Geometry: An Interactive Journey to Mastery



2. D௘ͽ $UHD ELJVTXDUHíVPDOOVTXDUH ͼ௘x௘ͽ^2 íx^2.
   E௘ͽ )RXUUHFWDQJOHVDVVKRZQLQFigure S.21.2.
   F௘ͽ )RXURYHUODSSLQJUHFWDQJOHVDQGVXEWUDFWLQJWKH
   area of four 3 × 3 squares to compensate,
   as shown in Figure S.21.3.
   G௘ͽ 7ZRUHFWDQJOHVRIRQHVL]HSOXVWZRUHFWDQJOHV
   of a smaller size, as shown in Figure S.21.4.
   H௘ͽ &RPSOHWHDOJHEUDRQHDFKRIWKHPWRVHHWKDW
   they all equal 12x + 36.


3. D௘ͽ x^2  î VRWKHVLGHOHQJWKLVx = 6.
   E௘ͽ x^2 = 96, so x 96.
   F௘ͽ xab.


4. The triangle on the left is a right isosceles triangle.
   ௘6HHFigure S.21.5௘ͽ
   By the Pythagorean theorem, we have h^2 + h^2 = 8^2 ,
   giving h 82.
   By Example 1 in the lesson, the area of the parallelogram
   is 10 h 802.


5. Area =^12 ˜˜ u 20 hh 10 10 10 tan 72° 100 tan 72° |307.8.
   ͼ௘6HHFigure S.21.6௘ͽ

33

3

3

3

3

xx

x

xx

x

x

33 x

Figure S.21.2

33

3

3

3

3

xx

x

xx

x

x

33 x

Figure S.21.3

33

3

3

3

3

xx

x

xx

x

x

33 x

Figure S.21.4

45°

8

10

h

Figure S.21.5

72°

10 10

h

Figure S.21.6