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Date November 2014 Marks available 2 Reference code 14N.1.sl.TZ0.2
Level SL only Paper 1 Time zone TZ0
Command term Calculate Question number 2 Adapted from N/A

Question

The four points A(−6, −11) , B(−2, 7) , C(4, 9) and D(6, 3) define the vertices of a kite.

Calculate the distance between points \({\text{B}}\) and \({\text{D}}\).

[2]
a.

The distance between points \({\text{A}}\) and \({\text{C}}\) is \(\sqrt {500} \).

Calculate the area of the kite \({\text{ABCD}}\).

[4]
b.

Markscheme

\({\text{BD}} = \sqrt {\left( {{4^2} + {8^2}} \right)} \)     (M1)

Note:     Award (M1) for correct substitution into the distance formula.

 

\( = 8.94\left( {8.94427 \ldots ,{\text{ }}\sqrt {80} ,{\text{ }}4\sqrt 5 } \right)\)     (A1)     (C2)

a.

Area \({\text{ABCD}} = 2 \times \left( {0.5 \times \frac{{{\text{their BD}}}}{2} \times \sqrt {500} } \right)\)     (M1)(M1)(M1)

Note: Award (M1) for dividing their BD by 2, (M1) for correct substitution into the area of triangle formula, (M1) for adding two triangles (or multiplied by 2).

Accept alternative methods:

Area of kite \( = 0.5 \times \sqrt {500}  \times \) their part (a).

Award (M1) for stating kite formula.

Award (M1) for correctly substituting in \(\sqrt {500} \).

Award (M1) for correctly substituting in their part (a).

 

\( = 100\)     (A1)     (C4)

Note: Accept 99.9522 if 3 sf answer is used from part (a).

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.0 » Coordinates in two dimensions.

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