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⬅ Back to Plane Geometry
📐 Key Concepts: Triangles

Special Right Triangles:
30-60-90: sides in ratio $1 : \sqrt{3} : 2$ (short leg : long leg : hypotenuse)
45-45-90: sides in ratio $1 : 1 : \sqrt{2}$

Equilateral Triangle (side $s$): height $h = \dfrac{s\sqrt{3}}{2}$; area $A = \dfrac{s^2\sqrt{3}}{4}$; inradius $r = \dfrac{s\sqrt{3}}{6}$; circumradius $R = \dfrac{s\sqrt{3}}{3}$.

General Triangle:
Inradius: $r = \dfrac{A}{s}$ where $s = \dfrac{a+b+c}{2}$ (semiperimeter)
Circumradius: $R = \dfrac{abc}{4A}$
For right triangle: $r = \dfrac{a+b-c}{2}$ and $R = \dfrac{c}{2}$ (hypotenuse/2)

Median length from vertex $A$ to midpoint of $BC$: $m_a = \dfrac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$

Centroid (intersection of medians): divides each median in ratio 2:1 from vertex.

Problem 1: 30-60-90 Special Triangle

The hypotenuse of a 30-60-90 triangle is 20 cm. Find:
a. The shorter leg (opposite 30°)
b. The longer leg (opposite 60°)
c. The area of the triangle
d. The perimeter of the triangle

In a 30-60-90 triangle the sides are in ratio $1 : \sqrt{3} : 2$. With hypotenuse $c = 20$ cm:

a. Shorter leg (opposite 30°):

$$a = \frac{c}{2} = \frac{20}{2} = \boxed{10 \text{ cm}}$$

b. Longer leg (opposite 60°):

$$b = \frac{c\sqrt{3}}{2} = \frac{20\sqrt{3}}{2} = 10\sqrt{3} \approx \boxed{17.32 \text{ cm}}$$

c. Area:

$$A = \frac{1}{2}ab = \frac{1}{2}(10)(10\sqrt{3}) = 50\sqrt{3} \approx \boxed{86.6 \text{ cm}^2}$$

d. Perimeter:

$$P = 10 + 10\sqrt{3} + 20 = 30 + 10\sqrt{3} \approx \boxed{47.32 \text{ cm}}$$

Problem 2: Equilateral Triangle Properties

An equilateral triangle has a perimeter of 36 cm. Find:
a. The height
b. The area
c. The inradius (radius of the inscribed circle)
d. The circumradius (radius of the circumscribed circle)

Side length $s = 36/3 = 12$ cm.

a. Height:

$$h = \frac{s\sqrt{3}}{2} = \frac{12\sqrt{3}}{2} = 6\sqrt{3} \approx \boxed{10.39 \text{ cm}}$$

b. Area:

$$A = \frac{s^2\sqrt{3}}{4} = \frac{144\sqrt{3}}{4} = 36\sqrt{3} \approx \boxed{62.35 \text{ cm}^2}$$

c. Inradius ($r = A/s_{\text{semi}} = A/(P/2)$):

$$r = \frac{36\sqrt{3}}{18} = 2\sqrt{3} \approx \boxed{3.46 \text{ cm}}$$

d. Circumradius ($R = 2r$ for equilateral triangle, or $R = s/\sqrt{3}$):

$$R = \frac{s}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx \boxed{6.93 \text{ cm}}$$

Problem 3: Isosceles Triangle — Altitude to Base

An isosceles triangle has equal legs of 13 cm each and an altitude from the vertex angle to the base of 5 cm. Find:
a. The base length
b. The vertex angle
c. The base angles
d. The area

The altitude bisects the base. Let half the base = $x$.

a. Base length (right triangle: leg = 13, altitude = 5):

$$x = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$$
$$\text{Base} = 2x = \boxed{24 \text{ cm}}$$

b. Vertex angle (altitude bisects the vertex angle; $\tan(\alpha/2) = x/h = 12/5$):

$$\frac{\alpha}{2} = \arctan\!\left(\frac{12}{5}\right) \approx 67.38° \Rightarrow \alpha \approx \boxed{134.8°}$$

c. Base angles:

$$\beta = \frac{180° - 134.8°}{2} \approx \boxed{22.6°}$$

d. Area:

$$A = \frac{1}{2} \times \text{base} \times h = \frac{1}{2}(24)(5) = \boxed{60 \text{ cm}^2}$$

Problem 4: Triangle Given AAS — Law of Sines

In triangle ABC, angle A = 55°, angle B = 78°, and side c (the side opposite angle C, i.e., AB) = 20 cm. Find:
a. Angle C
b. Sides a and b
c. The area of the triangle

a. Angle C:

$$C = 180° - 55° - 78° = \boxed{47°}$$

b. Using Law of Sines ($a/\sin A = c/\sin C$):

$$\frac{20}{\sin47°} = \frac{a}{\sin55°} = \frac{b}{\sin78°}$$
$$a = \frac{20\sin55°}{\sin47°} = \frac{20(0.8192)}{0.7314} \approx \boxed{22.4 \text{ cm}}$$
$$b = \frac{20\sin78°}{\sin47°} = \frac{20(0.9781)}{0.7314} \approx \boxed{26.75 \text{ cm}}$$

c. Area:

$$A = \frac{1}{2}ab\sin C = \frac{1}{2}(22.4)(26.75)\sin47° \approx \frac{1}{2}(599.2)(0.7314) \approx \boxed{219.1 \text{ cm}^2}$$

Problem 5: Inscribed and Circumscribed Circles of a Right Triangle

A right triangle has legs of 6 cm and 8 cm. Find:
a. The hypotenuse
b. The inradius (radius of the inscribed circle)
c. The circumradius (radius of the circumscribed circle)
d. The area of the inscribed circle
e. The area of the circumscribed circle

a. Hypotenuse:

$$c = \sqrt{6^2 + 8^2} = \sqrt{100} = \boxed{10 \text{ cm}}$$

b. Inradius (for any right triangle: $r = (a + b - c)/2$):

$$r = \frac{a + b - c}{2} = \frac{6 + 8 - 10}{2} = \frac{4}{2} = \boxed{2 \text{ cm}}$$

c. Circumradius (for a right triangle: $R = c/2$, the circumscribed circle has the hypotenuse as diameter):

$$R = \frac{c}{2} = \frac{10}{2} = \boxed{5 \text{ cm}}$$

d. Area of inscribed circle:

$$A_{\text{in}} = \pi r^2 = \pi(2)^2 = 4\pi \approx \boxed{12.57 \text{ cm}^2}$$

e. Area of circumscribed circle:

$$A_{\text{circ}} = \pi R^2 = \pi(5)^2 = 25\pi \approx \boxed{78.54 \text{ cm}^2}$$