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Triangles

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

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Solution

$A E ⊥ B C$
$B D = C D$
$B C = a$ , $A C = b$ , $A B = c$ , $E D = x$ , $A D = p$ , $A E = h$
To prove: $c^{2} = p^{2} - a x + \frac{a 2}{4}$
Proof:-
In $△ A B E$
$A B^{2} = A E^{2} + B E^{2}$
$A B^{2} = A E^{2} + (B C - E D - C D)^{2}$
$A B^{2} = A E^{2} + (B C - E D - \frac{B C}{2})^{2}$
$\Rightarrow A B^{2} = A E^{2} + (\frac{B C}{2} - E D)^{2}$
$\Rightarrow c^{2} = h^{2} + (\frac{a}{2} - x)^{2}$
$\Rightarrow c^{2} = h^{2} + \frac{a^{2}}{4} + x^{2} - a x$ ------------(1)

In $△ A E D$
$A E^{2} + E D^{2} = A D^{2}$
$\Rightarrow h^{2} + x^{2} = p^{2}$ ------------------(2)

Using Equation (2) in (1), we get
$c^{2} = h^{2} + x^{2} + \frac{a^{2}}{4} - a x$
$c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

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Similar questions

⊥

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} = 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 Fig. 7.223, D is the mid-point of side BC and $A E ⊥ B C$ . If BC = a AC=b, AB=c, ED=z, 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) $b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

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