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

1 Angle sums / 3 pts

(a) \((11-2)\cdot180^\circ=1620^\circ\). (1 pt)

(b) \(n=360\div15=24\) sides. (1 pt) Each interior angle is \(180^\circ-15^\circ=165^\circ\). (1 pt)

2 A parallelogram / 4 pts

(a) Consecutive angles are supplementary: \(2x+20+3x-10=180\), so \(5x=170\) and \(x=34\). (1 pt) \(\angle W=88^\circ\), \(\angle X=92^\circ\); then \(\angle Y=88^\circ\), \(\angle Z=92^\circ\) (opposite angles). (1 pt)

(b) The diagonals bisect each other: \(3t+1=5t-9\), so \(t=5\). (1 pt) \(WO=16\), so \(WY=32\). (1 pt)

3 True or false? / 3 pts

(a) True: opposite sides of a rhombus are congruent, which proves a parallelogram. (1 pt)

(b) False: in a \(6\) by \(8\) rectangle the diagonals are congruent but not perpendicular (only in a square). (1 pt)

(c) False: a kite with sides \(3,3,5,5\) in order has no pair of parallel sides. (1 pt)

4 A rhombus / 4 pts

(a) Half-diagonals \(9\) and \(12\) form a right triangle: \(\sqrt{81+144}=\sqrt{225}=15\text{ cm}\). (2 pts)

(b) \(4\cdot15=60\text{ cm}\). (1 pt)

(c) \(\dfrac{18\cdot24}{2}=216\text{ cm}^2\). (1 pt)

5 An isosceles trapezoid / 3 pts

(a) \(m=\dfrac{10+24}{2}=17\text{ in}\). (1 pt)

(b) Each leg covers a horizontal offset of \(\dfrac{24-10}{2}=7\text{ in}\), so \(h=\sqrt{25^2-7^2}=\sqrt{576}=24\text{ in}\). (1 pt)

(c) Area \(=17\cdot24=408\text{ in}^2\). (1 pt)

6 Coordinate proof / 3 pts

(a) Midpoint of \(\overline{AC}\): \((3.5,\,3.5)\); midpoint of \(\overline{BD}\): \((3.5,\,3.5)\). The diagonals bisect each other. (1 pt)

(b) \(AB=\sqrt{16+1}=\sqrt{17}\) and \(BC=\sqrt{1+16}=\sqrt{17}\): adjacent sides are congruent, so the parallelogram is a rhombus. (1 pt)

(c) Slope of \(\overline{AB}\) is \(\dfrac14\) and slope of \(\overline{BC}\) is \(4\). The product is \(1\ne-1\), so the sides are not perpendicular. (1 pt)