Question

Point D is on side BC of equilateral ▲ABC. From point D, perpendicular line segments with lengths 4 and 8 inches are drawn meeting sides AB and AC at points R and T. Find the number of inches in the height of ▲ABC. [Hint: Draw a line segment from A to D.]

Answer 1

Height = 12 in

Explanation:

From inspection of the diagram of the given information:

• ΔCTD and ΔBRD are both right angles with given bases TD and RD.
• ΔABC is an equilateral triangle ⇒ ∠TCD = ∠RBD = 60°
• Side length of ΔABC is equal to the sum of the hypotenuses of ΔCTD and ΔBRD.

Use the sine trigonometric ratio to calculate the hypotenuse of ΔCTD (CD):

`\implies \sf \sin(\theta)=\frac{\textsf{side opposite the angle}}{hypotenuse}`

`\implies \sf \sin 60^(\circ)=(8)/(CD)`

`\implies \sf CD=(16 √(3))/(3)`

As ΔCTD and ΔBRD are similar triangles, and RD is half TD, then BD is half CD:

`\implies \sf BD=(1)/(2)CD`

`\implies \sf BD=(1)/(2) \cdot (16 √(3))/(3)`

`\implies \sf BD=(8 √(3))/(3)`

Therefore, the side length of the equilateral triangle ABC is:

`\implies \sf CB=CD+BD`

`\implies \sf CB=(16 √(3))/(3)+(8 √(3))/(3)`

`\implies \sf CB=8 √(3)`

Height of an equilateral triangle formula

`\sf h=(√(3))/(2)s \quad \textsf{(where s is the side length)}`

Substitute the found side length CB into the formula and solve for h:

`\implies \sf h=(√(3))/(2) \cdot 8√(3)`

`\implies \sf h=(8√(3)√(3))/(2)`

`\implies \sf h=(8(3))/(2)`

`\implies \sf h=(24)/(2)`

`\implies \sf h=12\:\:in`

Answer 2

• 12 inches

Explanation:

Let the side of the triangle be s, since it is equilateral, all three angles measure s.

Now, if we connect A and D, we get two triangles:

• ΔABD and ΔACD

Find the area of each triangle:

• A = 1/2bh
• A(ΔABD) = 1/2s(RD) = 1/2s*4 = 2s
• A(ΔACD) = 1/2s(DT) = 1/2s*8 = 4s

The sum of the areas is the area of ΔABC:

• A = 2s + 4s = 6s

Now, assume the height of ABC is h, find its area:

• A(ΔABC) = 1/2sh

Compare this with the sum of areas and solve for h:

• 6s = 1/2sh
• 6 = 1/2h
• 12 = h
• h = 12