Hello from MrBillDoesMath!
Answer:
108 cm^2
Discussion:
The area of the logo =
area of Trapezoid on top (I) + area of triangle on bottom (II)
I = (1/2) h ( b1 + b2) h = altitude, b1, b2 = bases
= (1/2) 6 ( 6 + 12)
= (1/2) 6 ( 18)
= 54
II = (1/2) b h
= (1/2) (12) ( 15 - 6) => altitude of triangle = 15 - 6 NOT (1/2) 15
= 54
I + II= 54 + 54 = 108
Thank you,
MrB
Sin C draw a line from C to the opposite side 21 and hypotenuse is the long side
Answer 21/29
Answer:
Step-by-step explanation:
4) ΔSTW ≅ ΔBFN . So, corresponding parts of congruent triangles are congruent.
a) BN = SW d) m∠W = m∠N
BN = 9 cm m∠W = 82°
b) TW = FN e) m∠B = m∠S
TW = 14 cm m∠B = 67°
c) BF = ST f) m∠B + m∠N + m∠F = 180°
BF = 17 cm 67 + 82 + m∠F = 180
149 + m∠F = 180
m∠F = 180 - 149
m∠F = 31°
5) ΔUVW ≅ ΔTSR
UV = TS
12x - 7 = 53
12x = 53+7
12x = 60
x = 60/12
x = 5
UW =TR
3z +14 = 50
3z = 50 - 14
3z = 36
z = 36/3
z = 12
SR =VW
5y - 33 = 57
5y = 57 + 33
5y = 90
y = 90/5
y = 18
7) ΔPHS ≅ ΔCNF
∠C = ∠P
4z - 32 = 36
4z = 36 + 32
4z = 68
z = 68/4
z = 17
∠H = ∠N
6x - 29 = 115
6x = 115 + 29
6x = 144
x = 144/6
x = 24
∠P + ∠H + ∠S = 180 {Angle sum property of triangle}
36 +115 + ∠S = 180
151 + ∠S = 180
∠S = 180 - 151
∠S = 29°
∠F = ∠S
3y - 1 = 29
3y = 29 + 1
3y = 30
y = 30/3
y = 10
8) ΔDEF ≅ ΔJKL
DE = 18 ; EF = 23
DF = 9x - 23
JL= 7x- 11
DF = JL {Corresponding parts of congruent triangles}
9x - 23 = 7x - 11
9x - 7x - 23 = -11
2x - 23 = -11
2x = -11 + 23
2x = 12
x = 12/2
x = 6
JK = DE {Corresponding parts of congruent triangles}
3y - 21 = 18
3y = 18 + 21
3y = 39
y = 39/3
y = 13
Y = 3x - 7 . . . . . (1)
6x - 2y = 12 . . . (2)
Putting (1) into (2) gives
6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
14 = 12
Therefore, there is no solution.
