Math Problem Statement

Question 1 involves finding the surface area and volume of a rectangular box with dimensions 6 cm, 5 cm, and 7 cm. Question 2 asks for the surface area and volume of a sphere with a radius of 12 cm. Question 3 involves calculating the lengths of sidewalks connecting buildings in a campus quadrangle using a distance formula.

Solution

Let's break down the questions in the images:

Question 1:

  • We have a rectangular box with dimensions: 6 cm, 5 cm, and 7 cm.
    • a) To find the surface area, we can use the formula for the surface area of a rectangular box: Surface Area=2lw+2lh+2wh\text{Surface Area} = 2lw + 2lh + 2wh where l=6l = 6, w=5w = 5, and h=7h = 7 cm.

    • b) To find the volume, we use the formula for the volume of a rectangular box: Volume=l×w×h\text{Volume} = l \times w \times h

Question 2:

  • A sphere with a radius of 12 cm.
    • a) The surface area of a sphere is given by the formula: Surface Area=4πr2\text{Surface Area} = 4\pi r^2 where r=12r = 12 cm.

    • b) The volume of a sphere is given by the formula: Volume=43πr3\text{Volume} = \frac{4}{3}\pi r^3

Question 3:

  • A campus quadrangle with dimensions 40 m by 30 m.
    • The buildings are positioned at the corners of a triangle, and we need to find the length of the sidewalks (gold lines) between the Library, Chemistry, and Humanities buildings. We can apply the distance formula to calculate the length of the sidewalks, given the coordinates for the locations of the buildings.
    • The distance formula is: Distance=(x2x1)2+(y2y1)2\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

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Math Problem Analysis

Mathematical Concepts

Surface Area
Volume
Geometry
Distance Formula

Formulas

Surface Area of a rectangular box = 2lw + 2lh + 2wh
Volume of a rectangular box = l × w × h
Surface Area of a sphere = 4πr^2
Volume of a sphere = 4/3πr^3
Distance = √((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Distance Formula

Suitable Grade Level

Grades 6-9