Mathematics 10 — Exercise 9.2 · Q12

Full question

A sector has perimeter 14 cm and area 10 cm². Find its radius.

Step-by-step solution

1. The perimeter contains BOTH radii and the arc: 2r+l=14.

   $$2\times x+y=14$$

2. Subtract 2r: l=14−2r.

   $$y=14-2\times x$$

3. Area=½rl=10. Substitute l=14−2r.

   $$\frac{x\times \left(14-2\times x\right)}{2}=10$$

4. Multiply ½r through: 7r−r²=10.

   $$-x^2+7\times x=10$$

5. Move all terms and multiply by −1: r²−7r+10=0.

   $$x^2-7\times x+10=0$$

6. Split −7r into −5r−2r.

   $$x^2-7\times x+10=0$$

7. Group and factor: r(r−5)−2(r−5)=(r−2)(r−5).

   $$\left(x-5\right)\times \left(x-2\right)=0$$

8. A product is zero when a factor is zero: r−2=0 or r−5=0. Add 2 or 5 respectively.

9. For r=2: l=14−4=10, θ=10/2=5 rad. For r=5: l=14−10=4, θ=4/5 rad. Both lengths are positive.

10. Both sectors satisfy the data: the first is a major sector, the second a minor sector. If a minor sector is required, choose r=5 cm.

Answer

r=2 cm (major sector) or r=5 cm (minor sector).