D ...

Question

In figure, 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 $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$ .

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Solution

AE⊥BC (Hence, proved).

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Similar questions

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c^{2} = p^{2} - a x + \frac{a^{2}}{4}

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B C = a

A C = b

E D = x

A D = p

A E = h

c^{2} = p^{2} - a x + \frac{a^{2}}{4}

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b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{2}

b^{2} = p^{2} + a x + \frac{a^{2}}{4}

c^{2} = p^{2} - a x + \frac{a^{2}}{4}

b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{2}

In the given figure, D is the midpoint of side BC and AE $⊥$ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$

(ii) $c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

(iii) $(b^{2} + c^{2}) = 2 p^{2} + \frac{1}{2} a^{2}$

(iv) $(b^{2} - c^{2}) = 2 a x$

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