Area of a triangle Problem set B One question at a time

In triangle DEF, DE = 7.9 cm, FD = 10.3 cm and angle D = 49°. Find the area of the triangle.

• The angle 49° is between the sides 7.9 cm and 10.3 cm
• \(\text{Area}=\frac12\times7.9\times10.3\times\sin 49°\)
• From the calculator: area = 30.7 cm²

In triangle KLM, KL = 7.4 cm, LM = 9.1 cm and angle KLM = 115°. Find the area of the triangle.

• The angle 115° at L is between the sides 7.4 cm and 9.1 cm
• \(\text{Area}=\frac12\times7.4\times9.1\times\sin 115°\)
• From the calculator: area = 30.5 cm²

Triangle XYZ has XY = 9.6 cm, ZX = 6.8 cm and area 27.3 cm². Angle X is acute. Find angle X.

• The angle is between the two given sides
• \(27.3=\frac12\times9.6\times6.8\times\sin\theta\)
• \(\sin\theta=\frac{2\times27.3}{9.6\times6.8}\) = 0.8364
• From the calculator: θ = sin⁻¹(0.8364) = 56.8°

Triangle PQR has PQ = 8.0 cm, angle P = 42° and area 36 cm². Find the length of RP.

• The angle 42° is between the sides 8.0 cm and x
• \(36=\frac12\times8.0\times x\times\sin 42°\)
• \(x=\frac{2\times36}{8.0\times\sin 42°}\), so RP = 13.5 cm

In triangle DEF, EF = 8.3 cm, FD = 6.1 cm and DE = 11.5 cm. Find the smallest angle.

• The smallest angle is opposite the shortest side, FD: it is angle E
• \(\cos\theta=\frac{8.3^2+11.5^2-6.1^2}{2\times8.3\times11.5}\) = 0.8587
• θ = cos⁻¹(0.8587) = 30.8°

In triangle XYZ, the perpendicular from Y meets XZ at S and YS = h. The sides are x = YZ and y = ZX. Show that the area of the triangle is \(\frac12 xy\sin Z\).

• In triangle ZSY: \(\sin Z=\frac{h}{x}\), so \(h=x\sin Z\)
• The base is ZX = y and the height is h: \(\text{Area}=\frac12\times y\times h\)
• Substitute: \(\text{Area}=\frac12\times y\times x\sin Z=\frac12 xy\sin Z\)

A triangular flower bed has two sides of 6.2 m and 5.1 m. The angle between them is 66°. Find the area of the flower bed.

• The angle 66° is between the sides 6.2 m and 5.1 m
• \(\text{Area}=\frac12\times6.2\times5.1\times\sin 66°\)
• From the calculator: area = 14.4 m²

A triangular garden bed LMN has area 9.2 m². LM = 4.6 m and angle L = 47°. Find the length of NL.

• The angle 47° is between the sides 4.6 m and x
• \(9.2=\frac12\times4.6\times x\times\sin 47°\)
• \(x=\frac{2\times9.2}{4.6\times\sin 47°}\), so NL = 5.47 m

A triangular car park has sides of 48 m, 71 m and 90 m.

1. (a) Find the largest angle of the car park.

   • \(\cos\theta=\frac{48^2+71^2-90^2}{2\times48\times71}\) = −0.1108
   • θ = cos⁻¹(−0.1108) = 96.4°

2. (b) Find the area of the car park.

   • θ is between the sides 48 m and 71 m
   • \(\text{Area}=\frac12\times48\times71\times\sin 96.4°\) = 1690 m²

Triangle DEF has DE = 5.5 cm, FD = 8.0 cm and area 18 cm². Find the two possible sizes of angle D.

• \(18=\frac12\times5.5\times8.0\times\sin D\)
• \(\sin D=\frac{2\times18}{5.5\times8.0}\) = 0.8182
• From the calculator: D = 54.9°
• A sine is also positive for an obtuse angle: D = 180° − 54.9° = 125°

A paving slab WXYZ has sides WX = 6.4 m, XY = 8.2 m, YZ = 7.9 m and ZW = 7.0 m. Angle WXY = 72°.

1. (a) Find the area of triangle WXY.

   • \(\text{Area}=\frac12\times6.4\times8.2\times\sin 72°\) = 25.0 m²

2. (b) Find the length of WY.

   • \(WY^2=6.4^2+8.2^2-2\times6.4\times8.2\times\cos 72°\) = 75.77
   • WY = 8.70 m

3. (c) Find angle WZY.

   • \(\cos\theta=\frac{7.0^2+7.9^2-8.704^2}{2\times7.0\times7.9}\) = 0.3223
   • θ = cos⁻¹(0.3223) = 71.2°

4. (d) Find the area of the quadrilateral WXYZ.

   • Area of triangle WZY = \(\frac12\times7.0\times7.9\times\sin 71.20°\) = 26.2 m²
   • Total = 24.96 + 26.17 = 51.1 m²