Use the Law of Cosines to find the length of side c. The formula is c2 = a2 + b2 - 2ab (C).

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Here are the solutions to questions 4 to 8:

4. In $\triangle ABC$ , $m\angle C = 42^\circ$ , $a = 19$ and $b = 26$ . Find the length of side $c$ , to the nearest integer.

Use the Law of Cosines to find the length of side $c$ .
The formula is $c^2 = a^2 + b^2 - 2ab \cos(C)$ .
$c^2 = 19^2 + 26^2 - 2(19)(26) \cos(42^\circ)$
Calculate the squares and the product.
$c^2 = 361 + 676 - 988 \cos(42^\circ)$ $c^2 = 1037 - 988 \cos(42^\circ)$
Calculate the value of $\cos(42^\circ)$ and substitute.
$\cos(42^\circ) \approx 0.74314$ $c^2 = 1037 - 988(0.74314)$ $c^2 = 1037 - 734.19872$ $c^2 = 302.80128$
Take the square root to find $c$ and round to the nearest integer.
$c = \sqrt{302.80128}$ $c \approx 17.401$
Rounding to the nearest integer, $c = 17$ .
Answer:
The length of side $c$ is $\boxed{\mathbf{17}}$ .

5. In a rhombus whose side measures 22 cm and the smaller angle is $55^\circ$ , find the length of the larger diagonal, to the nearest tenth.
Determine the larger interior angle of the rhombus.
The sum of consecutive angles in a rhombus is $180^\circ$ .
Larger angle $= 180^\circ - 55^\circ = 125^\circ$ .
Consider one of the right-angled triangles formed by the diagonals.
The diagonals of a rhombus bisect each other at right angles and bisect the angles of the rhombus.
The hypotenuse of this right triangle is the side of the rhombus, which is 22 cm.
The angle opposite half of the larger diagonal is half of the larger angle, so $\frac{125^\circ}{2} = 62.5^\circ$ .
Use the sine function to find half the length of the larger diagonal ( $d_L/2$ ).
$\sin(62.5^\circ) = \frac{d_L/2}{22}$ $d_L/2 = 22 \sin(62.5^\circ)$ $d_L/2 = 22 \times 0.88701$ $d_L/2 = 19.51422$
Calculate the full length of the larger diagonal and round to the nearest tenth.
$d_L = 2 \times 19.51422$ $d_L = 39.02844$
Rounding to the nearest tenth, $d_L = 39.0$ cm.
Answer:
The length of the larger diagonal is $\boxed{\mathbf{39.0 \text{ cm}}}$ .

6. Use Heron's formula to find the area of a triangle of lengths 7, 8 and 9. Give your answer in surd form.
Calculate the semi-perimeter ( $s$ ).
$s = \frac{a+b+c}{2} = \frac{7+8+9}{2} = \frac{24}{2} = 12$
Apply Heron's formula
Area $= \sqrt{s(s-a)(s-b)(s-c)}$ .
$\text{Area} = \sqrt{12(12-7)(12-8)(12-9)}$ $\text{Area} = \sqrt{12(5)(4)(3)}$
Multiply the terms under the square root.
$\text{Area} = \sqrt{12 \times 5 \times 4 \times 3}$ $\text{Area} = \sqrt{60 \times 12}$ $\text{Area} = \sqrt{720}$
Simplify the surd.
Find the largest perfect square factor of 720.
$720 = 144 \times 5$ $\text{Area} = \sqrt{144 \times 5}$ $\text{Area} = \sqrt{144} \times \sqrt{5}$ $\text{Area} = 12\sqrt{5}$
Answer:
The area of the triangle is $\boxed{\mathbf{12\sqrt{5}}}$ .

7. What is the area of a triangle where every side is 5 units long correct to the nearest unit?
Recognize that a triangle with all sides equal is an equilateral triangle.
The side length $s = 5$ units.
Use the formula for the area of an equilateral triangle
Area $= \frac{\sqrt{3}}{4} s^2$ .
$\text{Area} = \frac{\sqrt{3}}{4} (5^2)$ $\text{Area} = \frac{\sqrt{3}}{4} (25)$ $\text{Area} = \frac{25\sqrt{3}}{4}$
Calculate the numerical value and round to the nearest unit.
$\sqrt{3} \approx 1.73205$ $\text{Area} = \frac{25 \times 1.73205}{4}$ $\text{Area} = \frac{43.30125}{4}$ $\text{Area} = 10.8253125$
Rounding to the nearest unit, the area is 11.
Answer:
The area of the triangle is $\boxed{\mathbf{11 \text{ units}^2}}$ .

8. Calculate the exact area of $\triangle ABC$ .
Identify the given information from the diagram.
Side $AC = 5$ units.
Side $BC = 6$ units.
Included angle $C = 60^\circ$ .
Use the formula for the area of a triangle given two sides and the included angle
Area $= \frac{1}{2}ab \sin(C)$ .
$\text{Area} = \frac{1}{2} (AC)(BC) \sin(C)$ $\text{Area} = \frac{1}{2} (5)(6) \sin(60^\circ)$
Substitute the exact value of $\sin(60^\circ)$ .
$\sin(60^\circ) = \frac{\sqrt{3}}{2}$ $\text{Area} = \frac{1}{2} (30) \left(\frac{\sqrt{3}}{2}\right)$
Simplify the expression.
$\text{Area} = 15 \left(\frac{\sqrt{3}}{2}\right)$ $\text{Area} = \frac{15\sqrt{3}}{2}$
Answer:
The exact area of $\triangle ABC$ is $\boxed{\mathbf{\frac{15\sqrt{3}}{2}}}$ .

Use the Law of Cosines to find the length of side c. The formula is c2 = a2 + b2 − 2ab (C).

Step-by-step solution

Use the Law of Cosines to find the length of side $c$ .

Calculate the squares and the product.

Calculate the value of $\cos(42^\circ)$ and substitute.

Take the square root to find $c$ and round to the nearest integer.

Determine the larger interior angle of the rhombus.

Consider one of the right-angled triangles formed by the diagonals.

Use the sine function to find half the length of the larger diagonal ( $d_L/2$ ).

Calculate the full length of the larger diagonal and round to the nearest tenth.

Calculate the semi-perimeter ( $s$ ).

Apply Heron's formula

Multiply the terms under the square root.

Simplify the surd.

Recognize that a triangle with all sides equal is an equilateral triangle.

Use the formula for the area of an equilateral triangle

Calculate the numerical value and round to the nearest unit.

Identify the given information from the diagram.

Use the formula for the area of a triangle given two sides and the included angle

Substitute the exact value of $\sin(60^\circ)$ .

Simplify the expression.

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