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The diagram shows a section of a garden divided into three parts - Leaving Cert Mathematics - Question 8 - 2018

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The diagram shows a section of a garden divided into three parts. In the diagram: $|PR| = 3.3 \, m$, $|PQ| = 6.5 \, m$, $|QT| = 8 \, m$, $|\angle QPR| = 90^{\circ}$,... show full transcript

Worked Solution & Example Answer:The diagram shows a section of a garden divided into three parts - Leaving Cert Mathematics - Question 8 - 2018

Step 1

Use the theorem of Pythagoras to find |RQ|

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Answer

Using the Pythagorean theorem:

∣RQ∣2=∣PR∣2+∣PQ∣2|RQ|^2 = |PR|^2 + |PQ|^2 ∣RQ∣2=(3.3)2+(6.5)2|RQ|^2 = (3.3)^2 + (6.5)^2 ∣RQ∣2=10.89+42.25|RQ|^2 = 10.89 + 42.25 ∣RQ∣2=53.14|RQ|^2 = 53.14

Thus, ∣RQ∣=53.14≈7.3 m|RQ| = \sqrt{53.14} \approx 7.3 \, m

Step 2

Show that \alpha = 31^{\circ}, correct to the nearest degree.

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Answer

Using the sine function for angle \alpha:

sin⁡α=∣PR∣∣PQ∣=3.36.5\sin \alpha = \frac{|PR|}{|PQ|} = \frac{3.3}{6.5}

Calculating \alpha:

α=sin⁡−1(3.36.5)≈30.51∘\alpha = \sin^{-1}\left(\frac{3.3}{6.5}\right) \approx 30.51^{\circ}

Rounding to the nearest degree, we have:

α≈31∘\alpha \approx 31^{\circ}

Step 3

Use the value of \alpha given in part (b) to find the value of \beta.

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Answer

Using the angle sum of a triangle:

β=180∘−90∘−α\beta = 180^{\circ} - 90^{\circ} - \alpha β=180∘−90∘−31∘\beta = 180^{\circ} - 90^{\circ} - 31^{\circ} β=180∘−121∘=59∘\beta = 180^{\circ} - 121^{\circ} = 59^{\circ}

Step 4

Use the Cosine Rule to find the length of |RS|. Give your answer correct to the nearest metre.

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Answer

Using the Cosine Rule:

∣RS∣2=∣RQ∣2+∣QS∣2−2∣RQ∣∣QS∣cos⁡(β)|RS|^2 = |RQ|^2 + |QS|^2 - 2 |RQ| |QS| \cos(\beta)

Substituting the known values:

∣RS∣2=(7.3)2+(8)2−2(7.3)(8)cos⁡(59∘)|RS|^2 = (7.3)^2 + (8)^2 - 2(7.3)(8)\cos(59^{\circ}) ∣RS∣2=53.29+64−2(7.3)(8)(0.515)|RS|^2 = 53.29 + 64 - 2(7.3)(8)(0.515)

Calculate:

∣RS∣2=117.29−60.14≈57.15|RS|^2 = 117.29 - 60.14 \\ \approx 57.15

Thus, ∣RS∣≈57.15≈7.6 m|RS| \approx \sqrt{57.15} \approx 7.6 \, m Rounding to the nearest metre: ∣RS∣≈8 m|RS| \approx 8 \, m

Step 5

Find the length of the arc TS. Give your answer in metres, correct to one decimal place.

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Answer

The formula for the length of the arc is given by:

Arc TS=2πrθ360\text{Arc } TS = 2 \pi r \frac{\theta}{360}

Substituting the values:

Arc TS=2π(8)31360\text{Arc } TS = 2 \pi (8) \frac{31}{360}

Calculating: Arc TS≈4.3 m\text{Arc } TS \approx 4.3 \, m

Step 6

Find the area of the sector SQT. Give your answer in square metres, correct to one decimal place.

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Answer

The area of the sector is given by:

Area=πr2θ360\text{Area} = \pi r^2 \frac{\theta}{360}

Substituting the values:

Area=π(8)231360\text{Area} = \pi (8)^2 \frac{31}{360}

Calculating: Area≈17.3 m2\text{Area} \approx 17.3 \, m^2

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