Answer:
2. D. [tex]\displaystyle 9\:in.[/tex]
1. C. [tex]\displaystyle 408\:in.^2[/tex]
Step-by-step explanation:
[tex]\displaystyle 2A_B + h[c + b + a] = S.A.[/tex]
Area of a Triangle [keep this in mind to find the base area]:
[tex]\displaystyle \frac{1}{2}bh = A, \frac{1}{2}hb = A, or\:\frac{hb}{2} = A[/tex]
Pythagorean Theorem [You will need this theorem to find the lengths of both hypotenuses on both triangular bases]:
[tex]\displaystyle a^2 + b^2 = c^2[/tex]
2. [tex]\displaystyle 2A_B + p[c + 3 + 4] = 120 → 2[6] + p[5 + 3 + 4] = 120 \\ \\ 12 + 12p = 120 → 108 = 12p; 9 = p[/tex]
1. [tex]\displaystyle 2A_B + 15[c + 6 + 8] = S.A. → 2[24] + 15[10 + 6 + 8] = S.A. \\ \\ 48 + 15[24] = S.A. → 48 + 360 = S.A.; 408 = S.A.[/tex]
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