VABCD is a pyramid with a horizontal rectangular base ABCD. AC meets BD at N and VN is perpendicular to ABCD. Given that VN=12cm, AB=DC=8cm and AD=BC=6cm, calculate:
(a) AC
(b) AV
(c) the volume of the pyramid
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i can solve this geometry but i don't know exactly how to present it. So please when you amswer me, solve it clearly with the reasons:x
P/S: im a vienamese and my grammar of math isnt well. Can anybody give me the yahoo ID so i can contact? I want to improve my math skills
(a) AC
(b) AV
(c) the volume of the pyramid
*****
i can solve this geometry but i don't know exactly how to present it. So please when you amswer me, solve it clearly with the reasons:x
P/S: im a vienamese and my grammar of math isnt well. Can anybody give me the yahoo ID so i can contact? I want to improve my math skills
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AC and AV can be found with the Pythagorean Theorem, which is just a special case of the law of cosines in which the cosine is equal to zero.
As for part c, the below formula is valid and reduces to the formula V = Ah/3
Vol (VABCD) = 2/(3!)*DET[AN,AB,AZ]+2/(3!)*DET[AD,AN,AZ…
2/(3!) =1/3
Expand the determinants as scalar triple products. Since the cross products of the vectors in the rectangular base are parallel to NV, the dot product becomes a multiplication of the magnitude of the vector product by the height of the volume. Lastly the vector products are equal to the area of the base, you can prove that by drawing a parallelogram with sides AZ and AB and showing its area to be |AZ|*|AB|sin(theta).
Thus Vol = Ah/3
where A is the area of the base
h is the height of the pyramid
As for part c, the below formula is valid and reduces to the formula V = Ah/3
Vol (VABCD) = 2/(3!)*DET[AN,AB,AZ]+2/(3!)*DET[AD,AN,AZ…
2/(3!) =1/3
Expand the determinants as scalar triple products. Since the cross products of the vectors in the rectangular base are parallel to NV, the dot product becomes a multiplication of the magnitude of the vector product by the height of the volume. Lastly the vector products are equal to the area of the base, you can prove that by drawing a parallelogram with sides AZ and AB and showing its area to be |AZ|*|AB|sin(theta).
Thus Vol = Ah/3
where A is the area of the base
h is the height of the pyramid