Find the area of each triangle below.
A collection of problems on calculating triangle areas using different formulas, with detailed solutions.
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times CD \)
\( \text{Area} = \frac{1}{2} \times CA \times CB \times \sin(C) \)
\( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \)
where \( s = \frac{1}{2}(a+b+c) \) is the semi-perimeter
For vertices \( A(x_A, y_A), B(x_B, y_B), C(x_C, y_C) \):
\[ \text{Area} = \frac{1}{2} \left| \det \begin{bmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{bmatrix} \right| \]where det represents the determinant of the 3×3 matrix.
Problem 1
Find the area of each triangle below.
Problem 2
Find the area of triangle CDB.
Problem 3
Find the area of the triangle below.
Problem 4
Find the area of an equilateral triangle with side length 6 cm.
Problem 5
Rectangle ABCD has length 60 and width 30. BF = 10, GC = 20. Find the shaded area.
Problem 6
Quadrilateral FGHE is inscribed in rectangle ABCD (length 100, width 50). E and G are midpoints of AB and DC respectively. Find area of FEHG.
Problem 7
Find the area of the triangle formed by the intersection of lines: \( y = x \), \( y = -x + 4 \), and \( y = 2x + 2 \).
Problem 8
Find the area of the shape below.
Solution 1
a) Base and height known: \( \text{Area} = \frac{1}{2} \times 5 \times 3 = 7.5 \text{ units}^2 \)
b) Two sides and included angle: \( \text{Area} = \frac{1}{2} \times 35 \times 40 \times \sin(70^\circ) \approx 657.8 \text{ mm}^2 \)
c) Heron's formula: \( s = \frac{5+3+6}{2} = 7 \), \( \text{Area} = \sqrt{7(7-5)(7-3)(7-6)} = \sqrt{7 \times 2 \times 4 \times 1} = \sqrt{56} \approx 7.5 \text{ units}^2 \)
d) Using coordinates:
\( \text{Area} = \frac{1}{2} \left| \det\begin{pmatrix}2&4&1\\-1&2&1\\3&-2&1\end{pmatrix} \right| = \frac{1}{2} |2(2+2) -4(-1-3) +1(2-6)| = \frac{1}{2} |8 + 16 - 4| = 10 \text{ units}^2 \)
Solution 2
Base \( CD = 20 \), height \( AB = 80 \):
\( \text{Area} = \frac{1}{2} \times 20 \times 80 = 800 \text{ units}^2 \)
Solution 3
Using sine law: \( \frac{\sin A}{6} = \frac{\sin 55^\circ}{5} \)
\( \sin A = \frac{6 \sin 55^\circ}{5} \approx 0.983 \)
\( A \approx 79.4^\circ \), so \( B \approx 180^\circ - 55^\circ - 79.4^\circ = 45.6^\circ \)
\( \text{Area} = \frac{1}{2} \times 5 \times 6 \times \sin 45.6^\circ \approx 10.7 \text{ units}^2 \)
Solution 4
Equilateral triangle side = 6 cm:
Height \( h = \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3} \)
\( \text{Area} = \frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3} \approx 15.6 \text{ cm}^2 \)
Solution 5
Shaded area = Rectangle area - Triangle EFG area
\( FG = 60 - 10 - 20 = 30 \), height = 30
Triangle area = \( \frac{1}{2} \times 30 \times 30 = 450 \)
Shaded area = \( 60 \times 30 - 450 = 1800 - 450 = 1350 \text{ units}^2 \)
Solution 6
Quadrilateral FEHG = Triangle FEG + Triangle EHG
Common base \( EG = 100 \), heights = 25 each
Each triangle area = \( \frac{1}{2} \times 100 \times 25 = 1250 \)
Total area = \( 1250 + 1250 = 2500 \text{ units}^2 \)
Solution 7
Intersection points:
A: \( 2x+2 = -x+4 \Rightarrow x=\frac{2}{3}, y=\frac{10}{3} \)
B: \( 2x+2 = x \Rightarrow x=-2, y=-2 \)
C: \( x = -x+4 \Rightarrow x=2, y=2 \)
Using determinant formula:
\( \text{Area} = \frac{1}{2} \left| \frac{2}{3}(-2-2) + (-2)(2-\frac{10}{3}) + 2(\frac{10}{3}+2) \right| = \frac{1}{2} \left| -\frac{8}{3} + \frac{8}{3} + \frac{32}{3} \right| = \frac{16}{3} \text{ units}^2 \)
Solution 8
Shape = Triangle ADC + Triangle ACB
First, find AC using cosine law in triangle ACB:
\( AC^2 = 400^2 + 800^2 - 2(400)(800)\cos45^\circ = 160000 + 640000 - 640000 \times \frac{\sqrt{2}}{2} \)
\( AC \approx 589.5 \)
For triangle ADC (sides 245, 432, 589.5), use Heron's:
\( s = \frac{245+432+589.5}{2} = 633.25 \)
\( \text{Area} = \sqrt{633.25 \times 388.25 \times 201.25 \times 43.75} \approx 46526 \text{ ft}^2 \)
Triangle ACB area = \( \frac{1}{2} \times 400 \times 800 \times \sin45^\circ \approx 113137 \text{ ft}^2 \)
Total area \( \approx 46526 + 113137 = 158663 \text{ ft}^2 \)