Please solve q 58 ?

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2 Answers
May 20, 2018

Choice 3 is correct

Explanation:

Diagram of Right Triangles

Given: # \frac{\overline{AB}}{\overline{BC}} = \frac{\overline{CD}}{\overline{AC}} = \frac{\overline{AD}}{\overline{DE}} = k #

Required: Find # (\frac{\overline{AE}}{\overline{BC}})^2 #

Analysis: use Pythagorean Theorem #c = \sqrt{a^2 + b^2} #

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Solution: Let, # \overline{BC} = x #, # \because \frac{\overline{AB}}{\overline{BC}} = k,#

# \overline{AB} = kx #, use Pythagorean Theorem to find the value of # \overline{AC} # :

# \overline{AC} = \sqrt{\overline{BC}^2 + \overline{AB}^2 } = \sqrt{ x^2 + k^2x^2} = \sqrt{(x^2)(1+k^2)} = x\sqrt{1+k^2} #

# \overline{AC} = x\sqrt{1+k^2} #
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# \because \frac{\overline{CD}}{\overline{AC}} = k,# # \overline{CD} = \overline{AC} * k = xk\sqrt{1+k^2}#

Use Pythagorean Theorem to find the value of # \overline{AD}# :

# \overline{AD} = \sqrt{\overline{CD}^2 + \overline{AC}^2 #

# = \sqrt{(xk\sqrt{1+k^2})^2 + (x\sqrt{1+k^2})^2 }#

# = \sqrt{x^2k^2(1+k^2) +x^2(1+k^2) } #

# = \sqrt{x^2 [k^2(1+k^2) + 1(1+k^2)] } #

# = x\sqrt{(k^2+1)(1+k^2)} #, thus

# \overline{AD} = x(1+k^2) #

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# \because \frac{\overline{AD}}{\overline{DE}} = k,#

# \overline{DE} = \frac{\overline{AD}}{k} = \frac{x}{k} * (1+k^2)#

Use Pythagorean Theorem to find the value of # \overline{AE}# :

# \overline{AE}^2 = \sqrt{\overline{DE}^2 +\overline{AD}^2 = #

# = \sqrt{(frac{x}{k} * (1+k^2))^2 + (x(1+k^2))^2 #

# = \sqrt{(x^2/k^2)(1+k^2)^2 + (x^2)(1+k^2)^2 #

# = x\sqrt{(1/k^2 + 1)(1+k^2)^2 #

# = x\sqrt{\frac{1+k^2}{k^2} (1+k^2)^2} #

Thus,
# \overline{AE} = x\sqrt{\frac{(1+k^2)^3}{k^2} #
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# (\frac{\overline{AE}}{\overline{BC}})^2 #

# = (\frac{x\sqrt{\frac{(1+k^2)^3}{k^2}}}{x} )^2 #

# = (\sqrt{\frac{(1+k^2)^3}{k^2}})^2 #

Thus,
# (\frac{\overline{AE}}{\overline{BC}})^2 = \frac{(1+k^2)^3}{k^2} #


May 21, 2018

I got #(k^2+1)^3/k^2# which is choice (3).

Explanation:

We're gonna do every problem in Rahul's book!

This one is weird though with a diagram with right angles that aren't. Is it supposed to be 3D? The middle fraction is upside down compared to the others; let's assume that's correct.

Rahul, you deserve a better book.

We'll renotate for sanity:

#b=AB, c=AC, d=AD, e=AE, p=BC, q=CD, r=DE#

We're given

#k = b/p = q/c = d/r #

We want to find #e^2/p^2,# a hint that we'll never have to write a square root.

#b=pk, quad quad q=kc, quad quad r=d/k#

#c^2 = b^2 + p^2 = p^2k^2+p^2 =p^2(1 + k^2) #

# d^2 = c^2 + q^2=c^2+(kc)^2=c^2(1+k^2) =p^2(1+k^2)^2#

#e^2 = d^2+r^2 = d^2(1+1/k^2)= p^2(1+k^2)^2(1+1/k^2)#

#e^2/p^2 = (1+k^2)^2(1+1/k^2) =(k^2+1)^3/k^2#

Choice (3)