Grade 8

Triangles and quadrilaterals: practice problems

Use properties of shapes to find angles and lengths, and use congruence conditions to write proofs.

Basic, Standard, Advanced: Grade 8 · Term 3

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Isosceles triangles

An isosceles triangle has two equal sides and equal base angles. With an apex angle of $40^\circ$ , each base angle is $(180^\circ - 40^\circ) \div 2 = 70^\circ$ .

Congruence conditions

Congruent triangles

Three pairs of equal sides (SSS)
Two pairs of equal sides and the included angle (SAS)
One pair of equal sides and the angles at both ends (ASA)

Right triangles are also congruent if the hypotenuse and one acute angle, or the hypotenuse and one other side, are equal.

Parallelograms and special quadrilaterals

In a parallelogram, opposite sides and opposite angles are equal and the diagonals bisect each other. Adding a right angle (or equal diagonals) gives a rectangle; adding equal adjacent sides (or perpendicular diagonals) gives a rhombus; both give a square.

Equal areas

If $\mathrm{AD} \parallel \mathrm{BC}$ , then $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DBC}$ share the base BC and have equal heights, so equal areas.

Writing a proof

Name the two triangles, list three equal parts with reasons (given, vertical angles, alternate angles, common side and so on), state the congruence condition, and then conclude using corresponding parts.

Q1BasicGr 8 · Term 3Triangles and quadrilaterals

In parallelogram ABCD the diagonals meet at O. $\mathrm{AB} = 7\text{ cm},\ \mathrm{BC} = 9\text{ cm},\ \mathrm{OA} = 4\text{ cm},\ \mathrm{OB} = 7\text{ cm}$ . Find the following lengths.

(1) $\mathrm{CD}$
(2) $\mathrm{AD}$
(3) $\mathrm{AC}$
(4) $\mathrm{OD}$

Answer

(1) $7\text{ cm}$
(2) $9\text{ cm}$
(3) $8\text{ cm}$
(4) $7\text{ cm}$

Solution

Opposite sides are equal:
$\mathrm{CD} = \mathrm{AB} = 7,\quad \mathrm{AD} = \mathrm{BC} = 9$
The diagonals bisect each other:
$\mathrm{AC} = 2\mathrm{OA} = 8,\quad \mathrm{OD} = \mathrm{OB} = 7$

Q2StandardGr 8 · Term 3Triangles and quadrilaterals

In parallelogram ABCD, the bisector of $\angle \mathrm{A}$ meets BC at E. If $\mathrm{AB} = 5\text{ cm}$ and $\mathrm{AD} = 9\text{ cm}$ , find EC.

Answer

4\text{ cm}

Solution

AD ∥ BC gives equal alternate angles, and AE is a bisector, so
$\angle \mathrm{BEA} = \angle \mathrm{DAE} = \angle \mathrm{BAE}$
$\triangle \mathrm{ABE}$ is isosceles, so
$\mathrm{BE} = \mathrm{AB} = 5$
Since BC = AD,
$\mathrm{EC} = 9 - 5 = 4\ (\text{cm})$

Q3AdvancedGr 8 · Term 3Triangles and quadrilaterals

Rectangle ABCD is folded along diagonal BD. C moves to E, and BE meets AD at F. Find $\angle x$ .

Answer

\angle x = 126^\circ

Solution

Folded angles are equal: $\angle \mathrm{FBD} = \angle \mathrm{DBC}$ . Since $\angle \mathrm{ABC} = 90^\circ$ ,
$\angle \mathrm{DBC} = (90^\circ - 36^\circ) \div 2 = 27^\circ$
AD ∥ BC gives $\angle \mathrm{FDB} = \angle \mathrm{DBC}$ . In $\triangle \mathrm{FBD}$ ,
$\angle x = 180^\circ - 27^\circ \times 2 = 126^\circ$

Q1BasicGr 8 · Term 3Triangles and quadrilaterals

In parallelogram ABCD the diagonals meet at O. $\mathrm{AB} = 9\text{ cm},\ \mathrm{BC} = 17\text{ cm},\ \mathrm{OA} = 11\text{ cm},\ \mathrm{OB} = 8\text{ cm}$ . Find the following lengths.

(1) $\mathrm{CD}$
(2) $\mathrm{AD}$
(3) $\mathrm{AC}$
(4) $\mathrm{OD}$

Q2StandardGr 8 · Term 3Triangles and quadrilaterals

If the condition $\mathrm{AC} \perp \mathrm{BD}$ is added to parallelogram ABCD, what kind of quadrilateral is it?

Aa rectangle
Ba rhombus
Ca square
Da trapezoid

Q3AdvancedGr 8 · Term 3Triangles and quadrilaterals

In the figure, $\mathrm{OA} = \mathrm{AB} = \mathrm{BC}$ . Find $\angle x$ .

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Isosceles triangles

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