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Further Geometry: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A sheet-metal blank

Advanced. This question goes beyond core Geometry. It is not required by the course.

The edges of a triangular sheet-metal blank measure $14$ cm, $25$ cm, and $29$ cm. Find the exact area and its value to the nearest tenth of a square centimeter.
1. Hint 1
  Check that the two shorter lengths can meet while spanning the longest edge.
2. Hint 2
  Find half the perimeter, then compare it with each of the three side lengths. The area uses the square root of a product.
Answer

$30\sqrt{34}$ square centimeters, approximately $174.9$ square centimeters.

Full solution

The two shorter sides sum to $39$ cm, which is more than $29$ cm.
The sides form a nondegenerate triangle.

Find the semiperimeter in centimeters.

$\begin{gathered} s=\frac{14+25+29}{2}\\ s=34 \end{gathered}$

The differences from it are 20, 9, and 5 cm.

Substitute in Heron's formula.

$K=\sqrt{34(20)(9)(5)}$

The radicand factors as $900(34)$ .
Take the positive square root.

$\begin{gathered} K=30\sqrt{34}\\ K\approx174.9 \end{gathered}$

Answer

$30\sqrt{34}$ square centimeters, approximately $174.9$ square centimeters.

Key idea

After checking closure, the semiperimeter and its three side differences determine area directly.
Problem 2 A matching panel

Advanced. This question goes beyond core Geometry. It is not required by the course.

The first triangular panel has two sides measuring $18$ cm and $32$ cm, with an interior angle of $120^\circ$ between them.

The second panel has one side of length $27$ cm. At one endpoint of that side, the interior angle measures $30^\circ$ . What exact length must the other side at this corner have for the panels to have equal areas?

Schematic. Use the measurements stated in the question.

Text description of this figure

The first panel has an interior angle of 120 degrees between its 18 cm and 32 cm sides. The second panel has an interior angle of 30 degrees at an endpoint of its 27 cm side.
1. Hint 1
  Each panel has two sides meeting at a known angle, so each area can be written using those sides.
2. Hint 2
  Let the unknown side of the second panel be x. Set the two area expressions equal and use the sines of the two marked angles.
Answer

$\dfrac{64\sqrt3}{3}$ cm.

Full solution

The first panel uses an obtuse included angle.

$\begin{gathered} \sin120^\circ=\sin60^\circ\\ \sin120^\circ=\tfrac{\sqrt3}{2} \end{gathered}$

$\begin{gathered} K_1=\tfrac12(18)(32)\sin120^\circ\\ K_1=144\sqrt3 \end{gathered}$

Let $x$ be the unknown side length in centimeters.
The second included angle has sine one half.

$\begin{gathered} K_2=\tfrac12(27)x\sin30^\circ\\ K_2=\tfrac{27x}{4} \end{gathered}$

Set the two areas equal.

$\tfrac{27x}{4}=144\sqrt3$

Solve for the positive length.

$x=\tfrac{64\sqrt3}{3}$

This is positive, and the included angle lies strictly between zero and a straight angle, so these sides form the required panel.

Answer

$\dfrac{64\sqrt3}{3}$ cm.

Key idea

Matching areas means matching the two products of adjacent lengths and their included sine factors.
Problem 3 A survey log

Advanced. This question goes beyond core Geometry. It is not required by the course.

The boundary of a simple parcel follows $A(2,-1)$ , $B(3,4)$ , $C(11,6)$ , $D(10,0)$ , $E(13,-4)$ , and $F(5,-5)$ , then closes at $A$ .

What signed area does this log give? State whether the log runs clockwise or counterclockwise, and give the ordinary area.

Schematic. Use the measurements stated in the question.

Text description of this figure

The outline joins A to B, B to C, C to D, D to E, E to F, and F back to A. The exact coordinates are listed in the statement.
1. Hint 1
  Each boundary edge is used once, including the last edge back to the start.
2. Hint 2
  Subtract the two ordered coordinate products for each edge. Keep each result signed until you have added all six.
Answer

Signed area $-77.5$ square units. The log runs clockwise. Ordinary area $77.5$ square units.

Full solution

Calculate each directed difference in boundary order.

$\begin{gathered} d_{AB}=2(4)-3(-1)\\ d_{AB}=11 \end{gathered}$

$\begin{gathered} d_{BC}=3(6)-11(4)\\ d_{BC}=-26 \end{gathered}$

$\begin{gathered} d_{CD}=11(0)-10(6)\\ d_{CD}=-60 \end{gathered}$

$\begin{gathered} d_{DE}=10(-4)-13(0)\\ d_{DE}=-40 \end{gathered}$

$\begin{gathered} d_{EF}=13(-5)-5(-4)\\ d_{EF}=-45 \end{gathered}$

$\begin{gathered} d_{FA}=5(-1)-2(-5)\\ d_{FA}=5 \end{gathered}$

Add all six differences before halving.

$2S=11-26-60-40-45+5$

$\begin{gathered} S=-155/2\\ S=-77.5 \end{gathered}$

The negative sign means the walk is clockwise.
The ordinary area is 77.5 square units.

Answer

Signed area $-77.5$ square units. The log runs clockwise. Ordinary area $77.5$ square units.

Key idea

The final sign gives the direction of the boundary, and its magnitude gives the ordinary area.
Problem 4 A shoreline record

Advanced. This question goes beyond core Geometry. It is not required by the course.

Two stations $A$ and $B$ are $69$ m apart. Both sight the same marker $C$ , off the baseline. All three locations lie in one horizontal plane.

At $A$ , the angle from the baseline ray pointing away from $B$ to ray $AC$ is $127^\circ$ . At $B$ , the angle from the baseline ray pointing away from $A$ to ray $BC$ is $104^\circ$ .

Find the distance $AC$ to the nearest tenth of a meter and the area of the triangle bounded by the two sight lines and the baseline to the nearest tenth of a square meter.

Schematic. Use the measurements stated in the question.

Text description of this figure

The baseline runs from A on the left to B on the right. The marker C is above it. The left extension points from A away from B and the right extension points from B away from A. The two measurements are from those outward extensions toward C.
1. Hint 1
  The observed angles start from the outward rays. Which angles inside the triangle share a straight angle with those readings?
2. Hint 2
  Once the interior angles are known, match the measured baseline with its opposite angle and find the length from A to C.
3. Hint 3
  The baseline and AC meet at A. The area can be found from those two sides and their included angle.
Answer

$AC\approx86.1$ m. The area is approximately $2373.7$ square meters.

Full solution

Each interior angle supplements its corresponding exterior angle.

$\begin{gathered} A=180^\circ-127^\circ\\ A=53^\circ \end{gathered}$

$\begin{gathered} B=180^\circ-104^\circ\\ B=76^\circ \end{gathered}$

The angles at $A$ and $B$ leave $C=51^\circ$ , so the triangle has three positive angles.

The measured baseline is opposite $C$ .
The requested length $AC$ is opposite $B$ .

$AC=\frac{69\sin76^\circ}{\sin51^\circ}$

Use the unrounded value of $AC$ for the area, with $AB$ and $AC$ meeting at $A$ .

$K=\tfrac12(69)(AC)\sin53^\circ$

The distance rounds to $86.1$ m.
The area rounds to $2373.7$ square meters.

Answer

$AC\approx86.1$ m. The area is approximately $2373.7$ square meters.

Key idea

First convert outward readings to interior angles, then match opposite pairs for lengths and adjacent pairs for area.
Problem 5 A ring and an anchor

Advanced. This question goes beyond core Geometry. It is not required by the course.

A circular ring has center $O$ and radius $225$ mm. An anchor $P$ lies outside the ring. A tangent from $P$ meets the ring at $T$ .

The lengths of the three sides of triangle $OPT$ , in millimeters, form a primitive Pythagorean triple. The segment $OP$ is $81$ mm longer than $PT$ .

Find $PT$ and $OP$ . Explain how to locate both contacts with a compass and straightedge. Give the compass opening for the circle that locates them, and justify why this finds every contact.
1. Hint 1
  The right angle at a contact identifies the hypotenuse. The radius is the odd leg of a primitive triple.
2. Hint 2
  Write the hypotenuse minus the even leg in terms of the generating parameters. Also factor the square difference that gives the odd leg.
3. Hint 3
  For the construction, ask which circle contains every vertex that sees the segment from the center to the anchor at a right angle.
Answer

$\begin{gathered} PT=272\text{ mm}\\ OP=353\text{ mm} \end{gathered}$
Bisect $OP$ and draw the circle centered at its midpoint with compass opening 176.5 mm. The two crossings with the ring are the contacts.

Full solution

The radius is perpendicular to the tangent at $T$ .
Thus $OP$ is the hypotenuse, $OT$ is the odd leg of length 225, and $PT$ is the even leg.
Use primitive parameters $m>n>0$ .

Subtract twice the parameter product from the sum of their squares to obtain the square of their difference.

$\begin{gathered} OP-PT=(m-n)^2\\ OP-PT=81 \end{gathered}$

The difference of the parameters is $m-n=9$ .
Factor the odd leg.

$225=(m-n)(m+n)$

The parameter sum is 25.
Solve using the known sum and difference.

$\begin{gathered} m=17\\ n=8 \end{gathered}$

The parameters are coprime and have opposite parity.

$\begin{gathered} PT=2(17)(8)\\ PT=272\text{ mm} \end{gathered}$

$\begin{gathered} OP=17^2+8^2\\ OP=353\text{ mm} \end{gathered}$

Use an equal compass opening greater than half of $OP$ to draw intersecting arcs centered at $O$ and $P$ .
Draw the line through the arc crossings.
It is the perpendicular bisector of $OP$ and meets it at its midpoint $M$ .

Draw the circle centered at $M$ through $O$ and $P$ .
Its radius is $353/2=176.5$ mm.
It crosses the ring twice because the ring has positive radius and $P$ lies outside it.

Each crossing lies on the circle with diameter $OP$ , so it sees $OP$ at a right angle.
The line from $P$ through that crossing is perpendicular to the ring radius there and is tangent.

Conversely, every contact of a tangent through $P$ makes a right angle with the radius.
The converse of the semicircle theorem puts that contact on the same diameter circle, so no contact is missed.

Answer

$\begin{gathered} PT=272\text{ mm}\\ OP=353\text{ mm} \end{gathered}$
Bisect $OP$ and draw the circle centered at its midpoint with compass opening 176.5 mm. The two crossings with the ring are the contacts.

Key idea

The contact triangles connect the integer side data with the diameter that determines both possible contacts.
Problem 6 A corner measurement

Advanced. This question goes beyond core Geometry. It is not required by the course.

The three edges of a triangular plate measure $17$ cm, $31$ cm, and $44$ cm. Let $\theta$ be the interior angle between the $17$ cm and $31$ cm edges.

Give $\cos\theta$ and $\sin\theta$ exactly and classify $\theta$ as acute, right, or obtuse.

Find the area of the plate to the nearest hundredth of a square centimeter.
1. Hint 1
  Check that the three edges can close. Then identify which edge is across from the named angle.
2. Hint 2
  Find the cosine from the three lengths. The sine is positive at every interior angle of a nondegenerate triangle.
3. Hint 3
  The two shorter edges meet at the named angle, so their lengths and its sine determine the area.
Answer

$\begin{gathered}\cos\theta=-\tfrac{343}{527}\\\sin\theta=\tfrac{4\sqrt{10005}}{527}\end{gathered}$
The angle is obtuse. The area is approximately $200.05$ square centimeters.

Full solution

The two shorter lengths have sum $48$ cm, which exceeds $44$ cm, so the triangle closes.

The side across from $\theta$ has length $44$ cm.
Use the three-side cosine quotient.

$\cos\theta=\frac{17^2+31^2-44^2}{2(17)(31)}$

This quotient simplifies to $-343/527$ .
It is negative, so the angle is obtuse.

Because $\theta$ is an interior angle, take the positive square root of $1-\cos^2\theta$ .

$\sin\theta=\frac{\sqrt{527^2-343^2}}{527}$

The radicand is $160080=16(10005)$ , so the sine is $4\sqrt{10005}/527$ .

Substitute in the included-angle area formula.
The product $17(31)$ cancels the denominator 527.

$\begin{gathered} K=2\sqrt{10005}\\ K\approx200.05 \end{gathered}$

Answer

$\begin{gathered}\cos\theta=-\tfrac{343}{527}\\\sin\theta=\tfrac{4\sqrt{10005}}{527}\end{gathered}$
The angle is obtuse. The area is approximately $200.05$ square centimeters.

Key idea

The three sides give the signed cosine, and a triangle's positive sine gives the perpendicular height factor.
Problem 7 Choosing a closing rail

Advanced. This question goes beyond core Geometry. It is not required by the course.

Two rails of lengths $22$ cm and $36$ cm will form two edges of a triangular opening. The opening must have area at least $180$ square centimeters.

Option A. The third rail measures $54$ cm.

Option B. The third rail measures $57$ cm.

Option C. The third rail measures $59$ cm.

Which is the longest third rail that meets the requirement? Justify your choice and give the area it produces to the nearest tenth of a square centimeter.
1. Hint 1
  The third rail must first allow the two shorter rails to meet while spanning it.
2. Hint 2
  For each valid choice, compute the area from the three lengths. To compare with the area requirement exactly, square both sides of the comparison.
Answer

Option A, the $54$ cm rail. The area is approximately $276.0$ square centimeters.

Full solution

The sum of the two shorter rails is $22+36=58$ cm.
Compare with the longest rail before using any area calculation.

Option A.
Since $54<58$ , the rails close.
The semiperimeter is $56$ cm.

$K_A^2=56(34)(20)(2)$

The product is $76160$ , greater than
$180^2=32400.$
Its square root rounds to $276.0$ square centimeters.

Option B.
Since $57<58$ , these rails also close.
The semiperimeter is $57.5$ cm.

$K_B^2=57.5(35.5)(21.5)(0.5)$

This product is $21943.4375$ , below $32400$ .
The opening falls short of the required area.

Option C.
Since $59>58$ , the shorter rails cannot meet.
This option gives no triangular opening.

Only Option A meets both requirements.

Answer

Option A, the $54$ cm rail. The area is approximately $276.0$ square centimeters.

Key idea

Check that the lengths close before comparing their exact squared areas with the required area.
Problem 8 An incomplete triangle record

Advanced. This question goes beyond core Geometry. It is not required by the course.

A record for triangle $ABC$ gives an angle at $A$ of $37^\circ$ . The lengths of $BC$ and $AC$ are 31 m and 42 m, respectively. Both side readings and the angle are exact.

Find the area of every triangle that fits the record, up to congruence. Give each area to the nearest tenth of a square meter and name the vertex with the largest angle in that triangle.
1. Hint 1
  One length is opposite the known angle, and the other is opposite an unknown angle. Pair the two lengths with those angles.
2. Hint 2
  Check the angle a calculator returns and its supplement. Each candidate needs a positive third angle.
3. Hint 3
  Recover AB separately for each valid triangle. It meets AC at A.
Answer

When $C$ is the largest angle, the area is approximately $650.7$ square meters. When $B$ is the largest angle, the area is approximately $197.1$ square meters.

Full solution

The side of length $31$ faces $A$ , and the side of length $42$ faces $B$ .

$\sin B=\frac{42\sin37^\circ}{31}$

The right side is between $0$ and $1$ .
In degree mode, its inverse sine is about $54.6232^\circ$ .
The supplementary candidate is about $125.3768^\circ$ .
Keep full precision in the checks and area calculations.

Acute candidate for $B$ .
The angle sum leaves $C\approx88.3768^\circ$ , which is positive and greater than both $A$ and $B$ .

The side $c=AB$ is opposite $C$ .

$c=\frac{31\sin C}{\sin37^\circ}$

The side measures about 51.4902 m, strictly between 11 and 73 m.

$\begin{gathered} K=\tfrac12(42)c\sin37^\circ\\ K\approx650.7 \end{gathered}$

Obtuse candidate for $B$ .
The angle sum leaves $C\approx17.6232^\circ$ , which is positive.
The obtuse angle $B$ is largest.

With this value of $C$ , the same opposite-side proportion gives $c\approx15.5952$ m, also strictly between $11$ and $73$ .

Use this shorter side at the same included angle.

$\begin{gathered} K=\tfrac12(42)c\sin37^\circ\\ K\approx197.1 \end{gathered}$

These are all the possibilities because the sine equation has only the two stated interior candidates.

Answer

When $C$ is the largest angle, the area is approximately $650.7$ square meters. When $B$ is the largest angle, the area is approximately $197.1$ square meters.

Key idea

The same recorded angle and two sides can leave two different bases and change which vertex has the largest angle.
Problem 9 Right-triangle frames

Advanced. This question goes beyond core Geometry. It is not required by the course.

A workshop builds right-triangle frames with positive integer side lengths that have no common factor greater than $1$ . One leg must be exactly $95$ units long.

Find all possible sets of side lengths and say which has the shorter hypotenuse. Explain why there are no other sets.
1. Hint 1
  In a primitive triple, an odd leg is the difference of the parameter squares.
2. Hint 2
  Factor the difference of squares and list all positive factor pairs of 95. The smaller factor is the difference of the parameters.
3. Hint 3
  Check each resulting pair for coprimality and opposite parity before generating the other sides.
Answer

The triples are $(95,168,193)$ and $(95,4512,4513)$ , with the hypotenuse last. The shorter hypotenuse is $193$ units.

Full solution

Every primitive triple comes from positive integers $m>n$ that are coprime and of opposite parity.
The given leg is odd, so

$95=(m-n)(m+n)$

Both factors are positive and the second exceeds the first.
Since $95=5\cdot19$ with both factors prime, the only ordered factor pairs are $(1,95)$ and $(5,19)$ .

Factor pair $(1,95)$ .
Solve the difference and sum equations.

$\begin{gathered} m-n=1\\ m+n=95 \end{gathered}$

$\begin{gathered} m=48\\ n=47 \end{gathered}$

The consecutive inputs are coprime and of opposite parity.

Find the even leg $b$ and hypotenuse $c$ .

$\begin{gathered} b=2(48)(47)\\ b=4512 \end{gathered}$

$\begin{gathered} c=48^2+47^2\\ c=4513 \end{gathered}$

Factor pair $(5,19)$ .
Solve the difference and sum equations.

$\begin{gathered} m-n=5\\ m+n=19 \end{gathered}$

$\begin{gathered} m=12\\ n=7 \end{gathered}$

The inputs are coprime and of opposite parity.

Find the even leg $b$ and hypotenuse $c$ .

$\begin{gathered} b=2(12)(7)\\ b=168 \end{gathered}$

$\begin{gathered} c=12^2+7^2\\ c=193 \end{gathered}$

Both pairs produce the required primitive triples.
The factor list was exhaustive, and $193<4513$ .

Answer

The triples are $(95,168,193)$ and $(95,4512,4513)$ , with the hypotenuse last. The shorter hypotenuse is $193$ units.

Key idea

The positive factor pairs of the odd leg exhaust the possible primitive parameter pairs after parity and common factors are checked.
Problem 10 A cable layout

Advanced. This question goes beyond core Geometry. It is not required by the course.

A circular rim has its center $O$ at the origin and radius $4\sqrt{65}$ units. The anchor $P$ is at $(65,0)$ . A technician marks $T$ at $(16,28)$ and proposes that line $PT$ touches the rim only at $T$ .

Verify the proposed contact. Find the coordinates of the other contact $U$ for a tangent through $P$ and explain why there are exactly two contacts.

Find the ordinary area of the quadrilateral that follows $O,T,P,U$ and closes at $O$ .

Text description of this figure

The circular rim is centered at O. The anchor P is outside the rim. The marked point T is on the rim above the horizontal level of O and P.
1. Hint 1
  Check the marked point's distance to the center. Then compare the three squared distances in the triangle formed with the anchor.
2. Hint 2
  Look at what reflection across line OP preserves. How does the diameter-circle construction account for every possible contact?
3. Hint 3
  For the area, follow the stated quadrilateral boundary and keep each coordinate contribution signed until the end.
Answer

Line $PT$ is tangent at $T$ . The other contact is $U=(16,-28)$ , and there are exactly two contacts. The quadrilateral's ordinary area is $1820$ square units.

Full solution

Verification.
Compute the squared distance from the center to $T$ .

$\begin{gathered} OT^2=16^2+28^2\\ OT^2=1040 \end{gathered}$

This equals the radius squared, so $T$ lies on the rim.

Compare the other two squared distances.

$\begin{gathered} PT^2=49^2+28^2\\ PT^2=3185 \end{gathered}$

$\begin{gathered} OP^2=65^2\\ OP^2=4225 \end{gathered}$

The two smaller squares add to the largest.

$OT^2+PT^2=OP^2$

Thus the angle at $T$ is right by the converse of the Pythagorean theorem.
The line perpendicular to the radius at $T$ touches the rim only there.

Other contact.
Line $OP$ is the horizontal axis.
Reflecting in it preserves the circle and fixes $P$ , so it takes the tangent through $T$ to a tangent through
$U=(16,-28).$

The two contacts are distinct.
Every contact must see $OP$ at a right angle and therefore lies on the circle with diameter $OP$ .
Two distinct circles meet in at most two points, so there can be no further contacts.

Area.
The edges $O$ to $T$ and $U$ to $O$ give zero cross-product differences.
Each of the other two differences is
$-65(28)=-1820.$

The doubled signed sum is

$2S=0-1820-1820+0$

Halving gives $S=-1820$ square units.

Taking its magnitude gives $1820$ square units.

Answer

Line $PT$ is tangent at $T$ . The other contact is $U=(16,-28)$ , and there are exactly two contacts. The quadrilateral's ordinary area is $1820$ square units.

Key idea

The right-angle condition restricts the contacts to the diameter circle, and a closed coordinate walk measures the resulting footprint.

Multiple choice

Core practice