Huxsjxjsjxwxbwhudnuwudwudywydbywbhbdwbhdjwndjwbidnjwjdejndunundeuxnennxuenjjxenjxnjrncrnjncndjcnjencjejnjcnjenfuejndjjnenfjnendjwnjdnjwnc
(3x + 1/2 ) + ( 7x - 4 1/2 )
3x+7x + 1/2 - 4 1/2
10x -4
Given:
8 tiles
radius from the center of the archway to the inner edge of the tile. 7 ft.
radius from the center of the archway to the outer edge of the tile. 8 ft = 7 ft + 1 ft.
Area of a semi circle = π r² / 2
A = (3.14 * (7ft)²) / 2 = (3.14 * 49ft²) / 2 = 153.86 ft² / 2 = 76.93 ft²
A = (3.14 * (8ft)²) / 2 = (3.14 * 64ft²) / 2 = 200.96 ft² / 2 = 100.48 ft²
100.48 ft² - 76.93 ft² = 23.55 ft²
23.55 ft² / 8 tiles = 2.94 ft² per tile.
Answer:
<u>Triangle ABC and triangle MNO</u> are congruent. A <u>Rotation</u> is a single rigid transformation that maps the two congruent triangles.
Step-by-step explanation:
ΔABC has vertices at A(12, 8), B(4,8), and C(4, 14).
- length of AB = √[(12-4)² + (8-8)²] = 8
- length of AC = √[(12-4)² + (8-14)²] = 10
- length of CB = √[(4-4)² + (8-14)²] = 6
ΔMNO has vertices at M(4, 16), N(4,8), and O(-2,8).
- length of MN = √[(4-4)² + (16-8)²] = 8
- length of MO = √[(4+2)² + (16-8)²] = 10
- length of NO = √[(4+2)² + (8-8)²] = 6
Therefore:
and ΔABC ≅ ΔMNO by SSS postulate.
In the picture attached, both triangles are shown. It can be seen that counterclockwise rotation of ΔABC around vertex B would map ΔABC into the ΔMNO.