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mamaluj [8]
3 years ago
5

A square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches. The area of the resul

ting rectangle is 6o in 2 what was the area of the
Mathematics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

area of the original square = 36 in²

Step-by-step explanation:

given data

one side is increased = 9 inches

other side is decreased = 2 inches

area of the resulting rectangle = 60 in²

solution

we consider here x as the length of any one side

and

longer side of the resulting rectangle = x + 9

and

the shorter side x - 2

so that area of this rectangle is (s+9) × (s-2) = 60 in²

standard form of equation is  x² + 7x - 18 = 60   .....................1

it will be

x² + 7x - 78 = 0

(s+13) (s-6) = 0

so that here x  = -13

and  x = 6

here we Discard x = -13 because the side length cannot be negative.

so x = 6

and area of the original square = 36 in²

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What is the eighth term of the arithmetic sequence? An=<br> 10n – 14?
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Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

  • \sin{A} = \sin{103\textdegree{}},
  • The opposite side of angle A a = BC = 26,
  • The angle C is to be found, and
  • The length of the side opposite to angle C c = AB = 6.

\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}.

\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}.

\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}.

Note that the inverse sine function here \sin^{-1}() is also known as arcsin.

<h3>17</h3>

By the law of cosine,

c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C},

where

  • a, b, and c are the lengths of sides of triangle ABC, and
  • \cos{C} is the cosine of angle C.

For triangle ABC:

  • b = 21,
  • c = 30,
  • The length of a (segment BC) is to be found, and
  • The cosine of angle A is \cos{123\textdegree}.

Therefore, replace C in the equation with A, and the law of cosine will become:

a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}.

\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}.

<h3>18</h3>

For triangle ABC:

  • a = 14,
  • b = 9,
  • c = 6, and
  • Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}.

\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}.

\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree.

<h3>15</h3>

For triangle DEF:

  • The length of segment DF is to be found,
  • The length of segment EF is 9,
  • The sine of angle E is \sin{64\textdegree}}, and
  • The sine of angle D is \sin{39\textdegree}.

Apply the law of sine:

\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}

\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9.

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