Proposition 19. Problem.

396. To construct a square which shall have to a given square the ratio of two given lines.

Given, the square S, and the ratio P: Q.

Required, to construct a square which shall be to S as P is to Q.

Cons. Draw the st. line AK;
P and DBQ.

take AD =

А

Upon AB as a diameter, describe a

At D erect the DC, and join AC,

E

D

Oce.

BC.

On CB, or CB produced, take CH a side of S.
Draw HE

to BA.

Then CE is the side of the required square.

S

a st. line || to a side of a ▲ cuts the other two sides proportionally (298),

R

S

397. SCH. To construct a polygon similar to a given polygon S, and having to it the given ratio of P to Q, we find, as in (396), a side x so that x2 shall be to s (where s is a side of S) as P is to Q, and upon a as a side homologous to s, construct the

Χ

S

P.

Q

polygon R similar to S (354); this will be the polygon

required.

(379)

Proposition 20. Problem.

398. To construct a polygon equivalent to a given polygon P, and similar to a given polygon Q.

Given, two polygons P and Q. Required, to construct a polygon equivalent to P and similar to Q.

Cons. Find p and the sides of squares equivalent respectively to P and Q.

(386)

Take any side of Q as AB, and

find a fourth proportional A'B' to

I, P, and AB.

(347)

A

B

p.

A'

ଓ

Upon A'B', homologous to AB, construct Q' similar

... Q' is equivalent to P and similar to Q.

Q.E. F.

APPLICATIONS.

1. To find the area of an equilateral triangle in terms of

its side a.*

denote the alt., and S the area, of the A.

h = √ a2

Let

Then

2

2. To find the area of a triangle in terms of its sides and the radius of the circumscribing circle.

Let a, b, c denote the sides and h the alt. of the ▲, and R the radius of the circumscribing O.

Therefore, the area of a triangle is equal to the product of the three sides divided by four times the radius of the circumscribing circle.

3. To find the area of a triangle in terms of its sides.

...S= √s(sa) (s — b) (s — c).

5. Find the area of an equilateral triangle if a side = 1 foot. Ans. 0.433 sq. ft. 6. Find (1) the area of the triangle whose sides are 3, 4, and 5 feet, and (2) the radius of the circumscribing circle. Ans. (1) 6 sq. ft.; (2) 2.5 ft.

*Rouché et Comberousse, p. 286,

EXERCISES.

THEOREMS.

1. A triangle is divided by each of its medians into two parts of equal area.

2. A parallelogram is divided by its diagonals into four triangles of equal area.

3. ABC is a triangle, and its base BC is bisected at D; if H be any point in the median AD, show that the triangles ABH, ACH are equal in area.

4. In AC, a diagonal of the parallelogram ABCD, any point H is taken, and HB, HD are drawn: show that the triangle BAH is equal to the triangle DAH.

5. If two triangles have two sides of one respectively equal to two sides of the other, and the included angles supplementary; show that the triangles are equal in area.

6. ABCD is a parallelogram, and E, H are the middle points of the sides AD, BC; if Z is any point in EH, or EH produced, show that the triangle AZB is one quarter of the parallelogram ABCD.

7. If ABCD is a parallelogram, and E, H any points in DC and AD respectively, show that the triangles AEB, BHC are equal in area.

8. ABCD is a parallelogram, and P is any point within it: show that the sum of the triangles PAB, PCD is equal to half the parallelogram.

9. Find the ratio of a rectangle 18 yards by 14 yards to a square whose perimeter is 100 feet.

10. The medians AD, BE of the triangle ABC are produced to meet the straight line, drawn through C parallel to AB, in F, G respectively: prove that the triangle AGC is equal in area to the triangle BCF.

AD DF (105), etc.

11. ABCD is a parallelogram; through A any two straight lines AE, AF are drawn meeting BC in E and CD in F;

through D, DG is drawn parallel to AE meeting AF in G. Show that the parallelogram having AE, AG as adjacent sides is equal to ABCD.

Draw EH || to AG meeting DG produced in H; produce BC to meet DH in K; AH = AK; why? etc.

12. If O is the centroid (172) of the triangle ABC, prove that the triangles AOB, BOC, COA are equal in area.

13. Show that the line joining the middle points of the parallel sides of a trapezoid divides the area into two equal parts.

14. If any point in one side of a triangle be joined to the middle points of the other sides, the area of the quadrilateral thus formed is one-half that of the triangle.

15. Prove that any straight line which bisects a parallelogram must pass through the intersection of its diagonals.

Let EF bisect ABCD meeting AB in E, DC in F, BD in O; join ED, FB, etc.

16. APB, ADQ are two straight lines such that the triangles PAQ, BAD are equal. If the parallelogram ABCD be completed, and BQ joined cutting CD in R, show that CRAP.

17. If the straight line joining the middle points of two opposite sides of any quadrilateral divide the area into two equal parts; show that the two bisected sides are parallel.

18. Through the vertices of a quadrilateral straight lines are drawn parallel to the diagonals: prove that the figure thus formed is a parallelogram which is double the quadrilateral.

19. Through D, E, the middle points of the sides BC, CA of a triangle ABC, any two parallel straight lines are drawn meeting AB or AB produced in F and G: prove that the parallelogram DEGF is half the triangle ABC.

20. In a trapezoid the straight lines, drawn from the middle point of one of the non-parallel sides to the ends of the opposite side, form with that side a triangle equal to half the trapezoid.

21. Prove that the points F, A, K, in the figure of Prop. 7, are collinear.