Please find the surface area of the pyramid... I will mark you brainliest

Mathematics

1 answer:

Brut [27]2 years ago

3 0

Answer:

S.A. = 704 in²

Step-by-step explanation:

Surface area (S.A. ) = 1/2 lp + B where l is the slant height , p is the perimeter of the base, and B is the area of the base .

Given l = 14; The Base figure is a square so the p = 4· side = 4 · 16 = 64 and the area of the Base = s² = 256

S.A = 1/2 · 14 · 64 + 256

S.A. = 448 + 256

S.A. = 704 in²

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The vertices of a triangle ABC are A(7, 5), B(4, 2), and C(9, 2). What is measure of angle ABC? 30° 45° 56.31° 78.69°

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Length of the side BC (a) $=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5$

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We need to find ∠ABC or ∠B . So using Cosine rule, we will get...

$cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree$