Question
The diagram shows an incomplete scale drawing of a market place, ABCD, where D is on CX.
The scale is 1 centimeter represents 5 meters.
Scale : 1cm to 5m
D lies on CX such that angle DAB = 75°.
(a) On the diagram, draw the line AD and mark the position of D. [2]
(b) Find the actual length of the side BC of the market place.
m [2]
(c) In this part, use a ruler and compasses only.
Street sellers are allowed in the part of the market place that is
- more than 35 meters from A
and
- nearer to C than to B
and
- nearer to CD than to BC.
On the diagram, construct and shade the region where street sellers are allowed. [7]
(d) Write the scale of the drawing in the form 1: n.
1 : [1]
Answer/Explanation
Ans:
4(a) Correct ruled line with D marked
4(b) 47.5
4(c) Correct arc radius 7 cm
Correct ruled perpendicular bisector of BC with correct pairs of arcs
Correct ruled bisector of angle BCD with correct pairs of arcs
correct region shaded
4(d) [1 :] 500
Question
(a)
The diagram shows triangle FGH, with FG = 14 cm, GH = 12 cm and FH = 6 cm.
(i) Calculate the size of angle HFG.
(ii) Calculate the area of triangle FGH.
(b) 
The diagram shows triangle PQR, with RP = 12 cm, RQ = 18 cm and angle RPQ = 117°.
Calculate the size of angle RQP.
Answer/Explanation
(a) We can use the cosine rule here to solve
Cosine rule: For any triangle with sides a, b, and c and angle A opposite side a,
\(a^2 = b^2 + c^2 – 2bc\cos{A}\)
Therefore,
cos(HFG) = \(\frac{FG^2 + FH^2 – GH^2}{2.FG.FH}\)
Substituting the given values,
cos(HFG) =\( \frac{14^2 + 6^2 – 12^2}{2 .14. 6}\) = \(\frac{26}{84}\)
= \(\frac{13}{42}\)
So, HFG = \(cos^{-1}(\frac{13}{42}) \approx{68.6 degrees}\)
Therefore, the size of angle HFG is approximately 68.6 degrees.
(b) We can use Heron’s formula to find the area of triangle FGH,
Herons formula : For any triangle with sides a, b, and c, the area is given by:
Area =\( \sqrt{s(s-a)(s-b)(s-c)}\)
where s is the semi perimeter of the triangle
\(s =\frac{a + b + c}{2}\)
Here,
a = FG = 14 cm
b = GH = 12 cm
c = FH = 6 cm
Therefore, \(s =\frac{a + b + c}{2}\)
\(\frac{14 + 12 + 6}{2}\)
=16
Now,
Area = \( \sqrt{16(16-14)(16-12)(16-6)}\)
\(\sqrt{1624.10}\)
= \(\sqrt{1280}\)
= \(8\sqrt{20}\)
Therefore, the area of triangle FGH is \(8\sqrt{20}\) square centimeters.
Question
The quadrilateral ABCD represents an area of land.
There is a straight road from A to C.
AB = 79m, AD = 120m and CD = 95m.
Angle BCA = 26° and angle CDA = 77°.
(a) Show that the length of the road, AC, is 135m correct to the nearest metre.
(b) Calculate the size of the obtuse angle ABC.
(c) A straight path is to be built from B to the nearest point on the road AC.
Calculate the length of this path.
d) Houses are to be built on the land in triangle ACD.
Each house needs at least \(180{m}^2\)
of land.
Calculate the maximum number of houses which can be built.
Show all of your working
Answer/Explanation
(a) To solve this, we can use the law of cosines for finding the length of AC:
\( {AC}^2 = {AB}^2 +{BC}^2\)- 2.AB.BC.Cos(BCA)
For BC, we can use the law of sines
\(\frac{BC}{\sin(BCA)}\)
= \(\frac{CD}{\sin(CAD)}\)
Plugging in the given values, we get:
\({AC}^2 =\frac{{79}^2+{95}^2}{{Sin}^2(26)}\)
=135(rounded to the nearest metre)
(b) 48.5
(c) 30.2
(d) Maximum number of houses : \(\frac{Area}{180}\)
=\(\frac{\frac{1}{2}\times 120 \times 95 \times \sin(77)}{180}\)
=30.8