7.4

Construction of Triangles and Quadrilaterals

AConstruction of Triangles

To construct a triangle with three given sides (e.g. $AB = 5$ cm, $BC = 3$ cm, $CA = 4$ cm), we use a ruler and a pair of compasses.

🎬 Demo: Construct $\triangle ABC$ with $AB = 5$ cm, $BC = 3$ cm, $CA = 4$ cm (SSS)

Step 1 / 4

STEP 1.Draw and label a line segment $AB$ of length 5 cm with your ruler.

📖 Worked Example 18 · 直角三角形

(a) Construct $\triangle ABC$ in which $AB = 4.5$ cm, $BC = 3.5$ cm and $\angle ACB = 90°$. (b) Is $\triangle ABC$ an isosceles triangle?

SOLUTION

Steps:

Draw $BC$ of length $3.5$ cm.
At $C$, draw a ray perpendicular to $BC$ using a protractor or set square.
With $B$ as centre and $4.5$ cm as radius, draw an arc to cut the ray at $A$.
Draw segments $AB$ and $AC$.

(b) Method 1: By measuring with a ruler, $AC \approx 2.8$ cm $\ne AB$ or $BC$. So $\triangle ABC$ is not isosceles.

Method 2: By measuring with a protractor, $\angle ABC = 39°$ and $\angle BAC = 51°$. No two angles are equal, so $\triangle ABC$ is not isosceles. In fact, it is a right-angled scalene triangle.

✍️ Try It Yourself 18

(a) Construct $\triangle PQR$ in which $PQ = 7.5$ cm, $QR = 6$ cm and $\angle PRQ = 90°$. (b) Is $\triangle PQR$ isosceles?

Is $\triangle PQR$ isosceles? 检查

💡 思路

By Pythagoras: $PR^2 + QR^2 = PQ^2$ → $PR^2 = 7.5^2 - 6^2 = 56.25 - 36 = 20.25$ → $PR = 4.5$ cm. Sides $7.5, 6, 4.5$ — all different. Not isosceles. Right-angled scalene.

📖 Worked Example 19 · 已知两角夹边 (ASA)

Construct $\triangle XYZ$ with $XY = 5$ cm, $\angle YXZ = 40°$ and $\angle ZYX = 70°$. Measure and estimate $YZ$.

SOLUTION

Steps:

Draw $XY$ of length $5$ cm.
At $X$, draw a ray making $40°$ with $XY$.
At $Y$, draw a ray making $70°$ with $XY$.
The intersection is $Z$.
Measure $YZ$.

Length of $YZ \approx \mathbf{3.4}$ cm

✍️ Try It Yourself 19

Construct $\triangle LMN$ with $LM = 8$ cm, $\angle LMN = 30°$ and $\angle NLM = 75°$. Measure $LN$.

$LN \approx $ cm 检查

💡 思路

Apply law of sines: $\angle N = 180 - 30 - 75 = 75°$. $LN/\sin(\angle M) = LM/\sin(\angle N)$ → $LN = 8 \sin(30°)/\sin(75°) = 4/0.966 \approx 4.14$ cm.

BConstruction of Quadrilaterals

With enough information about sides and angles, we can construct any quadrilateral using ruler, protractor and compass.

📖 Worked Example 20 · 四边形作图

Construct quadrilateral $PQRS$ with $PQ = 3$ cm, $QR = 2.5$ cm, $SP = 3.5$ cm, $\angle SPQ = 90°$ and $\angle PQR = 105°$. Measure $\angle QRS$.

SOLUTION

Construction steps:

Draw $PQ$ of length $3$ cm.
At $Q$, draw a ray making $105°$ with $PQ$.
With $Q$ as centre and $2.5$ cm as radius, mark $R$ on this ray.
At $P$, draw a ray perpendicular to $PQ$ (so $\angle SPQ = 90°$).
With $P$ as centre and $3.5$ cm as radius, mark $S$ on this ray.
Draw $QR$, $RS$, $SP$. $PQRS$ is the required quadrilateral.
Using a protractor, $\angle QRS \approx \mathbf{90°}$.

✍️ Try It Yourself 20

Construct quadrilateral $XYZT$ in which $XY = 6$ cm, $YZ = 7.5$ cm, $TX = 6$ cm, $\angle TXY = 130°$ and $\angle XYZ = 90°$. Measure $\angle XTZ$.

$\angle XTZ \approx $ $°$ 检查

📖 Worked Example 21 · 梯形作图（含辅助线）

Construct trapezium $ABCD$ in which $AB \parallel DC$, $AB = 5$ cm, $DC = 2$ cm, $\angle DAB = 60°$ and $\angle ABC = 45°$.

ANALYSIS: Drawing $DC$ first is tricky because we don't yet know exactly where $D$ goes. Instead, draw an auxiliary segment $ED$ parallel to $BC$ with $E$ on $AB$ and $ED = BC$. Then $BCDE$ is a parallelogram, so $EB = DC = 2$ cm and $AE = 5 - 2 = 3$ cm. Triangle $ADE$ has known $AE = 3$ cm, $\angle DAE = 60°$, $\angle AED = \angle ABC = 45°$ (corresponding angles since $ED \parallel BC$) — easy to draw.

SOLUTION

Construction steps:

Draw $AB$ of length $5$ cm and mark point $E$ on it with $AE = 3$ cm.
At $A$ and $E$, draw rays making angles $60°$ and $45°$ respectively with $AB$.
Let $D$ be the intersection of the two rays from step 2. Draw $DC$ parallel to $AB$, with $DC = 2$ cm.
Draw side $BC$. $ABCD$ is the required trapezium.

✍️ Try It Yourself 21

Construct trapezium $PQRS$ in which $PQ \parallel SR$, $PQ = 7$ cm, $SR = 3$ cm, $\angle SPQ = \angle PQR = 50°$. (Note: this is an isosceles trapezium since the two base angles are equal.)

无需填写答案；按步骤画出图形即可。

PPractice Exercise 7.4

BASIC MASTERY

Q1. For each set of side lengths, determine if a triangle can be drawn. If yes, classify by sides.
(a) $\triangle ABC$ with $AB = 7$ cm, $BC = 4$ cm, $CA = 5$ cm: possible? 检查 ; type by sides: 检查
(b) $\triangle DEF$ with $DE = 6$ cm, $EF = 3$ cm, $FD = 2$ cm: possible? 检查 (since $2 + 3 = 5 < 6$, triangle inequality fails)
Q2. Construct $\triangle DEF$ with $DE = 7$ cm, $\angle FDE = 60°$, $\angle DEF = 45°$. (b) Classify by angles. (c) Which angle is opposite the longest side?
(b) Type: triangle 检查
(c) Opposite the longest side is: 检查
💡 思路
$\angle F = 180 - 60 - 45 = 75°$. All three angles < 90°, so acute-angled. The largest angle (75°) is opposite the longest side, which is $DE = 7$. The angle $\angle DFE$ is opposite $DE$.
Q3. Construct $\triangle GHK$ with $GH = 5$ cm, $GK = 4$ cm, $\angle HGK = 110°$.
(b) Measure $HK$ (to nearest $0.1$ cm): cm 检查
(c) Classify by sides: 检查
💡 思路
Use law of cosines: $HK^2 = 5^2 + 4^2 - 2(5)(4)\cos(110°) = 41 + 13.68 ≈ 54.68$ → $HK ≈ 7.4$ cm. Three different sides + one angle > 90° → obtuse-angled scalene.
Q4. Construct parallelogram $ABCD$ with $AB = 3$ cm, $AD = 2$ cm, $\angle BAD = 70°$. (No numeric answer — verify by drawing.)
Q5. Construct trapezium $KLMN$ with $KL \parallel NM$, $KL = 4$ cm, $LM = 3$ cm, $NK = 2.5$ cm, $\angle LKN = 90°$. (No numeric answer.)

INTERMEDIATE

Q6. Construct rhombus $PQRS$ in which diagonals $PR = 4$ cm and $QS = 3$ cm. (No numeric answer.)
Q7. (a) Construct $\triangle XYZ$ with $YZ = 2.5$ cm, $XZ = 6.5$ cm, $\angle XYZ = 90°$. (b) Measure $XY$ (to nearest $0.1$ cm). (c) Classify by sides.
(b) $XY = $ cm 检查
(c) Classification: 检查
💡 思路
By Pythagoras: $XY^2 + YZ^2 = XZ^2$ → $XY^2 = 6.5^2 - 2.5^2 = 36$ → $XY = 6.0$ cm. The triangle has a right angle at $Y$ — so right-angled.
Q8. (a) Construct $\triangle ABC$ with $AB = 4$ cm, $AC = 4$ cm, $\angle ABC = 30°$. (b) Measure $\angle BAC$ and $\angle ACB$ to the nearest degree. (c) Classify by sides.
(b) $\angle BAC = $ $°$ ; $\angle ACB = $ $°$ 检查 BAC 检查 ACB
(c) Classification by sides: 检查
💡 思路
$AB = AC$ → isos with $\angle ABC = \angle ACB$. Given $\angle ABC = 30°$, so $\angle ACB = 30°$. $\angle BAC = 180 - 60 = 120°$.
Q9. Construct quadrilateral $ABCD$ with $AD = 4$ cm, $BC = 2$ cm, $CD = 3$ cm, $\angle BCD = 120°$, $\angle ADC = 100°$. (No numeric answer.)

ADVANCED

Q10. (a) Draw an equilateral $\triangle ABC$ of any size. (b) Plot midpoints $D, E, F$ of sides $AB, BC, CA$. (c) Draw $\triangle DEF$. (d) Type of $\triangle DEF$? (e) Find $DE/AB$.
(d) Type: 检查
(e) $DE / AB = $ 检查
💡 思路
Midsegment of any triangle is parallel to the opposite side and equal to half its length. Applied to all 3 sides of an equilateral triangle → midpoint triangle is also equilateral, with side ratio $1:2$.
Q11. In $\triangle ABC$, $AB = 4$ cm, $BC = 3$ cm, $\angle BAC = 30°$. (a) Construct $\triangle ABC$. (b) How many different triangles satisfy the given measurements? (c) What can you conclude about SSA?
(b) Number of triangles: 检查
(c) Conclusion (one-sentence): When two sides $BC < AB$ and a non-included angle are given, triangles. 检查
💡 思路
This is the classic SSA ambiguous case. $\sin(\angle C)/AB = \sin(\angle A)/BC$ → $\sin(\angle C) = 4 \cdot \sin(30°)/3 = 2/3$. Two values for $\angle C$ ≈ 41.8° or 138.2°, both leading to valid triangles. Hence SSA does not uniquely determine a triangle.
Q12. (a) Draw two different quadrilaterals with $AB = CD = 3$ cm and $AD = BC = 2$ cm. (b) What types of quadrilateral can $ABCD$ be?
(b) Type(s): 检查
💡 思路
Opposite sides equal → parallelogram (always). If additionally the angles are right, it's a rectangle. So $ABCD$ is a parallelogram or, as a special case, a rectangle.

A pair of compasses can be used to check if $AC \ne BC$ in a constructed triangle: set the compass leg span to $AC$, then place the centre at $C$ and see if the other leg reaches $B$. If not, $AC \ne BC$.