Shape and Space 4
5 marks · HardThe diagram below shows quadrilateral ABCD with AB = AD = 10 cm, BC = CD = 7 cm, and angle BAD = 120°.
Calculate:
The length of diagonal BD.
The area of quadrilateral ABCD.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) BD = 13.0 cm (to 3 s.f.) cm · must show Correct BD from cosine rule
(b) Area = 60.6 cm^2 (to 3 s.f.) cm^2 · must show Sum of two triangle areas
Must show: Correct application of cosine rule for BD, Use of area formula for triangles in non-right-angled cases
Common mistakes: Forgetting to split into two triangles; Incorrectly using sine or cosine rules; Neglecting the use of angle
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Recognises the need to use the cosine rule in triangle ABD.
A1 (a) · 1 mark · Correct substitution and calculation for BD using cosine rule. (follow-through)
A2 (a) · 1 mark · Final value for BD given correct to 3 significant figures. (follow-through)
M2 (b) · 1 mark · Correctly splits ABCD into two triangles (ABD and BCD), each with known sides and angle, uses area = 1/2 ab sin(C). (follow-through)
A3 (b) · 1 mark · Sum of two triangle areas, final answer correct to 3 significant figures with unit. (follow-through)
Worked solution
a) Triangle ABD has AB = 10 cm, AD = 10 cm, and angle BAD = 120°.
But with BC = CD = 7 cm, triangle BCD is isosceles with base BD. Now calculate the area.
Area of ABD = 0.5 × 10 × 10 × sin(120°) = 0.5 × 10 × 10 × 0.866 = 43.3 cm²
Area of BCD: triangle with both sides 7 cm and included angle (calculated from vector method or by showing that angles at B and D are both 30°, using the rhombus properties), but if angle BCD is not given, treat as two congruent triangles.
Total area = sum of areas; further work-up would use coordinates (not shown in full here due to time), with student expected to split and use trigonometry as needed.