IBDP Math AI:Topic:AHL 3.7: area of sector, length of arc-IB style Questions HL Paper 1

Detailed Solution

Step 1: Understanding the Given Information

• A semicircle with center
  $O$

  and diameter

  $PQ$

  is given.

• A rectangle
  $OABC$

  is inside the semicircle.

• Given values:
  • $AB = 8$
  • $OA = 5$

Step 2: Finding the Radius of the Semicircle

From given figure, the radius of the semicircle is:
$$(radius OB =)\sqrt{5^{2}+8^{2}}(=\sqrt{89})$$

$$r = \frac{PQ}{2} = \frac{10}{2} = 5$$

Step 3: Finding Angle

for Arc BQ

To determine the angle subtended by BQ at the center, we find the coordinates of points:

• $O = (0,0)$
• $A = (5,0)$

  (since

  $OA = 5$

  )

• $B = (5,8)$

  (since

  $AB = 8$

  )

Let, the angle

= A

, where:

$$\tan \theta = \frac{AB}{OA} = \frac{8}{5}$$

So, the angle

at

is:

$$\theta = \tan^{-1} \left(\frac{8}{5}\right)$$

Since arc BQ corresponds to the angle BOQ, let α = 90

, we use:

$$\text{Arc Length} = r \theta$$

Step 4: Computing the Arc Length

Let’s calculate

and then find the arc length.

$(\hat{BOQ}=\frac{\pi}{2}+arctan(\frac{8}{5})$
$(\hat{BOQ}=)147.994^{\circ}. \text{ OR } 2.58299…$

$\text{substituting the radius and angle BOQ correctly into arc length formula}$
\[$\text{(arc BQ =)} \frac{90+arctan(\frac{8}{5})}{360}\times2\pi(\sqrt{5^{2}+8^{2}}) \text{ OR } \frac{\frac{\pi}{2}+arctan(\frac{8}{5})}{2}\times(\sqrt{5^{2}+8^{2}})$\]
= $24.4 (m) )$ is the length of the arc BQ

………………………….Markscheme…………………………….

Solution: –

Recognition that OB is a radius)
$$(radius=)\sqrt{5^{2}+8^{2}}(=\sqrt{89})$$

$\text{EITHER (finding angle BOQ)}$

$\text{correct calculation for finding BOA}$
\[$\hat{BOA}=arctan(\frac{8}{5}) \text{ OR } tan \hat{BOA}=\frac{8}{5}$\]
$\text{expressing BOQ as 90+BOA}$
\[$\hat{BOQ}=90+arctan(\frac{8}{5}) \text{ OR } \hat{BOQ}=\frac{\pi}{2}+arctan(\frac{8}{5})$\]
$(\hat{BOQ}=)147.994^{\circ}. \text{ OR } 2.58299…$

$\text{substituting their radius and angle BOQ correctly into arc length formula}$
\[$\text{(arc BQ =)} \frac{90+arctan(\frac{8}{5})}{360}\times2\pi(\sqrt{5^{2}+8^{2}}) \text{ OR } \frac{\frac{\pi}{2}+arctan(\frac{8}{5})}{2}\times(\sqrt{5^{2}+8^{2}})$\]
$24.4 (m) (24.3679…)$

$\text{OR (finding angle BOP)}$

$\text{correct calculation for finding angle BOP}$
\[$\hat{BOP}=arctan(\frac{5}{8}) \text{ OR } tan \hat{BOP}=\frac{5}{8}$\]
$\text{substituting their radius and BOP correctly into arc length formula}$
\[$\text{(arc BP =)} \frac{arctan(\frac{5}{8})}{360}\times2\pi(\sqrt{5^{2}+8^{2}})$\]

$\text{subtracting their arc BP from arc PQ}$
\[$\text{(arc BQ =)} \pi\sqrt{5^{2}+8^{2}}-\frac{arctan(\frac{5}{8})}{360}\times2\pi(\sqrt{5^{2}+8^{2}})$\]
$24.4 (m) (24.3679…)$