There is no perpendicular height here, so \(\frac{1}{2} \times \text{base} \times \text{height}\) cannot be used. Use the two sides that meet at the known angle:

\(\text{Area} = \frac{1}{2}ab\sin C\)

\(\text{Area} = \frac{1}{2} \times 8 \times 8 \times \sin 33^\circ = \frac{1}{2} \times 64 \times \sin 33^\circ\)

\(\text{Area} = 17.4\) cm²

Start from the area formula with the two sides that meet at \(A\), one of which is \(x\):

\(\frac{1}{2} \times 11 \times x \times \sin 44^\circ = 39\)

Multiply both sides by \(2\):

\(11 \times x \times \sin 44^\circ = 78\)

Then divide by everything that is multiplying \(x\):

\(x = \frac{ 78 }{ 11 \times \sin 44^\circ } = 10.2\)

\(x = 10.2\) cm

The diagonal \(BD\) cuts the parallelogram into two triangles, \(ABD\) and \(CBD\), which are congruent — so the parallelogram is exactly twice one of them.

Triangle \(ABD\) has sides \(AB = 14\) cm and \(AD = 11\) cm with the angle of \(80^\circ\) between them:

\(\text{Area of } ABD = \frac{1}{2} \times 14 \times 11 \times \sin 80^\circ = 75.83\) cm²

Double it, using the unrounded value:

\(\text{Area} = 14 \times 11 \times \sin 80^\circ = 154\sin 80^\circ = 151.7\)

The area is \(151.7\) cm².

The two sides given meet at \(x\), so they are the \(a\) and \(b\) of the area formula:

\(\frac{1}{2} \times 10 \times 8 \times \sin x = 35.6\)

\(\frac{1}{2} \times 80 \times \sin x = 35.6\)

\(\sin x = \frac{2 \times 35.6}{ 80 }\)

\(x = \sin^{-1}\left( \frac{2 \times 35.6}{ 80 } \right) = 62.9^\circ\)

There is a second angle with the same sine, the obtuse one, and a triangle of this area could equally have that. The question asks for the acute one.

The diagonal splits the quadrilateral into two triangles, and each one has two sides with the angle between them given. Work out the two areas separately and add.

Triangle \(ABC\): the sides \(AB = 12\) cm and \(AC = 12\) cm meet at the angle of \(26^\circ\).

\(\text{Area} = \frac{1}{2} \times 12 \times 12 \times \sin 26^\circ = 31.56\) cm², to 2 d.p.

Triangle \(ACD\): the sides \(CD = 13\) cm and \(CA = 12\) cm meet at the angle of \(39^\circ\).

\(\text{Area} = \frac{1}{2} \times 13 \times 12 \times \sin 39^\circ = 49.09\) cm², to 2 d.p.

Now add them. Keep both areas in the calculator at full accuracy and round only the total — adding the two rounded values above can leave the last digit one out.

\(\text{Area of } ABCD = 80.6\) cm²

\(\frac{1}{2}ab\sin C\) needs two sides and the angle between them, and no angle is given at all. So the cosine rule has to come first.

Take the angle at \(A\), which lies between \(AB\) and \(AC\), the two sides the area formula will then use.

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{ 169 + 289 - 144 }{ 442 } = \frac{ 314 }{ 442 }\)

\(A = \cos^{-1}\left( \frac{ 314 }{ 442 } \right) = 44.73^\circ\), to 2 d.p.

Now use the two sides that meet at \(A\), and keep the unrounded angle in the calculator.

\(\text{Area} = \frac{1}{2} \times 13 \times 17 \times \sin A = \frac{1}{2} \times 221 \times \sin A\)

\(\text{Area} = 77.8\) cm²

The area formula \(\frac{1}{2}ab\sin C\) needs two sides with the angle between them. Only one side is given, so a second one has to be found first.

The angle at \(R\) is the one opposite the side that is known, so start with the angle sum:

\(\angle PRQ = 180 - 39 - 56 = 85^\circ\)

Now \(PQ\) with the angle at \(R\) is a complete pair, and \(QR\) is opposite the angle at \(P\):

\(QR = \frac{ 14 \times \sin 39^\circ }{ \sin 85^\circ } = 8.84\) m, to 2 d.p.

\(PQ\) and \(QR\) meet at \(Q\), and the angle there is given, so those are the two sides to use. Keep the unrounded \(QR\) in the calculator.

\(\text{Area} = \frac{1}{2} \times 14 \times QR \times \sin 56^\circ = 51\) m², to the nearest square metre

Each of the six triangles has two sides that are radii, so both are \(13\) cm, and they meet at \(O\). That is two sides with the angle between them, which is what \(\frac{1}{2}ab\sin C\) needs.

The six angles at \(O\) fill a full turn and are all equal:

\(\text{angle at } O = \frac{360}{ 6 } = 60^\circ\)

\(\text{Area of one triangle} = \frac{1}{2} \times 13 \times 13 \times \sin 60^\circ = 73.179\) cm², to 3 d.p.

All six triangles are congruent, so multiply by 6. Keep the unrounded value in the calculator.

\(\text{Area} = 6 \times \frac{1}{2} \times 169 \times \sin 60^\circ = 439.1\) cm²

The two sides given are \(BC\) and \(CA\), and they meet at \(C\) — but the angle given is at \(A\), so \(\frac{1}{2}ab\sin C\) cannot be used straight away. Find the angle at \(C\) first.

\(CA\) is opposite the angle at \(B\) and \(BC\) is opposite the angle at \(A\), so the sine rule joins them:

\(\sin B = \frac{ 13 \times \sin 77^\circ }{ 15 }\), so \(\angle ABC = 57.61^\circ\), to 2 d.p.

\(\angle ACB = 180 - 77 - 57.61 = 45.39^\circ\)

\(\angle ABC\) is opposite the shorter of the two given sides, so it is the acute answer rather than its obtuse partner.

Now \(BC\) and \(CA\) meet at \(C\) and the angle there is known. Keep the unrounded angle in the calculator.

\(\text{Area} = \frac{1}{2} \times 15 \times 13 \times \sin C = \frac{1}{2} \times 195 \times \sin C\)

\(\text{Area} = 69.4\) cm²

Both sides that meet at \(A\) are unknown, but they are not independent: \(AB = x\) and \(AC = 4x\). Put those into the area formula.

\(\frac{1}{2} \times x \times 4x \times \sin 119^\circ = 62\)

The two lengths multiply together, so this gives an \(x^2\) rather than an \(x\):

\(\frac{ 4x^2 \sin 119^\circ }{2} = 62\)

\(4x^2 \sin 119^\circ = 124\)

\(x^2 = \frac{ 124 }{ 4 \times \sin 119^\circ }\)

A length is positive, so take the positive square root:

\(x = 6.0\)

So \(AB = 6.0\) cm, and \(AC\) is four times that.