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📐 Key Concepts: Area of Triangles

Heron's Formula — given three sides $a$, $b$, $c$:

$$s = \frac{a+b+c}{2}, \qquad A = \sqrt{s(s-a)(s-b)(s-c)}$$

SAS Formula — given two sides and the included angle $C$:

$$A = \frac{1}{2}ab\sin C$$

Bisector property: A median from a vertex to the midpoint of the opposite side divides the triangle into two triangles of equal area.

Problem 1: Heron's Formula

Two sides of a triangle measure 8 cm and 12 cm. Find its area if its perimeter is 26 cm.

Heron's Formula – Diagram Heron's Formula – Diagram Heron's Formula – Diagram
Solution Solution Solution Solution

Step 1 — Find the third side: Perimeter = 26, sides 8 and 12, so third side $c = 26 - 8 - 12 = 6$ cm.

Step 2 — Heron's Formula:

$$s = \frac{8 + 12 + 6}{2} = 13$$
$$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{13 \times 5 \times 1 \times 7} = \sqrt{455} \approx \boxed{21.3 \text{ cm}^2}$$

Problem 2: Worded Problem (Unknown Altitudes)

Two triangles have equal bases. The altitude of one triangle is 3 units more than its base, and the altitude of the other is 3 units less than its base. Find the altitudes if the areas of the triangles differ by 21 square units.

Unknown Altitudes – Diagram Unknown Altitudes – Diagram Unknown Altitudes – Diagram
Solution Solution Solution Solution

Let the equal base = $b$. Altitude of Triangle 1: $h_1 = b + 3$. Altitude of Triangle 2: $h_2 = b - 3$.

$$A_1 - A_2 = \frac{1}{2}b(b+3) - \frac{1}{2}b(b-3) = \frac{1}{2}b\bigl[(b+3)-(b-3)\bigr] = \frac{1}{2}b(6) = 3b = 21$$
$$b = 7 \Rightarrow h_1 = 7 + 3 = \boxed{10 \text{ units}}, \quad h_2 = 7 - 3 = \boxed{4 \text{ units}}$$

Problem 3: Bisected Longest Side

A triangular piece of wood with dimensions 130 cm, 180 cm, and 190 cm is to be divided by a line bisecting the longest side drawn from its opposite vertex. Find the area of the part adjacent to the 180 cm side.

Bisected Side – Diagram Bisected Side – Diagram Bisected Side – Diagram
Solution Solution Solution Solution

Step 1 — Total area (sides 130, 180, 190 cm; $s = 250$):

$$A_{\text{total}} = \sqrt{250(250-130)(250-180)(250-190)} = \sqrt{250 \times 120 \times 70 \times 60} = \sqrt{126{,}000{,}000} \approx 11{,}225 \text{ cm}^2$$

Step 2 — A median from the vertex opposite the longest side (190 cm) to its midpoint divides the triangle into two triangles of equal area.

$$A_{\text{part adjacent to 180 cm side}} = \frac{A_{\text{total}}}{2} \approx \frac{11{,}225}{2} \approx \boxed{5{,}612.5 \text{ cm}^2}$$

Problem 4: Area from Two Sides and an Included Angle

A triangular plot has two sides of 9 m and 12 m with an included angle of 65°. Find the area of the triangular plot.

Using the SAS area formula with $a = 9$ m, $b = 12$ m, and included angle $C = 65°$:

$$A = \frac{1}{2}ab\sin C = \frac{1}{2}(9)(12)\sin 65°$$
$$A = 54 \times 0.9063 \approx \boxed{48.9 \text{ m}^2}$$