Hard A-Level Radians, arcs and sectors Questions

Challenging, exam-style A-Level Radians, arcs and sectors questions with worked solutions. Stretch yourself on the hardest radians, arc length, sector area, sector perimeter problems.

radiansarc lengthsector areasector perimetersegment areachord length
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A sector has radius r=12r=12 and angle θ=π2\theta=\frac{\pi}{2} radians. Which of the following is the exact area of the triangle formed by the radii?
Show worked solution

Worked solution

  1. State the given radius and angle

    r=12, θ=π2 radr=12,\ \theta=\frac{\pi}{2}\text{ rad}

    Write down the information provided in the question.

  2. Express the angle in degrees

    θ=π2=90∘\theta=\frac{\pi}{2}=90^\circ

    Converting helps you picture the size of the sector.

  3. Recall the area of the triangle between the two radii

    T=12r2sin⁡θT=\tfrac{1}{2}r^2\sin\theta

    The two radii and the chord form a triangle.

  4. Calculate the triangle area

    T=12×122×sin⁡π2=72T=\tfrac{1}{2}\times 12^2\times\sin \frac{\pi}{2}=72

    Use the included-angle area formula.

  5. Recall the arc length formula

    s=rθs=r\theta

    The arc length equals the radius times the angle in radians.

  6. Calculate the arc length

    s=12×π2=6πs=12\times \frac{\pi}{2}=6 \pi

    Substitute the radius and angle.

  7. Recall the sector area formula

    A=12r2θA=\tfrac{1}{2}r^2\theta

    The area of a sector uses the square of the radius.

  8. Calculate the sector area

    A=12×122×π2=36πA=\tfrac{1}{2}\times 12^2\times \frac{\pi}{2}=36 \pi

    Substitute the values and simplify.

  9. Recall the perimeter of a sector

    P=2r+rθP=2r+r\theta

    The perimeter is the two radii plus the arc.

  10. Calculate the perimeter

    P=2×12+6π=6π+24P=2\times 12+6 \pi=6 \pi + 24

    Add the two straight edges to the arc length.

  11. Recall the area of a segment

    S=12r2(θ−sin⁡θ)S=\tfrac{1}{2}r^2(\theta-\sin\theta)

    The segment is the sector minus the triangle.

  12. Calculate the segment area

    S=12×122(π2−sin⁡π2)=−72+36πS=\tfrac{1}{2}\times 12^2\left(\frac{\pi}{2}-\sin \frac{\pi}{2}\right)=-72 + 36 \pi

    Subtract the triangle from the sector.

  13. Recall the chord length

    c=2rsin⁡θ2c=2r\sin\tfrac{\theta}{2}

    The chord subtends the angle at the centre.

  14. Calculate the chord length

    c=2×12×sin⁡π4=122c=2\times 12\times\sin \frac{\pi}{4}=12 \sqrt{2}

    Halve the angle and use the chord formula.

  15. Compare with the options and select

    T=72T=72

    The matching option is the answer.

Answer
7272
Question 2
8 markschallenging
A sector has radius r=8r=8 and angle θ=2π3\theta=\frac{2 \pi}{3} radians. Which of the following is the exact length of the chord?
Show worked solution

Worked solution

  1. State the given radius and angle

    r=8, θ=2π3 radr=8,\ \theta=\frac{2 \pi}{3}\text{ rad}

    Write down the information provided in the question.

  2. Express the angle in degrees

    θ=2π3=120∘\theta=\frac{2 \pi}{3}=120^\circ

    Converting helps you picture the size of the sector.

  3. Recall the chord length

    c=2rsin⁡θ2c=2r\sin\tfrac{\theta}{2}

    The chord subtends the angle at the centre.

  4. Calculate the chord length

    c=2×8×sin⁡π3=83c=2\times 8\times\sin \frac{\pi}{3}=8 \sqrt{3}

    Halve the angle and use the chord formula.

  5. Recall the arc length formula

    s=rθs=r\theta

    The arc length equals the radius times the angle in radians.

  6. Calculate the arc length

    s=8×2π3=16π3s=8\times \frac{2 \pi}{3}=\frac{16 \pi}{3}

    Substitute the radius and angle.

  7. Recall the sector area formula

    A=12r2θA=\tfrac{1}{2}r^2\theta

    The area of a sector uses the square of the radius.

  8. Calculate the sector area

    A=12×82×2π3=64π3A=\tfrac{1}{2}\times 8^2\times \frac{2 \pi}{3}=\frac{64 \pi}{3}

    Substitute the values and simplify.

  9. Recall the perimeter of a sector

    P=2r+rθP=2r+r\theta

    The perimeter is the two radii plus the arc.

  10. Calculate the perimeter

    P=2×8+16π3=16+16π3P=2\times 8+\frac{16 \pi}{3}=16 + \frac{16 \pi}{3}

    Add the two straight edges to the arc length.

  11. Recall the area of the triangle between the two radii

    T=12r2sin⁡θT=\tfrac{1}{2}r^2\sin\theta

    The two radii and the chord form a triangle.

  12. Calculate the triangle area

    T=12×82×sin⁡2π3=163T=\tfrac{1}{2}\times 8^2\times\sin \frac{2 \pi}{3}=16 \sqrt{3}

    Use the included-angle area formula.

  13. Recall the area of a segment

    S=12r2(θ−sin⁡θ)S=\tfrac{1}{2}r^2(\theta-\sin\theta)

    The segment is the sector minus the triangle.

  14. Calculate the segment area

    S=12×82(2π3−sin⁡2π3)=−163+64π3S=\tfrac{1}{2}\times 8^2\left(\frac{2 \pi}{3}-\sin \frac{2 \pi}{3}\right)=- 16 \sqrt{3} + \frac{64 \pi}{3}

    Subtract the triangle from the sector.

  15. Compare with the options and select

    c=83c=8 \sqrt{3}

    The matching option is the answer.

Answer
838 \sqrt{3}
Question 3
8 markschallenging
A sector has radius r=5r=5 and angle θ=π2\theta=\frac{\pi}{2} radians. Which of the following is the exact perimeter of the sector?
Show worked solution

Worked solution

  1. State the given radius and angle

    r=5, θ=π2 radr=5,\ \theta=\frac{\pi}{2}\text{ rad}

    Write down the information provided in the question.

  2. Express the angle in degrees

    θ=π2=90∘\theta=\frac{\pi}{2}=90^\circ

    Converting helps you picture the size of the sector.

  3. Recall the perimeter of a sector

    P=2r+rθP=2r+r\theta

    The perimeter is the two radii plus the arc.

  4. Calculate the perimeter

    P=2×5+5π2=5π2+10P=2\times 5+\frac{5 \pi}{2}=\frac{5 \pi}{2} + 10

    Add the two straight edges to the arc length.

  5. Recall the arc length formula

    s=rθs=r\theta

    The arc length equals the radius times the angle in radians.

  6. Calculate the arc length

    s=5×π2=5π2s=5\times \frac{\pi}{2}=\frac{5 \pi}{2}

    Substitute the radius and angle.

  7. Recall the sector area formula

    A=12r2θA=\tfrac{1}{2}r^2\theta

    The area of a sector uses the square of the radius.

  8. Calculate the sector area

    A=12×52×π2=25π4A=\tfrac{1}{2}\times 5^2\times \frac{\pi}{2}=\frac{25 \pi}{4}

    Substitute the values and simplify.

  9. Recall the area of the triangle between the two radii

    T=12r2sin⁡θT=\tfrac{1}{2}r^2\sin\theta

    The two radii and the chord form a triangle.

  10. Calculate the triangle area

    T=12×52×sin⁡π2=252T=\tfrac{1}{2}\times 5^2\times\sin \frac{\pi}{2}=\frac{25}{2}

    Use the included-angle area formula.

  11. Recall the area of a segment

    S=12r2(θ−sin⁡θ)S=\tfrac{1}{2}r^2(\theta-\sin\theta)

    The segment is the sector minus the triangle.

  12. Calculate the segment area

    S=12×52(π2−sin⁡π2)=−252+25π4S=\tfrac{1}{2}\times 5^2\left(\frac{\pi}{2}-\sin \frac{\pi}{2}\right)=- \frac{25}{2} + \frac{25 \pi}{4}

    Subtract the triangle from the sector.

  13. Recall the chord length

    c=2rsin⁡θ2c=2r\sin\tfrac{\theta}{2}

    The chord subtends the angle at the centre.

  14. Calculate the chord length

    c=2×5×sin⁡π4=52c=2\times 5\times\sin \frac{\pi}{4}=5 \sqrt{2}

    Halve the angle and use the chord formula.

  15. Compare with the options and select

    P=5π2+10P=\frac{5 \pi}{2} + 10

    The matching option is the answer.

Answer
5π2+10\frac{5 \pi}{2} + 10
Question 4
8 markschallenging
A sector has radius r=6r=6 and angle θ=2π3\theta=\frac{2 \pi}{3} radians. Which of the following is the exact arc length?
Show worked solution

Worked solution

  1. State the given radius and angle

    r=6, θ=2π3 radr=6,\ \theta=\frac{2 \pi}{3}\text{ rad}

    Write down the information provided in the question.

  2. Express the angle in degrees

    θ=2π3=120∘\theta=\frac{2 \pi}{3}=120^\circ

    Converting helps you picture the size of the sector.

  3. Recall the arc length formula

    s=rθs=r\theta

    The arc length equals the radius times the angle in radians.

  4. Calculate the arc length

    s=6×2π3=4πs=6\times \frac{2 \pi}{3}=4 \pi

    Substitute the radius and angle.

  5. Recall the sector area formula

    A=12r2θA=\tfrac{1}{2}r^2\theta

    The area of a sector uses the square of the radius.

  6. Calculate the sector area

    A=12×62×2π3=12πA=\tfrac{1}{2}\times 6^2\times \frac{2 \pi}{3}=12 \pi

    Substitute the values and simplify.

  7. Recall the perimeter of a sector

    P=2r+rθP=2r+r\theta

    The perimeter is the two radii plus the arc.

  8. Calculate the perimeter

    P=2×6+4π=12+4πP=2\times 6+4 \pi=12 + 4 \pi

    Add the two straight edges to the arc length.

  9. Recall the area of the triangle between the two radii

    T=12r2sin⁡θT=\tfrac{1}{2}r^2\sin\theta

    The two radii and the chord form a triangle.

  10. Calculate the triangle area

    T=12×62×sin⁡2π3=93T=\tfrac{1}{2}\times 6^2\times\sin \frac{2 \pi}{3}=9 \sqrt{3}

    Use the included-angle area formula.

  11. Recall the area of a segment

    S=12r2(θ−sin⁡θ)S=\tfrac{1}{2}r^2(\theta-\sin\theta)

    The segment is the sector minus the triangle.

  12. Calculate the segment area

    S=12×62(2π3−sin⁡2π3)=−93+12πS=\tfrac{1}{2}\times 6^2\left(\frac{2 \pi}{3}-\sin \frac{2 \pi}{3}\right)=- 9 \sqrt{3} + 12 \pi

    Subtract the triangle from the sector.

  13. Recall the chord length

    c=2rsin⁡θ2c=2r\sin\tfrac{\theta}{2}

    The chord subtends the angle at the centre.

  14. Calculate the chord length

    c=2×6×sin⁡π3=63c=2\times 6\times\sin \frac{\pi}{3}=6 \sqrt{3}

    Halve the angle and use the chord formula.

  15. Compare with the options and select

    s=4πs=4 \pi

    The matching option is the answer.

Answer
4π4 \pi
Question 5
8 markschallenging
A sector has radius r=10r=10 and angle θ=π2\theta=\frac{\pi}{2} radians. Which of the following is the exact area of the sector?
Show worked solution

Worked solution

  1. State the given radius and angle

    r=10, θ=π2 radr=10,\ \theta=\frac{\pi}{2}\text{ rad}

    Write down the information provided in the question.

  2. Express the angle in degrees

    θ=π2=90∘\theta=\frac{\pi}{2}=90^\circ

    Converting helps you picture the size of the sector.

  3. Recall the sector area formula

    A=12r2θA=\tfrac{1}{2}r^2\theta

    The area of a sector uses the square of the radius.

  4. Calculate the sector area

    A=12×102×π2=25πA=\tfrac{1}{2}\times 10^2\times \frac{\pi}{2}=25 \pi

    Substitute the values and simplify.

  5. Recall the arc length formula

    s=rθs=r\theta

    The arc length equals the radius times the angle in radians.

  6. Calculate the arc length

    s=10×π2=5πs=10\times \frac{\pi}{2}=5 \pi

    Substitute the radius and angle.

  7. Recall the perimeter of a sector

    P=2r+rθP=2r+r\theta

    The perimeter is the two radii plus the arc.

  8. Calculate the perimeter

    P=2×10+5π=5π+20P=2\times 10+5 \pi=5 \pi + 20

    Add the two straight edges to the arc length.

  9. Recall the area of the triangle between the two radii

    T=12r2sin⁡θT=\tfrac{1}{2}r^2\sin\theta

    The two radii and the chord form a triangle.

  10. Calculate the triangle area

    T=12×102×sin⁡π2=50T=\tfrac{1}{2}\times 10^2\times\sin \frac{\pi}{2}=50

    Use the included-angle area formula.

  11. Recall the area of a segment

    S=12r2(θ−sin⁡θ)S=\tfrac{1}{2}r^2(\theta-\sin\theta)

    The segment is the sector minus the triangle.

  12. Calculate the segment area

    S=12×102(π2−sin⁡π2)=−50+25πS=\tfrac{1}{2}\times 10^2\left(\frac{\pi}{2}-\sin \frac{\pi}{2}\right)=-50 + 25 \pi

    Subtract the triangle from the sector.

  13. Recall the chord length

    c=2rsin⁡θ2c=2r\sin\tfrac{\theta}{2}

    The chord subtends the angle at the centre.

  14. Calculate the chord length

    c=2×10×sin⁡π4=102c=2\times 10\times\sin \frac{\pi}{4}=10 \sqrt{2}

    Halve the angle and use the chord formula.

  15. Compare with the options and select

    A=25πA=25 \pi

    The matching option is the answer.

Answer
25π25 \pi

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