CAIE P1 2016 June — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeArea of triangle or parallelogram using vectors
DifficultyStandard +0.8 This question requires recognizing that triangle ABC is right-angled (3-4-5 Pythagorean triple), calculating the area of ABC, then finding the areas of three circular sectors using the angles of the triangle, and finally subtracting to find the shaded region. It combines multiple concepts (Pythagoras, triangle area, sector area, angle sum) in a non-routine geometric configuration that requires careful spatial reasoning about where the touching points lie.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)
6 \includegraphics[max width=\textwidth, alt={}, center]{8c358a10-a3e1-47b5-ae62-30ba6b76c167-3_655_1011_255_566} The diagram shows triangle \(A B C\) where \(A B = 5 \mathrm {~cm} , A C = 4 \mathrm {~cm}\) and \(B C = 3 \mathrm {~cm}\). Three circles with centres at \(A , B\) and \(C\) have radii \(3 \mathrm {~cm} , 2 \mathrm {~cm}\) and 1 cm respectively. The circles touch each other at points \(E , F\) and \(G\), lying on \(A B , A C\) and \(B C\) respectively. Find the area of the shaded region \(E F G\).
Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(BAC=\sin^{-1}(3/5)\) or \(\cos^{-1}(4/5)\) or \(\tan^{-1}(3/4)\)B1 Accept \(36.8(7)°\)
\(ABC=\sin^{-1}(4/5)\) or \(\cos^{-1}(3/5)\) or \(\tan^{-1}(4/3)\)B1 Accept \(53.1(3)°\)
\(ACB=\pi/2\) (Allow 90°)B1
Shaded area \(=\triangle ABC -\) sectors \((AEF+BEG+CFG)\)M1
\(\triangle ABC=\frac{1}{2}\times 4\times 3\) oeB1
Sum sectors \(=\frac{1}{2}\left[3^2(0.6435)\right]+2^2(0.9273)+1^2(1.5708)\)M1
OR \(\frac{\pi}{360}\left[3^2(36.8(7))+2^2(53.1(3))+1^2(90)\right]\)
\(6-5.536=0.464\)A1 [7] 
## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $BAC=\sin^{-1}(3/5)$ or $\cos^{-1}(4/5)$ or $\tan^{-1}(3/4)$ | B1 | Accept $36.8(7)°$ |
| $ABC=\sin^{-1}(4/5)$ or $\cos^{-1}(3/5)$ or $\tan^{-1}(4/3)$ | B1 | Accept $53.1(3)°$ |
| $ACB=\pi/2$ (Allow 90°) | B1 | |
| Shaded area $=\triangle ABC -$ sectors $(AEF+BEG+CFG)$ | M1 | |
| $\triangle ABC=\frac{1}{2}\times 4\times 3$ oe | B1 | |
| Sum sectors $=\frac{1}{2}\left[3^2(0.6435)\right]+2^2(0.9273)+1^2(1.5708)$ | M1 | |
| OR $\frac{\pi}{360}\left[3^2(36.8(7))+2^2(53.1(3))+1^2(90)\right]$ | | |
| $6-5.536=0.464$ | A1 [7] | |

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{8c358a10-a3e1-47b5-ae62-30ba6b76c167-3_655_1011_255_566}

The diagram shows triangle $A B C$ where $A B = 5 \mathrm {~cm} , A C = 4 \mathrm {~cm}$ and $B C = 3 \mathrm {~cm}$. Three circles with centres at $A , B$ and $C$ have radii $3 \mathrm {~cm} , 2 \mathrm {~cm}$ and 1 cm respectively. The circles touch each other at points $E , F$ and $G$, lying on $A B , A C$ and $B C$ respectively. Find the area of the shaded region $E F G$.

\hfill \mbox{\textit{CAIE P1 2016 Q6 [7]}}
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