AI SL · 12.10
0 of 60 skills done
Area of a triangle
12.10.3
Screenshot and video to come
Screenshot and video to come
Drag the corners to make any triangle. Tick the three measurements you are given, then choose what to find. The applet names the rule and writes the working.
Tick three measurements.
In triangle PQR, PQ = 9.2 cm, PR = 6.5 cm and angle P = 38°. Find the area of the triangle.
Have a go first. Then open the working one line at a time.
Angle P is between PQ and PR
The formula needs the angle between the two sides you use.
\(\text{Area}=\frac12 ab\sin C\)
a and b are the two sides, C the angle between them.
\(\text{Area}=\frac12\times9.2\times6.5\times\sin 38°\)
Degree mode.
From the calculator: area = 18.4 cm²
Area is in square units.
Triangle ABC has AB = 5.6 cm, AC = 9.4 cm and area 21.5 cm². Angle A is acute. Find angle A.
Have a go first. Then open the working one line at a time.
\(21.5=\frac12\times5.6\times9.4\times\sin A\)
A is between AB and AC.
\(\sin A=\frac{2\times21.5}{5.6\times9.4}\) = 0.8169
Multiply by 2, then divide by 5.6 × 9.4.
From the calculator: A = sin⁻¹(0.8169) = 54.8°
A second angle, 125°, has the same sine. The question says acute, so choose 54.8°.
Triangle ABC has AB = 11.2 cm, angle A = 63° and area 38.6 cm². Find the length of AC.
Have a go first. Then open the working one line at a time.
\(38.6=\frac12\times11.2\times x\times\sin 63°\)
Call AC = x. The 63° is between AB and AC.
\(x=\frac{2\times38.6}{11.2\times\sin 63°}\)
Multiply by 2, then divide by 11.2 sin 63°.
From the calculator: AC = 7.74 cm
Check: put x back in the formula and you get 38.6.
Triangle ABC has BC = 7.8 cm, CA = 9.5 cm and AB = 12.1 cm. (a) Find angle C. (b) Find the area of the triangle.
Have a go first. Then open the working one line at a time.
\(\cos C=\frac{a^2+b^2-c^2}{2ab}\)
C is opposite AB = 12.1, the longest side.
\(\cos C=\frac{7.8^2+9.5^2-12.1^2}{2\times7.8\times9.5}\) = 0.03158
From the calculator.
C = cos⁻¹(0.03158) = 88.2°
Keep the full value on the calculator.
\(\text{Area}=\frac12\times7.8\times9.5\times\sin \)88.2° = 37.0 cm²
C is between the sides 7.8 and 9.5.
In triangle ABC, BD is the perpendicular from B to AC, with BD = h. Side a = BC = 8.0 cm and angle C = 35°. (a) Find h. (b) Use area = ½ × base × height with base b to show that area = ½ab sin C.
Have a go first. Then open the working one line at a time.
\(\sin C=\frac{h}{a}\), so \(h=a\sin C\) = 8.0 × sin 35° = 4.59 cm
BD is opposite C and BC is the hypotenuse.
\(\text{Area}=\frac12\times b\times h\)
Base AC = b and the height is BD.
\(\text{Area}=\frac12\times b\times a\sin C=\frac12 ab\sin C\)
The angle C is the one between a and b.