FURTHER PROPERTIES OF TRIANGLES.

436. Definition. A segment AB is said to be projected upon a line if perpendiculars from A and B are drawn tol. If these meet 7 in points C and D, then CD is the projection of AB upon l.

437. THEOREM. The square of a side opposite an acute angle of a triangle is equal to the sum of the squares of the other two sides minus twice the product of one of these sides and the projection of the other upon it.

Outline of Proof : In either figure let B be the given acute angle, and in each case BD is the projection of BC upon AB.

Call this projection m.

We are to prove that b2 = a2 + c2 - 2 cm.

In the left figure, b2 = h2 + (cm)2.

In the right figure, b2 = h2 + (m — c)2.

In either case h2 = a2 — m2.

(1)

(2)

(3)

Substitute (3) in (1) or in (2), and complete the proof. Modify each figure so as to draw the projection of AB upon BC and call this n. Then give the proof to show that b2 = a2 + c2 - 2 an.

1. The area of a polygon may be found by drawing its longest diagonal and letting fall perpendiculars upon this diagonal from each of the remaining vertices.

Draw a figure like the one in the margin, only on a much larger scale, measure the necessary lines, compute the areas of the various parts (see § 314), and thus find its total area.

439. THEOREM. The square of the side opposite an obtuse angle of a triangle is equal to the sum of the squares of the other two sides plus twice the product of one of these sides and the projection of the other upon it.

Outline of Proof: Let B be the given obtuse angle and BD the projection of BC upon AB. Call this projection m.

As in the preceding theorem show that

b2 = a2 + c2 + 2 cm.

Also modify the figure so as to show the projection of AB on BC and call this n. Then show that

b2 = a2 + c2 + 2 an.

440. THEOREM. The sum of the squares of two sides of any triangle is equal to twice the square

of half the third side plus twice the

square of the median drawn to that side. A

B

Suggestion. Make use of the two preceding theorems.

1. Compute the medians of a triangle in terms of the sides.

2. Show that the difference of the squares of two sides of a triangle is equal to twice the product of the third side and the projection of the median upon that side.

3. The base of a triangle is 40 feet and the altitude is 30 feet. Find the area of the triangle cut off by a line parallel to the base and 10 feet from the vertex.

or

Solution.

m

The area of ▲ ABC = 1⁄2 ABX CD = hc.

It is first necessary to express h in terms of a, b, c.

Then

4 c2

c)(b − a + c).

or ac-b=28-2b.

a + c − b = 2(s — b)

bac=2(8-e)

b+ca=2(8-a).

Hence area of ▲ ABC = √ 8( s − a)(s — b)(s — c).

443. PROBLEM. To express the area of a triangle in terms of the three sides and the radius of the circumscribed circle.

b

ha

=

E

In the figure CE 2 r, where r is the radius of the circumscribed circle.

But area of ▲ ABC = √8(8 − a) ( s − b ) (8 − c). (§ 442) Hence,

1. Compute the areas of the triangles whose sides are (1) 7, 9, 12; (2) 11, 9, 7; (3) 3, 4, 5.

2. Express the radius of the circumscribed circle of a triangle in terms of the three sides.

3. Find the area of an equilateral triangle whose side is a by § 443, and also without this theorem, and compare results.

r.

PROBLEMS AND APPLICATIONS.

1. Given a regular dodecagon (twelve-sided polygon) with radius Connect alternate vertices.

(a) Prove that the resulting figure is a regular hexagon.

(b) Find the apothem of the hexagon.

(c) Find the area of each of the triangles formed by joining the alternate vertices of the dodecagon.

2. Find the area of a regular dodecagon of radius r.

3. Find the side of a regular dodecagon of radius r.

4. Find the apothem of a regular dodecagon of radius r,

(a) by means of Exs. 2 and 3 and § 343.

(b) by means of Ex. 3 and § 319.

5. Given the side 8 of a regular dodecagon, find apothem. Also find the apothem if the side is s. See Exs. 9-12, page 235.

6. Given a circle of radius 6:

(a) Find the area of a regular hexagon circumscribed about it. (b) Find the area of a regular octagon inscribed in it, also of one circumscribed about it.

7. Using the formula obtained in Ex. 5, find the side of a regular dodecagon whose apothem is b.

8. Find the radius of a circle circumscribed about a dodecagon whose apothem is b.

9. Find the difference between the radii of the regular dodecagons inscribed in and circumscribed about a circle of radius 10 inches. 10. Solve Ex. 9 if the radius of the circle is r.

11. What is the radius of a circle if the difference between the areas of the inscribed and circumscribed regular dodecagons is 12 square inches?

12. What is the radius of a circle if the difference between the areas of the inscribed and circumscribed regular hexagons is 8 square inches? See Ex. 11, page 259.

13. A regular hexagon and a regular dodecagon have equal sides. Find these sides if the area of the dodecagon is six square inches more than twice the area of the hexagon.

14. Solve Ex. 13 if the area of the dodecagon exceeds twice the area of the hexagon by b square inches.