The width of a rectangle is 38 centimeters. The perimeter is at least 692 centimeters. Write an inequality that represents all p
ossible values for the length of the rectangle. Then solve the inequality.
1 answer:
Answer:
See bolded / underlined / italicized below -
Step-by-step explanation:
This is a great question!
If x were the length of this rectangle, then we could conclude the following,
2( 38 ) + 2( x ) > 692,
As you can see there is a greater than sign present, as the perimeter is at least 692 centimeters. In this case the perimeter is given to be at least 692 centimeters, but can also be calculated through double the width and double the length together. And of course we are given the width to be 38 cm -
2( 38 ) + 2x > 692,
76 + 2x > 692,
2x > 616,
x > 308
<u><em>Solution = Length should be at least 308 cm</em></u>
( The attachment below is not drawn to scale )
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Answer:
A = 20sinθ(6 + 5 cosθ) cm²
Step-by-step explanation:
Drop perpendiculars DE and CF to AB.
Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.
The formula for the area of the trapezium is
A = ½(a + b)h
DE = 10sinθ
AE = 10cosθ
BF = 10cosθ
EF = CD = 12 cm
AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ
A = ½(a + b)h
= ½(12 +12 + 20 cosθ) × 10 sinθ
=(24 + 20 cosθ) × 5 sinθ
= 4(6 + 5cosθ) × 5sinθ
= 20sinθ(6 + 5 cosθ) cm²
