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Radda [10]
3 years ago
14

Lara wants to inscribe a square in a circle using only a compass and straightedge. She will first construct a circle and locate

its center. She will then perform the following steps.
Step 1: Use the straightedge to draw a diameter of the circle. Label the points at which the diameter touches the circle as A and B.
Step 2: Draw a second diameter through the center. Label the points at which the second diameter touches the circle as C and D.
Step 3: Join the end points of the two diameters to form a square.

What is the error in Lara's construction?

A. Show a diameter that did not pass through the center of the circle
B. She uses straightedge and not a compass to draw the diameter of the circle
C. She did not ensure that the second diameter was perpendicular to the first diameter
D. She did not ensure that the second diameter had the same length as the first diameter

Mathematics
1 answer:
astraxan [27]3 years ago
4 0
The answer is C. You can make a rectangle by putting the diagonals through the center anywhere, but if you want a square, you have to put the two diagonals perpendicular to each other and join their ends. See the diagram. 

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Lines k and n intersect on the y-axis
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a) The equation of line k is:

y = -\frac{202}{167}x + \frac{598}{167}

b) The equation of line j is:

y = \frac{167}{202}x + \frac{1546}{202}

The equation of a line, in <u>slope-intercept formula</u>, is given by:

y = mx + b

In which:

  • m is the slope, which is the rate of change.
  • b is the y-intercept, which is the value of y when x = 0.

Item a:

  • Line k intersects line m with an angle of 109º, thus:

\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}

In which m_1 and m_2 are the slopes of <u>k and m.</u>

  • Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}

  • The tangent of 109 degrees is \tan{109^{\circ}} = -\frac{29}{10}
  • Thus, the slope of line m is found solving the following equation:

\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}

-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}

m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2

m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}

\frac{167m_2}{80} = \frac{-202}{80}

m_2 = -\frac{202}{167}

Thus:

y = -\frac{202}{167}x + b

It goes through point (-2,6), that is, when x = -2, y = 6, and this is used to find b.

y = -\frac{202}{167}x + b

6 = -\frac{202}{167}(-2) + b

b = 6 - \frac{404}{167}

b = \frac{6(167)-404}{167}

b = \frac{598}{167}

Thus. the equation of line k, in slope-intercept formula, is:

y = -\frac{202}{167}x + \frac{598}{167}

Item b:

  • Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

-\frac{202}{167}m = -1

m = \frac{167}{202}

Then

y = \frac{167}{202}x + b

Also goes through point (-2,6), thus:

6 = \frac{167}{202}(-2) + b

b = \frac{(2)167 + 202(6)}{202}

b = \frac{1546}{202}

The equation of line j is:

y = \frac{167}{202}x + \frac{1546}{202}

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