Geometry Homework: Area, Volume, and Coordinate Geometry Solutions

This document presents solutions to a variety of geometry problems. It includes calculations for the area of parallelograms and triangles, the area of composite figures, and the volume and surface area of cuboids. The solutions are step-by-step, providing clear explanations for each calculation. Furthermore, the assignment covers a problem involving coordinate geometry, calculating the area of a polygon given its vertices. Additional examples include real-world applications, like calculating the painting area and water volume of a swimming pool. This resource is designed to aid students in understanding and solving geometry problems related to area, volume, and coordinate geometry.

Solution (1):
Given figure is a parallelogram whose base is 40 ft. and height is 20ft.
Area of parallelogram = base x height = 40 x 20 = 800 ft2
Solution (2):
Given figure is a triangle whose base length is 0.5 m + 0.8 m = 1.3 m and height is 1.2 m
Area of parallelogram = (1/2) x base x height = (1/2) x 1.3 x 1.2 = 1.56/2 = 0.78 m2
Solution (3):
In given figure 2 rectangles are there, first is ABGF and second is EGCD.
Area of rectangle = length x width
Area of rectangle ABGF = (18 + 7) x 12 = 25 x 12 = 300 in2
Area of rectangle EGCD = (32 – 12) x 7 = 20 x 7 = 140 in2
Area of figure = 300 + 140 = 440 in2

Solution (6):
Given figure is a cuboid whose length is (7/2) m, breadth is (7/8) m and height (5/4) m.
Volume of cuboid = length x breadth x height = (7/2) x (7/8) x (5/4) = 245/64 m2
Solution (7):
Given figure is a cuboid whose net is shown above. Above net has 6 rectangle. Total area of these 6
rectangle is equal to surface area of cuboid.
Surface area of cuboid = ar ABCN + ar CDEF + ar NCFK + ar MNKL + ar KFGJ + ar JGHI
Surface area of cuboid = (12 x 10) + (12 x 7) + (10 x 7) + (12 x 7) + (12 x 10) + (12 x 10)
Surface area of cuboid = 120 + 84 + 70 + 84 + 120 + 120 = 598 cm2

Solution (8):
Length of swimming pool = 8 m
Width of swimming pool = 6 m
Depth of swimming pool = 2 m
Painting area of swimming pool = (8 x 6) + 2(6 x 2) + 2(8 x 2) = 48 + 24 + 32 = 104 m2
Cost to paint the pool = 6 x 104 = $624
Solution (8) (a):
Solution (8) (b):
There are 5 faces of the pool to paint, one bottom face and 4 side faces.
Solution (8) (c):
Paint required = total area of paint = 104 m2
Solution (8) (d):
Cost to paint the pool = 6 x 104 = $624
Solution (8) (e):
Amount of water required in pool = volume of pool = length x width x depth = 8 x 6 x 2 = 96m3

Solution (4):
Base of parallelogram = 4.5 cm and area of parallelogram = 13.95 cm2
Let height of parallelogram = x cm
Area of parallelogram = (length) (height) = (4.5) (x) = 4.5x
4.5x = 13.95
Solution (4) (a):
Given equation: 4.5x = 13.95
So, x represent the height of parallelogram
Solution (4) (b):
4.5x = 13.95
X = 13.95/4.5 = 3.1 cm

Solution (5):
Area of triangle whose points are (x1, y1), (x2, y2) and (x3, y3)
= (1/2) [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Given points are (-3, 5), (0, -5) and (4, 3)
Area of polygon = (1/2) [-3(-5 -3) + 0(3 – 5) + 4(5 – (-5))]
Area of polygon = (1/2) [24 +0 + 40] = 64/2 = 32 unit2