(a) In triangle DEF, DE = 6 cm and DF = 4.8 cm.
Using a ruler and compasses only, construct triangle DEF.
Leave in your construction arcs.
The line EF has been drawn for you.
(b)
(i) Write down the letter of the triangle that is congruent to triangle T.
(ii) Write down the letter of the triangle that is similar but not congruent to triangle T.
(c)
The diagram shows an isosceles triangle.
(i) Show that the perpendicular height, h, is 6.58cm, correct to 3 significant figures.
(ii) Calculate the area of the triangle. Give the units of your answer.
(iii) Kalpit tries to arrange some of these triangles to make a regular polygon with centre O.
Show that Kalpit cannot make a regular polygon.
▶️ Answer/Explanation
(a) Ans: Correct construction with arcs
1. Place compass at E, draw arc with radius 6 cm
2. Place compass at F, draw arc with radius 4.8 cm
3. Intersection point is D
4. Connect D to E and F to complete ΔDEF
(b)(i) Ans: G
Triangle G is congruent to T as it has identical side lengths and angles, just rotated.
(b)(ii) Ans: H
Triangle H is similar to T (same angles, proportional sides) but smaller in size.
(c)(i) Ans: h = 6.58 cm
Using right triangle trigonometry:
$\tan(62°) = \frac{h}{3.5} \Rightarrow h = 3.5 \times \tan(62°) ≈ 6.58 \text{ cm}$
(c)(ii) Ans: 23.0 cm²
Area = ½ × base × height = ½ × 7 × 6.58 ≈ 23.0 cm²
(c)(iii) Ans: Cannot form regular polygon
Each triangle has vertex angle = 180° – 2×62° = 56°
Number needed = 360°/56° ≈ 6.43 (not integer)
Since we can’t have a fraction of a triangle, regular polygon is impossible.
(a) Town S is 44km from town R on a bearing of 117°.
(i) Using a scale of 1 cm represents 8km, mark the position of town S.
Scale: 1cm to 8km
(ii) Anvi cycles the 44km from R to S. She leaves R at 1315 and cycles at a speed of 12km/h. Work out the time she arrives at S.
(b) A tower has a height of 16 metres. When Jai makes a scale drawing of the tower it has a height of 20cm. Work out the scale Jai uses, giving your answer in the form 1 : n.
(c) X, Y and Z are three towns.
X is on a bearing of 288° from Y. Z is on a bearing of 018° from Y.
(i) Show that angle XYZ is 90°.
(ii) XY = 6 km and YZ = 9.7 km. Calculate XZ.
▶️ Answer/Explanation
(a)(i) Ans: Position marked 5.5 cm from R at 117° bearing
Scale conversion: 44 km ÷ 8 km/cm = 5.5 cm
(a)(ii) Ans: 16:55
Time = Distance ÷ Speed = 44 ÷ 12 = 3⅔ hours = 3h40m
Departure: 13:15 + 3h40m = 16:55
(b) Ans: 1:80
Scale ratio: Drawing height ÷ Actual height = 20 cm ÷ 16 m = 0.2 m ÷ 16 m = 1:80
(c)(i) Proof:
Angle between bearings: 288° – 18° = 270°
Internal angle XYZ = 360° – 270° = 90°
(c)(ii) Ans: 11.4 km
Using Pythagoras’ theorem:
XZ = √(6² + 9.7²) = √(36 + 94.09) = √130.09 ≈ 11.4 km