- The numerical equation from Qin Jiushao’s Shushu jiuzhang analyzed in Section 7.4.2 came from the geometrical problem of finding the area of a pointed field. If the sides and one diagonal are labeled as in Figure 7.22, show that the area of the lower triangle is given by and that the upper triangle by . Then the area of the entire field is given by . Show that satisfies the fourth-degree polynomial equation . If show that this equation becomes the one solved by Qin in the text.
.
2) Brahmagupta asserts that if
ABCD
is a quadrilateral inscribed in a circle, as in Exercise 7, then if s = ½ (
a+
b+
c+
d), the area of the quadrilateral is given by
S= (Fig. 8.12). Prove this result as follows:
a. In triangle
ABC
, drop a perpendicular from
Bto point
Eon
AC. Use the law of cosines applied to that triangle to show that
b. Let
Mbe the midpoint of
AC, so
x= 2
AM. Use the result of part a to show that
c. In triangle
ADC, drop a perpendicular from
Dto point
Fon
AC. Use arguments similar to those in parts a and b to show that
d. Denote the area of quadrilateral
ABCDby
P. Show that and therefore that
e. Extend
B Eto
K
such that angle
B K Dis a right angle, and complete the right triangle
B K D.
Then
BE+
DF=
BK.
Substitute this value in your expression from part d; then use the Pythagorean Theorem to conclude that
f. Since
EF=
EM+
FM ,conclude that Substitute this value into the expression for found in part e, along with the values for and found in Exercise 7. Conclude that
g. Since , show that
h. To prove the theorem, it is necessary to show that the final expression for given in part f is equal to the product of the four expressions in part g. it is clear that the denominators are both equal to 16. To prove that the numerator are equal involves a lot of algebraic manipulation. Work carefully and show that the two numerators are in fact equal.
3) Use a graphing calculator and/or calculus techniques to show that the algebraic formula of Bhaskara I approximates the Sine between 0 and 180 degrees with an error of no more than 1%. Find the values that are most in error.