What is the volume of a box with a length of 3 and 3/4, width of 3, and height of 3 and 1/4?
1 answer:
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Answer: x = 15, ∠K = 45.7°, ∠L = 45.7°, ∠M = 88.6°
<u>Step-by-step explanation:</u>
Since ∠K ≅ ∠L, then ΔKLM is an isoceles triangle with base KL
KM ≅ LM
3x + 23 = 7x - 37
23 = 4x - 37
60 = 4x
15 = x
KM = LM = 3x + 23
= 3(15) + 23
= 45 + 23
= 68
KL = 9x - 40
= 9(15) - 40
= 135 - 40
= 95
Next, draw a perpendicular bisector KN from K to KL. Thus, N is the midpoint of KL and ΔMNL is a right triangle.
- Since N is the midpoint of KL and KL = 95, then NL = 47.5
- Since ∠N is 90°, then NL is adjacent to ∠l and ML is the hypotenuse
Use trig to solve for ∠L (which equals ∠K):
cos ∠L =
cos ∠L =
∠L = cos⁻¹
∠L = 45.7
Triangle sum Theorem:
∠K + ∠L + ∠M = 180°
45.7 + 45.7 + ∠M = 180
91.4 + ∠M = 180
∠M = 88.6