## Engage NY Eureka Math Geometry Module 5 End of Module Assessment Answer Key

### Eureka Math Geometry Module 5 End of Module Assessment Task Answer Key

Question 1.

Let C be the circle in the coordinate plane that passes though the points (0,0), (0,6), and (8,0).

a. What are the coordinates of the center of the circle?

Answer:

Since the angle formed by the points (0,6) , (0,0) , and (8,0) is a right angle, the line segment connecting (0,6) to (8,0) must be the diameter of the circle. Therefore, the center of the circle is (4,3) , the midpoint of this diameter.

b. What is the area of the portion of the interior of the circle that lies in the first quadrant? (Give an exact answer in terms of Ï€.)

Answer:

The distance between (0,6) and (8,0) is 10: d=\(\sqrt{6^{2}+8^{2}}\)=10

So, the circle has radius 5. The area in question is composed of half a circle and a right triangle.

Area=(\(\frac{1}{2}\) âˆ™ 8 âˆ™ 6 ) + (\(\frac{1}{2}\) Ï€5^{2}) = \(\frac{25\pi}{2}\) + 24

Therefore, its area is \(\frac{25\pi}{2}\) + 24 square units.

c. What is the area of the portion of the interior of the circle that lies in the second quadrant? (Give an approximate answer correct to one decimal place.)

Answer:

We seek the area of the region shown. We have a chord of length 6 in a circle of radius 5.

Label the angle x as shown and distance d. By the Pythagorean theorem, d = 4. We also know that sin(x) = \(\frac{3}{5}\) , so x â‰ˆ 36.9Â° .

The shaded area is the difference of the area of a sector and of a triangle. We have

area=(\(\frac{2x}{360}\) Ï€ 5^{2})-(\(\frac{1}{2}\) âˆ™ 6âˆ™ 4 )

â‰ˆ (\(\frac{73.8}{360}\) âˆ™ 25Ï€ ) – 12

â‰ˆ 4.1

The area is 4.1 units^{2}.

d. What is the length of the arc of the circle that lies in the first quadrant with endpoints on the axes? (Give an exact answer in terms of Ï€.)

Answer:

Since this arc is a semicircle, it is half the circumference of the circle in length:

\(\frac{1}{2}\) âˆ™ 2Ï€ âˆ™ 5 = 5Ï€

The length is 5Ï€ units.

e. What is the length of the arc of the circle that lies in the second quadrant with endpoints on the axes? (Give an approximate answer correct to one decimal place.)

Answer:

Using the notation of part (c), this length is calculated as follows:

\(\frac{2x}{360}\) â‹… 2Ï€ â‹… 5 â‰ˆ \(\frac{73.8}{360}\) â‹…10Ï€ â‰ˆ 6.4

The length of the arc is approximately 6.4 units .

f. A line of slope – 1 is tangent to the circle with point of contact in the first quadrant. What are the coordinates of that point of contact?

Answer:

Draw a radius from the center of the circle, (4,3) , to the point of contact, which we will denote (x,y) .

This radius is perpendicular to the tangent line and has slope 1. Consequently, \(\frac{y-3}{x-4}\) = 1 ; that is, y – 3 = x – 4 .

Also, since (x,y) lies on the circle, we have

(x – 4 )^{2} + (y – 3 )^{2} = 25.

For both equations to hold, we must have

(x – 4 )^{2} + (x – 4 )^{2} = 25 , giving x = 4 + \(\frac{5}{\sqrt{2}}\) , or x = 4 – \(\frac{5}{\sqrt{2}}\) . It is clear from the diagram that the point of contact we seek has its x-coordinate to the right of the x-coordinate of the center of the circle. So, choose x = 4 + \(\frac{5}{\sqrt{2}}\) . The matching y-coordinate is

y = x – 4 + 3 = x – 1 = 3 + \(\frac{5}{\sqrt{2}}\) , so the point of contact has coordinates (4 + \(\frac{5}{\sqrt{2}}\), 3 + \(\frac{5}{\sqrt{2}}\) ).

g. Describe a sequence of transformations that show circle C is similar to a circle with radius one centered at the origin.

Answer:

Circle C has center (4, 3 ) and radius 5.

First, translate the circle four units to the left and three units downward. This gives a congruent circle with the origin as its center. (The radius is still 5.)

Perform a dilation from the origin with scale factor \(\frac{1}{5}\) . This will produce a similar circle centered at the origin with radius 1.

h. If the same sequence of transformations is applied to the tangent line described in part (f), will the image of that line also be a line tangent to the circle of radius one centered about the origin? If so, what are the coordinates of the point of contact of this image line and this circle?

Answer:

Translations and dilations map straight lines to straight lines. Thus, the tangent line will still be mapped to a straight line. The mappings will not alter the fact that the circle and the line touch at one point. Thus, the image will again be a line tangent to the circle.

Under the translation described in part (g), the point of contact, (4 + \(\frac{5}{\sqrt{2}}\), 3 + \(\frac{5}{\sqrt{2}}\)), is mapped to (\(\frac{5}{\sqrt{2}}\), \(\frac{5}{\sqrt{2}}\)). Under the dilation described, this is then mapped to (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\)).

Question 2.

In the figure below, the circle with center O circumscribes â–³ABC.

Points A, B, and P are collinear, and the line through P and C is tangent to the circle at C. The center of the circle lies inside â–³ABC.

a. Find two angles in the diagram that are congruent, and explain why they are congruent.

Answer:

Draw two radii as shown. Let mâˆ BAC = x . Then by the inscribed/central angle theorem, we have mâˆ BOC = 2x .

Since â–³BOC is isosceles, it follows that

mâˆ OCB = \(\frac{1}{2}\) (180Â° – 2x ) = 90Â°- x .

By the radius/tangent theorem, mâˆ OCP = 90Â° , so mâˆ BCP = x .

We have âˆ BAC â‰… âˆ BCP because they intercept the same arc and have the same measure.

b. If B is the midpoint of \(\overline{A P}\) and PC = 7, what is the length of \(\overline{P B}\)?

Answer:

By the previous question, â–³ ACP and â–³ CBP each have an angle of measure x and share the angle at P. Thus, they are similar triangles.

Since â–³ ACP and â–³ CBP are similar, matching sides come in the same ratio. Thus,

\(\frac{PB}{PC}\) =\(\frac{PC}{AP}\). Now, AP = 2 â‹… PB , and PC = 7 , so \(\frac{PB}{7}\) = \(\frac{7}{2PB}\). This gives PB = \(\frac{7}{\sqrt{2}}\).

c. If mâˆ BAC = 50Â°, and the measure of the arc AC is 130Â°, what is mâˆ P?

Answer:

By the inscribed/central angle theorem, arc BC has measure 100Â°. By the secant/tangent angle theorem,

mâˆ P = \(\frac{130^{\circ}-100^{\circ}}{2}\) = 15Â°.

(One can also draw in radii and base angles in triangles to obtain the same result.)

Question 3.

The circumscribing circle and the inscribed circle of a triangle have the same center.

a. By drawing three radii of the circumscribing circle, explain why the triangle must be equiangular and, hence, equilateral.

Answer:

Draw the three radii as directed, and label six angles a, b, c, d, e, and f as shown.

We have a = f because they are base angles of an isosceles triangle. (We have congruent radii.) In the same way,

b = c and d = e.

From the construction of an inscribed circle, we know that each radius drawn is an angle bisector of the triangle. Thus, we have a = b, c = d, and e = f.

It now follows that a = b = c = d = e = f. In particular, a + b = c + d = e + f, and the triangle is equiangular. Therefore, the triangle is equilateral.

b. Prove again that the triangle must be equilateral, but this time by drawing three radii of the inscribed circle.

Answer:

By the construction of the circumscribing circle of a triangle, each radius in this picture is the perpendicular bisector of a side of the triangle. If we label the lengths a, b, c, d, e, and f as shown, it follows that b = c, d = e, and a = f.

By the two-tangents theorem, we also have a = b, c = d, and e = f.

Thus, a = b = c = d = e = f, and in particular,

b + c = d + e = a + f; therefore, the triangle is equilateral.

c. Describe a sequence of straightedge and compass constructions that allows you to draw a circle inscribed in a given equilateral triangle.

Answer:

The center of an inscribed circle lies at the point of intersection of any two angle bisectors of the equilateral triangle.

To construct an angle bisector:

- Draw a circle with center at one vertex P of the triangle intersecting two sides of the triangle. Call those two points of intersection A and B.
- Setting the compass at a fixed width, draw two congruent intersecting circles, one centered at A and one centered at B. Call a point of intersection of these two circles Q. (We can assume Q is different from P.)
- The line through P and Q is an angle bisector of the triangle.

Next, construct two such angle bisectors and call their point of intersection O. This is the center of the inscribed circle. Finally, draw a line through O perpendicular to one side of the triangle. To do this:

- Draw a circle centered at O that intersects one side of the triangle at two points.

Call those points C and D. - Draw two congruent intersecting circles, one with center C and one with center D.
- Draw the line through the points of intersection of those two congruent circles. This is a line through O perpendicular to the side of the triangle.

Suppose this perpendicular line through O intersects the side of the triangle at the point R. Set the compass to have width equal to OR. This is the radius of the inscribed circle; so, drawing a circle of this radius with center O produces the inscribed circle.

Question 4.

Show that

(x – 2)(x – 6) + (y – 5)(y + 11) = 0

is the equation of a circle. What is the center of this circle? What is the radius of this circle?

Answer:

We have

(x – 2 )(x – 6 ) + (y – 5 )(y + 11 ) = 0

x^{2} – 8x + 12 + y^{2} + 6y – 55 = 0

x^{2} – 8x + 16 + y^{2} + 6y + 9 = 4 + 64

(x – 4 )^{2} + (y + 3 )^{2} = 68.

This is the equation of a circle with center (4, -3 ) and radius \(\sqrt{68}\).

b. A circle has diameter with endpoints (a, b) and (c, d). Show that the equation of this circle can be written as

(x – a)(x – c) + (y – b)(y – d) = 0.

Answer:

The midpoint of the diameter, which is (\(\frac{a+c}{2}\), \(\frac{b+d}{2}\)), is the center of the circle; half the distance between the endpoints, which is \(\sqrt{(c-a)^{2}+(d-b)^{2}}\) \(\frac{1}{2}\), is the radius of the circle. Thus, the equation of the circle is

(x – \(\frac{a+c}{2}\))^{2} + (y – \(\frac{b-d}{2}\))^{2} = \(\frac{1}{4}\) ((c -a )^{2} + (d -b )^{2} ).

Multiplying through by 4 gives

(2x – a -c )^{2} + (2y – b -d )^{2} = (c – a )^{2} + (d – b )^{2} .

This becomes

4 x^{2} + a^{2} + c^{2}-4xa-4xc + 2ac + 4 y^{2} + b^{2} + d^{2} – 4yb – 4yd + 2bd = c^{2} + a^{2}– 2ac + d^{2} + b^{2} – 2bd.

That is,

4 x^{2} – 4xa – 4xc + 4ac + 4 y^{2} – 4yb – 4yd + 4bd = 0.

Dividing through by 4 gives

x^{2} – xa – xc + ac + y^{2} – yb – yd + bd = 0.

That is,

(x – a )(x – c ) + (y – b )(y – d ) = 0,

as desired.

Question 5.

Prove that opposite angles of a cyclic quadrilateral are supplementary.

Answer:

Consider a cyclic quadrilateral with two interior opposite angles of measures x and y, as shown.

The vertices of the quadrilateral divide the circle into four arcs. Suppose these arcs have measures a, b, c, and d, as shown.

By the inscribed/central angle theorem, we have

a + d = 2y and b + c = 2x. So, a + b + c + d = 2(x + y).

But, a + b + c + d = 360Â°. Thus, it follows that

x + y = \(\frac{360^{\circ}}{2}\) = 180Â° .

By analogous reasoning, the angles in the second pair of interior opposite angles are supplementary as well. (This also follows from the fact that the interior angles of a quadrilateral add to 360Â°. The second pair of interior angles have measures adding to

360Â° – x – y = 360Â° – 180Â° = 180Â°.)