+0

helpppppp urgent plzzzzz

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433
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Mar 5, 2015

#3
+95360
+5

Well, XY is 10 you are right about that.

So long as the method is correct I am not too concerned.

I am sure you can fix the careless errors I thought there was likely to be some.

I take it you were happy with the method ?

There is probably a much easier way,  I have a reputation for taking the scenic route.  :))

Mar 5, 2015

#1
+95360
+5

I am not going to do this very elegently but perhaps i can do it.

First I am going to cut this square pyramid in half so I need a few more points.

X can be the midpoint of AB

Y is the midpoint of CD and

Z is the midpoint of EF

Consider triangle VXB       <VXB=90,      XB=5    VB=8    find VX

vx= sqrt( 64-25) = sqrt(39)

Now consider triangle  VXY          VX=VY=sqrt(39),    XY=8     Find <VXY (I'm going to call it $$\theta$$ )

$$\\39=39+64-(2*\sqrt{39}*8cos\theta)\\\\ 64=16\sqrt{39}*cos\theta\\\\ 4/\sqrt{39}=cos\theta\\\\$$

$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{4}}}{{\sqrt{{\mathtt{39}}}}}}\right)} = {\mathtt{50.169\: \!945\: \!446\: \!964^{\circ}}}$$

so <VXY= 50.17°

NOW lets consider triangle XZY       <ZYX=30°,    <ZXY=50.17°      XY=8      Find XZ

$$\\\frac{XZ}{sin30}=\frac{8}{sin(180-30-50.17)}\\\\ \frac{XZ}{sin30}=\frac{8}{sin(98.3)}\\\\ XZ=\frac{8}{sin(98.3)}\times sin30\\\\$$

$${\frac{{\mathtt{8}}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{98.3}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)} = {\mathtt{4.042\: \!340\: \!325\: \!221\: \!254\: \!6}}$$

so    $$XZ\approx 4.04 cm$$

NOW I am going to consider triangle AVB

VX=sqrt(39),   XZ=4.04      so     VZ=sqrt(39)-4.04 = 2.205 cm approx

NOW triangle  EFV is similar to triangle  ABV     so

$$\\\frac{EF}{AB}=\frac{VZ}{VX}\\\\ \frac{EF}{10}=\frac{2.205}{\sqrt{39}}\\\\ EF=\frac{2.205}{\sqrt{39}}\times 10\\\\$$

$${\frac{{\mathtt{2.205}}}{{\sqrt{{\mathtt{39}}}}}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{3.530\: \!825\: \!791\: \!402\: \!171\: \!3}}$$

So if I have not done anything incorrectly the answer is  EF= 3.53cm    (approximately)

BLAST, I FOUND THE WRONG ONE.  BUMMER!!!          (ﾉ °益°)ﾉ 彡

nevermind:

$$\\\frac{VE}{VA}=\frac{VZ}{VX}\\\\ \frac{VE}{8}=\frac{2.205}{\sqrt{39}}\\\\ VE=\frac{2.205}{\sqrt{39}}\times 8\\\\$$

$${\frac{{\mathtt{2.205}}}{{\sqrt{{\mathtt{39}}}}}}{\mathtt{\,\times\,}}{\mathtt{8}} = {\mathtt{2.824\: \!660\: \!633\: \!121\: \!737}}$$

AE=AV-EV

AE=8-2.825

$${\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2.825}} = {\frac{{\mathtt{207}}}{{\mathtt{40}}}} = {\mathtt{5.175}}$$

AE is approx 5.18 cm

Mar 5, 2015
#2
+5

The method looks good Melody, but,  almost at the top, shouldn't XY = 10 rather than 8 ?

Mar 5, 2015
#3
+95360
+5

Well, XY is 10 you are right about that.

So long as the method is correct I am not too concerned.

I am sure you can fix the careless errors I thought there was likely to be some.

I take it you were happy with the method ?

There is probably a much easier way,  I have a reputation for taking the scenic route.  :))

Melody Mar 5, 2015