Answer: 28,345.72
Step-by-step explanation:
46.93×60
Check the picture below.
now, we have a triangle with all three sides, thus we can use Heron's Area Formula on the triangle.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=10\\ b=26.695\\ c=22\\ s=29.3475 \end{cases} \\\\\\ A=\sqrt{29.3475(29.3475-10)(29.3475-26.695)(29.3475-22)} \\\\\\ A=\sqrt{29.3475(19.3475)(2.6525)(7.3475)}\implies A\approx \sqrt{11066.007} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill A\approx 105.195~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D10%5C%5C%20b%3D26.695%5C%5C%20c%3D22%5C%5C%20s%3D29.3475%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2829.3475-10%29%2829.3475-26.695%29%2829.3475-22%29%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B29.3475%2819.3475%29%282.6525%29%287.3475%29%7D%5Cimplies%20A%5Capprox%20%5Csqrt%7B11066.007%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20A%5Capprox%20105.195~%5Chfill)
Answer:
(D) -40
Step-by-step explanation:
Hope it helps.
According to me it is correct.
But please if it is wrong let me know.
Answer:
a) 49
b) 100 (I'm assuming that the 2 was supposed to be a square)
Step-by-step explanation:
Substitute the value in for x (the value being 7):
7² = 49
(7 + 3)² = (10)² = 100
Answer:
789 m²
Step-by-step explanation:
Consider the cross section created by a vertical plane through the apex of the pyramid and bisecting opposite sides. The cross section is an isosceles triangle with base 20 m and height 17 m. One side of this triangle is the slant height of the face of the pyramid.
The side of the triangle above can be found using the Pythagorean theorem. A median from the apex of the triangle will divide it into two right triangles, each with a base of 10 m and a height of 17 m. Then the hypotenuse is ...
s² = (10 m)² +(17 m)² = 389 m²
s = √389 m ≈ 19.723 m . . . . . slant height of one triangular face
__
The area of one triangular face is ...
A = (1/2)sb
where s is the slant height above, and b is the 20 m base of the face of the pyramid. There are 4 of these faces, so the total area is ...
total lateral area = 4A = 4(1/2)sb = 2sb = 2(19.723 m)(20 m)
total lateral area ≈ 789 m²
