Quadrilaterals and Polygons: practice solutions, Grade 10 – download the PDF

1 Interior angle sums ★★★

The formula is \(S=(n-2)\cdot180^\circ\).

1. \(n=6\): \(4\cdot180^\circ=720^\circ\).
2. \(n=8\): \(6\cdot180^\circ=1080^\circ\).
3. \(n=15\): \(13\cdot180^\circ=2340^\circ\).

2 A regular nonagon ★★★

Each exterior angle is \(360^\circ\div9=40^\circ\). Each interior angle is \(180^\circ-40^\circ=140^\circ\). Check: \((9-2)\cdot180^\circ\div9=1260^\circ\div9=140^\circ\).

3 Angles of a parallelogram ★★★

\(\angle R=\angle P=68^\circ\) because opposite angles are congruent. \(\angle Q\) and \(\angle P\) are consecutive, so they are supplementary: \(\angle Q=180^\circ-68^\circ=112^\circ\). Then \(\angle S=\angle Q=112^\circ\). Check: \(68+112+68+112=360\).

4 Opposite sides ★★★

Opposite sides of a parallelogram are congruent, so \(3x+2=5x-6\). Then \(8=2x\) and \(x=4\). \(AB=3\cdot4+2=14\text{ cm}\), and \(CD=5\cdot4-6=14\text{ cm}\), which matches.

5 True or false? ★★★

1. True. A square has four right angles, which is the definition of a rectangle.
2. False. A \(3\text{ cm}\) by \(5\text{ cm}\) rectangle has sides that are not all congruent.
3. False. In a non-rectangular parallelogram, such as one with sides \(5\) and \(8\) and a \(60^\circ\) angle, the two diagonals have different lengths. Only rectangles have congruent diagonals.

6 Diagonals that bisect ★★★

The diagonals bisect each other, so \(JN=NL\): \(2x+3=4x-7\), giving \(10=2x\) and \(x=5\). Then \(JN=2\cdot5+3=13\) and \(NL=4\cdot5-7=13\). The diagonal \(JL=13+13=26\).

7 A garden bed ★★★

The fence is the midsegment: \(m=\dfrac{12+20}{2}=16\text{ ft}\) (about \(4.9\text{ m}\)).

8 Angles in a pentagon ★★★

The sum is \((5-2)\cdot180^\circ=540^\circ\). So \(5x+100=540\), \(5x=440\), \(x=88\). The angles are \(88^\circ,\ 98^\circ,\ 108^\circ,\ 118^\circ,\ 128^\circ\). Check: \(88+98+108+118+128=540\).

9 How many sides? ★★★

Each exterior angle is \(180^\circ-156^\circ=24^\circ\). Since the exterior angles add up to \(360^\circ\), \(n=360\div24=15\). The polygon is a regular 15-gon. Check: \(13\cdot180^\circ\div15=156^\circ\).

10 A rhombus from its diagonals ★★★

The diagonals are perpendicular and bisect each other, so a side is the hypotenuse of a right triangle with legs \(5\) and \(12\): \(\sqrt{25+144}=13\text{ cm}\). The perimeter is \(4\cdot13=52\text{ cm}\). The area is \(\dfrac{10\cdot24}{2}=120\text{ cm}^2\).

11 A rectangular screen ★★★

1. The diagonals of a rectangle are congruent: \(7x-5=4x+10\), so \(3x=15\) and \(x=5\). The diagonal is \(7\cdot5-5=30\text{ cm}\).
2. By the Pythagorean theorem, \(\ell=\sqrt{30^2-18^2}=\sqrt{900-324}=\sqrt{576}=24\text{ cm}\).

12 Angles of a kite ★★★

The angles between unequal sides are congruent, so \(\angle D=\angle B=112^\circ\). The sum is \(360^\circ\), so \(\angle C=360^\circ-48^\circ-112^\circ-112^\circ=88^\circ\). Diagonal \(\overline{AC}\) bisects \(\angle A\), so it makes an angle of \(48^\circ\div2=24^\circ\) with \(\overline{AB}\).

13 Isosceles trapezoid angles ★★★

\(\overline{AB}\parallel\overline{DC}\), so \(\angle D\) and \(\angle A\) are supplementary (same-side interior angles): \(\angle D=180^\circ-65^\circ=115^\circ\). Base angles are congruent, so \(\angle C=\angle D=115^\circ\). Check: \(65+65+115+115=360\).

14 Which condition is enough? ★★★

1. Yes: one pair of opposite sides is parallel and congruent.
2. Yes: both pairs of opposite sides are congruent.
3. No: here the congruent sides are adjacent, which describes a kite and does not force opposite sides to be parallel.
4. No: a trapezoid has one parallel pair and is not a parallelogram.

15 Coordinate proof: parallelogram ★★★

1. Midpoint of \(\overline{AC}\): \(\left(\dfrac{-2+6}{2},\dfrac{-1+6}{2}\right)=(2,\,2.5)\). Midpoint of \(\overline{BD}\): \(\left(\dfrac{4+0}{2},\dfrac{1+4}{2}\right)=(2,\,2.5)\). The diagonals bisect each other, so \(ABCD\) is a parallelogram.
2. Slope of \(\overline{AB}\): \(\dfrac{1-(-1)}{4-(-2)}=\dfrac13\). Slope of \(\overline{BC}\): \(\dfrac{6-1}{6-4}=\dfrac52\). The product is \(\dfrac56\ne-1\), so \(\overline{AB}\) and \(\overline{BC}\) are not perpendicular. It is not a rectangle.

16 Coordinate proof: rectangle ★★★

Slopes: \(PQ=\dfrac24=\dfrac12\); \(QR=\dfrac{4-2}{3-4}=-2\); \(RS=\dfrac{2-4}{-1-3}=\dfrac12\); \(SP=\dfrac{0-2}{0-(-1)}=-2\). So \(\overline{PQ}\parallel\overline{RS}\) and \(\overline{QR}\parallel\overline{SP}\): a parallelogram. Since \(\dfrac12\cdot(-2)=-1\), adjacent sides are perpendicular, so it is a rectangle. Dimensions: \(PQ=\sqrt{16+4}=\sqrt{20}\) and \(QR=\sqrt{1+4}=\sqrt5\). Area \(=\sqrt{20}\cdot\sqrt5=\sqrt{100}=10\) square units. (The diagonals agree: \(PR=\sqrt{9+16}=5\) and \(QS=5\).)

17 Coordinate proof: rhombus ★★★

1. \(AB=\sqrt{9+16}=5\), \(BC=\sqrt{9+16}=5\), \(CD=\sqrt{9+16}=5\), \(DA=\sqrt{9+16}=5\). Four congruent sides: a rhombus.
2. The diagonals are \(AC=6\) (horizontal) and \(BD=8\) (vertical), so the area is \(\dfrac{6\cdot8}{2}=24\) square units.
3. In a square the diagonals are congruent, but \(AC=6\ne BD=8\).

18 Coordinate proof: trapezoid ★★★

1. Slopes: \(AB\) and \(DC\) are both \(0\), so they are parallel. \(AD\) has slope \(\dfrac42=2\) and \(BC\) has slope \(\dfrac{4}{-2}=-2\), so the legs are not parallel: exactly one pair. Legs: \(AD=\sqrt{4+16}=\sqrt{20}\) and \(BC=\sqrt{4+16}=\sqrt{20}\), so it is isosceles.
2. Bases: \(AB=8\), \(DC=4\). Midsegment \(=\dfrac{8+4}{2}=6\). (Check: the midpoints of the legs are \((1,2)\) and \((7,2)\), which are \(6\) apart.) Height \(=4\), so the area is \(6\cdot4=24\) square units.

19 Algebra with a parallelogram ★★★

1. Opposite angles are congruent: \(3x+15=5x-25\), so \(40=2x\) and \(x=20\). Then \(\angle A=\angle C=75^\circ\), and \(\angle B=\angle D=180^\circ-75^\circ=105^\circ\).
2. Let the short side be \(s\). Then \(2(s+s+6)=76\), \(2s+6=38\), \(s=16\). The sides are \(16\text{ in}\) and \(22\text{ in}\) (about \(40.6\text{ cm}\) and \(55.9\text{ cm}\)). Check: \(2\cdot16+2\cdot22=76\).

20 A mystery polygon ★★★

1. \((n-2)\cdot180=1980\) gives \(n-2=11\), so \(n=13\).
2. From one vertex you cannot reach itself or its two neighbors: \(13-3=10\) diagonals.
3. Each of the 13 vertices has 10, and each diagonal is counted twice: \(\dfrac{13\cdot10}{2}=65\) diagonals.

21 Joining the midpoints ★★★

1. \(M(4,1)\), \(N(9,5)\), \(P(6,7)\), \(Q(1,3)\).
2. Move from \(M\) to \(N\): \((+5,+4)\). Move from \(Q\) to \(P\): \((+5,+4)\). Move from \(N\) to \(P\): \((-3,+2)\). Move from \(M\) to \(Q\): \((-3,+2)\). Opposite sides are parallel and congruent, so \(MNPQ\) is a parallelogram.
3. In triangle \(ABC\), \(\overline{MN}\) joins two midpoints, so it is parallel to \(\overline{AC}\) and half as long. In triangle \(ADC\), \(\overline{QP}\) is also parallel to \(\overline{AC}\) and half as long. So \(\overline{MN}\) and \(\overline{QP}\) are parallel and congruent, which proves a parallelogram.