7. Find the lateral and surface area.
This is a regular pyramid. A regular pyramid is a right pyramid whose base is a regular polygon and whose apex is directly above the centre of the base. The lateral surface area is the sum of the areas of all the lateral faces while the surface area is the sum of all the lateral faces plus its base. In this exercise, the base is a square so this is also a square pyramid. Next, we have:
LATERAL SURFACE AREA:
For the lateral sides, we have four identical triangles, so the area of a triangle can be found as:
[tex]A=\frac{bh}{2} \\ \\ Where: \\ \\ b:base \\ \\ h:height[/tex]
and the lateral surface will be four times this value:
[tex]S_{L}=4A[/tex]
The base of the triangle is the same as the base of the square. So:
[tex]b=9yd[/tex].
On the other hand, the height of the triangle is the slant height of the pyramid, which is:
[tex]h=10yd[/tex]
So the area of a triangle is:
[tex]A=\frac{(9)(10)}{2} \\ \\ A=45yd^2[/tex]
Therefore:
[tex]S_{L}=4(45)\\ \\ \boxed{S_{L}=180yd^2}[/tex]
SURFACE AREA:
The surface area can be found as:
[tex]S=S_{L}+A_{b} \\ \\ Where: \\ \\ A_{b}: Area \ of \ the \ base \\ \\ S_{L}: Lateral \ surface[/tex]
Calculating the area of the base, which is a square, we have:
[tex]A_{b}=b^2 \\ \\ A_{b}=9^2 \\ \\ A_{b}=81yd^2[/tex]
Therefore:
[tex]S=180+81 \\ \\ \boxed{261yd^2}[/tex]
8. Find the lateral and surface area.
In this case, we have another similar pyramid compared to the previous one, but we are given the height of the pyramid and we'll name it [tex]H[/tex] in capital letter. We know that the area of a triangle is:
[tex]A=\frac{bh}{2} \\ \\ Where: \\ \\ b:base \\ \\ h:height[/tex]
and the lateral surface will be:
[tex]S_{L}=4A[/tex]
To find [tex]h[/tex], which is the slant height of the pyramid, we need to use the Pythagorean theorem. Next, it is true that:
[tex]h=\sqrt{\left(\frac{b}{2}\right)^2+H^2} \\ \\ b=14 \\ H=12 \\ \\ h=\sqrt{\left(\frac{14}{2}\right)^2+12^2} \therefore h=\sqrt{193}[/tex]
So the area of a triangle is:
[tex]A=\frac{(14)(\sqrt{193})}{2} \\ \\ A=7\sqrt{193}ft^2[/tex]
Therefore:
[tex]S_{L}=4(7\sqrt{193})\\ \\ S_{L}=28\sqrt{193}ft^2 \approx 388.9884[/tex]
Rounding to the nearest tenth:
[tex]\boxed{S_{L}=389.0ft^2}[/tex]
SURFACE AREA:
We know that the surface area can be found as:
[tex]S=S_{L}+A_{b} \\ \\ Where: \\ \\ A_{b}: Area \ of \ the \ base \\ \\ S_{L}: Lateral \ surface[/tex]
Calculating the area of the base, which is a square, we have:
[tex]A_{b}=b^2 \\ \\ A_{b}=14^2 \\ \\ A_{b}=196ft^2[/tex]
Therefore:
[tex]S=28\sqrt{193}+196 \\ \\ S \approx 584.9884ft^2[/tex]
Rounding to the nearest tenth:
[tex]\boxed{S=585ft^2}[/tex]
9. Lateral surface area.
Here Patrick is making a paper model of castle. He has a net, so he can fold it to build up a pyramid. That's amazing, right? Well, recall that for a pyramid like that the lateral surface area is the area of the lateral faces, that are all triangles. Thus, for a triangle:
[tex]A=\frac{bh}{2} \\ \\ Where: \\ \\ h: \ slant \ height \ of \ the \ pyramid \\ \\ b: base \ of \ the \ pyramid[/tex]
The slant height of the pyramid is [tex]h=20cm[/tex] because this is the same height of the triangle. On the other hand, the base is [tex]b=15cm[/tex]. So:
[tex]A=\frac{15(20)}{2} \\ \\ h=150cm^2[/tex]
Next the lateral surface area is:
[tex]S_{L}=4(150) \\ \\ \boxed{S_{L}=150cm^2}[/tex]
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THE OTHER ANSWERS HAVE BEEN ATTACHED BELOW.
