Q03 Test 33 English

Test No. 33 – Complete Solutions

131

Convert to degrees:  (i) \(\dfrac{\pi}{16}\) rad    (ii) \(\dfrac{11\pi}{5}\) rad

🔑 Key Concept – Radians to Degrees: \[ \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180°}{\pi} \] When the angle contains \(\pi\), it cancels out giving an exact result.

(i) Convert \(\dfrac{\pi}{16}\) rad to degrees

Step 1 – Apply the formula: \[ \frac{\pi}{16} \times \frac{180°}{\pi} \]

Step 2 – Cancel \(\pi\) and simplify: \[ = \frac{180°}{16} = \frac{45°}{4} = 11.25° \]

\( \boxed{\dfrac{\pi}{16}\ \text{rad} = 11.25°} \)

(ii) Convert \(\dfrac{11\pi}{5}\) rad to degrees

Step 1 – Apply the formula: \[ \frac{11\pi}{5} \times \frac{180°}{\pi} \]

Step 2 – Cancel \(\pi\) and simplify: \[ = \frac{11 \times 180°}{5} = \frac{1980°}{5} = 396° \]

\( \boxed{\dfrac{11\pi}{5}\ \text{rad} = 396°} \)

132

Find the arc length and area of a sector given \(\theta = \dfrac{\pi}{3}\) rad, \(r = 6\) cm.

🔑 Key Concept – Arc Length and Area of a Sector: \[ \ell = r\theta \qquad\qquad A = \frac{1}{2}r^2\theta \] where \(\theta\) is already in radians — no conversion needed here.

Given: \(r = 6\) cm,  \(\theta = \dfrac{\pi}{3}\) rad

Part A – Arc Length \(\ell\)

Step 1 – Apply arc length formula: \[ \ell = r\theta = 6 \times \frac{\pi}{3} = \frac{6\pi}{3} = 2\pi \]

Step 2 – Evaluate: \[ \ell = 2 \times 3.14159 \approx 6.28\ \text{cm} \]

\( \boxed{\ell = 2\pi \approx 6.28\ \text{cm}} \)

Part B – Area of Sector \(A\)

Step 1 – Apply area formula: \[ A = \frac{1}{2}r^2\theta = \frac{1}{2} \times (6)^2 \times \frac{\pi}{3} \]

Step 2 – Simplify: \[ A = \frac{1}{2} \times 36 \times \frac{\pi}{3} = \frac{36\pi}{6} = 6\pi \]

Step 3 – Evaluate: \[ A = 6 \times 3.14159 \approx 18.85\ \text{cm}^2 \]

\( \boxed{A = 6\pi \approx 18.85\ \text{cm}^2} \)

133

Find the midpoint of the line joining \(A(-\sqrt{5},\,-5)\) and \(B(-3\sqrt{5},\,5)\).

🔑 Key Concept – Midpoint Formula: \[ M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \]

Given: \(A(x_1, y_1) = (-\sqrt{5},\ -5)\),  \(B(x_2, y_2) = (-3\sqrt{5},\ 5)\)

Step 1 – Apply the midpoint formula: \[ M = \left(\frac{-\sqrt{5} + (-3\sqrt{5})}{2},\ \frac{-5 + 5}{2}\right) \]

Step 2 – Simplify the x-coordinate: \[ \frac{-\sqrt{5} - 3\sqrt{5}}{2} = \frac{-4\sqrt{5}}{2} = -2\sqrt{5} \]

Step 3 – Simplify the y-coordinate: \[ \frac{-5 + 5}{2} = \frac{0}{2} = 0 \]

\( \boxed{M = \left(-2\sqrt{5},\ 0\right)} \)

134

Find the unknown quantities in the following figures (similar rectangles).

🔑 Key Concept – Ratio of Areas of Similar Figures: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] Rearranging to find unknown area: \[ A_2 = A_1 \times \left(\frac{l_2}{l_1}\right)^2 \]

Given: Rectangle 1: \(l_1 = 10\ \text{cm},\quad A_1 = 240\ \text{cm}^2\)
Rectangle 2: \(l_2 = 6\ \text{cm},\quad A_2 = ?\)

Step 1 – Write the formula: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \]

Step 2 – Substitute known values: \[ \frac{240}{A_2} = \left(\frac{10}{6}\right)^2 \]

Step 3 – Simplify the ratio and square it: \[ \frac{10}{6} = \frac{5}{3} \qquad \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]

Step 4 – Solve for \(A_2\): \[ \frac{240}{A_2} = \frac{25}{9} \] \[ A_2 = \frac{240 \times 9}{25} = \frac{2160}{25} = 86.4\ \text{cm}^2 \]

\( \boxed{A_2 = 86.4\ \text{cm}^2} \)