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Standard IX

Mathematics

Pythagoras Theorem

In the given ...

Question

In the given figure D is the midpiont of side BC and If BC = a, AC = b, AB = c, ED = x, AD = p, and AE = h.
If $b^{2} - p^{2} - a x = \frac{a^{2}}{k}$ then k is -

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Solution

The correct option is B $4$
In $△ A E C$
${A C}^{2} = {A E}^{2} + {E C}^{2} b^{2} = h^{2} + {(\frac{a}{2} + x)}^{2} b^{2} = h^{2} + \frac{a^{2}}{4} + x^{2} + a x - (i)$
In $△ A E D$ , $p^{2} = h^{2} + x^{2}$
Substituting this in eq. $(i)$
$b^{2} = p^{2} + \frac{a^{2}}{4} + a x b^{2} - p^{2} - a x = \frac{a^{2}}{4}$
Given, $b^{2} - p^{2} - a x = \frac{a^{2}}{k}$
$∴ k = 4$

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Similar questions

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

Q. 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}

Q. 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

Q. In given figure, D is the mid-point of side BC and AE

⊥

BC. If

B C = a

A C = b

E D = x

A D = p

and

A E = h

, prove that:

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

Q. 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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In the given figure D is the midpiont of side BC and If BC = a, AC = b, AB = c, ED = x, AD = p, and AE = h.If b2−p2−ax=a2k then k is -

The correct option is B 4In △AECAC2=AE2+EC2b2=h2+(a2+x)2b2=h2+a24+x2+ax−(i)In △AED, p2=h2+x2Substituting this in eq.(i)b2=p2+a24+axb2−p2−ax=a24Given, b2−p2−ax=a2k∴k=4

In the given figure D is the midpiont of side BC and If BC = a, AC = b, AB = c, ED = x, AD = p, and AE = h.
If $b^{2} - p^{2} - a x = \frac{a^{2}}{k}$ then k is -