Answer:
part 1:
1) AB = BC = 10
2) DE = sqrt(10² - 9²) = sqrt(19)
3) CED = 90⁰
4) BCD = (90⁰ - 53⁰).2 = 74⁰
part 2:
because FGHJ is a rhombus
=> GK = KJ
<=> 3x + 1 = x + 9
<=> 2x = 8
<=> x = 4
and we have:
GHK + FGK = 90⁰
<=> (4y - 8)/2 + y + 4 = 90⁰
<=> 2y - 4 + y + 4 = 90⁰
<=> 3y = 90⁰
<=> y = 30
5) with x = 4 => GJ = 2KJ = 2.(4 + 9) = 26
6) JK = 4 + 9 = 13
7) with y = 30, HFJ = 90 - FGK = 90 - 34 = 56⁰
8) HJF = 2.FGK = 2.34 = 68⁰
part 3:
because MNLO is rectangle
=> MP = PO
<=> 5a - 7 = 2a + 8
<=> 3a = 15
<=> a = 5
9) with a = 5, MO = 2.PO = 2.(2.5 + 8) = 2.18 = 36
10) NP = PO = 2.5 + 8 = 18
11) LM² = LN² - MN² = MO² - 26² = 36² - 26² = 620
=> LM = 2sqrt(156) = 24,9
12) tan NMO = NO/MN = LM/MN = 24.9/26
=> NMO = 43,8⁰
part 4:
because RSTU is square
=> SR = RU
=> 2k + 7 = 3k - 2
<=> k = 9
with k = 9 =>RU = SR = 2.9 + 7 = 25
13) RV = 1/2.SU = 1/2.sqrt(25² + 25²) = 17,7
14) SU = 2RV = 2.17,7 = 35,4
15) perimeter RSTU = 25.4 = 100
16) area RSTU = 25² = 625
