The left side was cut out but x will be on top and 5 in the lower square. The last one is the answer.
Answer:
<h3><u>Question 7</u></h3>
<u>Lateral Surface Area</u>
The bases of a triangular prism are the triangles.
Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).
![\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20L.A.%3D2%2810%20%5Ctimes%206%29%2B%283%20%5Ctimes%206%29%3D138%5C%3A%5C%3Am%5E2)
<u>Total Surface Area</u>
Area of the isosceles triangle:
![\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20A%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20base%20%5Ctimes%20height%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot3%20%5Ccdot%20%5Csqrt%7B10%5E2-1.5%5E2%7D%3D%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5C%3Am%5E2)
Total surface area:
![\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20T.A.%3D2%5C%3Abases%2BL.A.%3D2%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5Cright%29%2B138%3D167.66%5C%3A%5C%3Am%5E2%5C%3A%282%5C%3Ad.p.%29)
<u>Volume</u>
![\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20Vol.%3Darea%5C%3Aof%5C%3Abase%20%5Ctimes%20height%3D%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5Cright%29%20%5Ctimes%206%3D88.98%5C%3A%5C%3Am%5E3%5C%3A%282%5C%3Ad.p.%29)
<h3><u>Question 8</u></h3>
<u>Lateral Surface Area</u>
The bases of a hexagonal prism are the pentagons.
Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).
![\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20L.A.%3D5%285%20%5Ctimes%206%29%3D150%5C%3A%5C%3Acm%5E2)
<u>Total Surface Area</u>
Area of a pentagon:
![\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2](https://tex.z-dn.net/?f=%5Csf%20A%3D%5Cdfrac%7B1%7D%7B4%7D%5Csqrt%7B5%285%2B2%5Csqrt%7B5%7D%29%7Da%5E2)
where a is the side length.
Therefore:
![\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20A%3D%5Cdfrac%7B1%7D%7B4%7D%5Csqrt%7B5%285%2B2%5Csqrt%7B5%7D%29%7D%5Ccdot%205%5E2%3D43.01%5C%3A%5C%3Acm%5E2%5C%3A%282%5C%3Ad.p.%29)
Total surface area:
![\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20T.A.%3D2%5C%3Abases%2BL.A.%3D2%2843.01%29%2B150%3D236.02%5C%3A%5C%3Acm%5E2%5C%3A%282%5C%3Ad.p.%29)
<u>Volume</u>
![\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20Vol.%3Darea%5C%3Aof%5C%3Abase%20%5Ctimes%20height%3D43.011193...%20%5Ctimes%206%3D258.07%5C%3A%5C%3Acm%5E3%5C%3A%282%5C%3Ad.p.%29)
Answer:
si
Step-by-step explanation:
Answer:
the classmate is incorrect
Step-by-step explanation:
when you do the math for both inequalities they are equal to each other, despite the fact that one has multiplication and the other has addition.
For this case we must follow the steps below:
step 1:
We place each of the given points on a coordinate axis
Step 2:
We join the AC points (represented by the orange line)
We join the BD points (represented by the blue line)
It is observed that the resulting figure after placing the 4 points on a coordinate axis, turns out to be a rhombus.
In addition, the blue and orange lines turn out to be perpendicular, that is, they have an angle of 90 degrees between them. This can be verified by finding the slopes of each of the two straight lines (blue and orange), which must be opposite reciprocal, that is, they comply: ![m1 * m2 = -1](https://tex.z-dn.net/?f=m1%20%2A%20m2%20%3D%20-1)
In this case, the slope of the orange line is
and that of the blue line is ![m2 = -1](https://tex.z-dn.net/?f=m2%20%3D%20-1)
Then
, it is verified that they are perpendicular.
Thus, the conclusion is that ABCD is a rhombus and AC is perpendicular to BD.
Answer:
See attached image
Option D