Math Problem Statement

The ratio of slant height and a side of base of a square-based pyramid is 5:6 and its total surface area is 1536 sq. cm. Write the relation among base area (A), height (h), and volume (V) of the pyramid.

Solution

The problem provided involves a square-based pyramid. The given information includes:

  • The ratio of the slant height (ll) to the side length of the square base (aa) is 5:65:6.
  • The total surface area of the pyramid is 1536cm21536 \, \text{cm}^2.

You are asked to write the relation among the base area (AA), the height (hh), and the volume (VV) of the pyramid.

Step 1: Relation Among Base Area, Height, and Volume

  1. Base Area (AA): For a square base with side length aa, the base area is given by: A=a2A = a^2

  2. Height (hh): The height hh of the pyramid relates to the slant height ll and the base side length aa using the Pythagorean theorem. Since the height, half of the base (a/2a/2), and slant height form a right triangle: h2+(a2)2=l2h^2 + \left(\frac{a}{2}\right)^2 = l^2

  3. Volume (VV): The volume of a pyramid is given by: V=13Ah=13a2hV = \frac{1}{3} A h = \frac{1}{3} a^2 h

Step 2: Given Ratios and Surface Area

  • The ratio l:a=5:6l:a = 5:6 implies l=56al = \frac{5}{6}a.
  • The total surface area is the sum of the base area and the area of the four triangular faces: Surface Area=A+4×Area of one triangular face\text{Surface Area} = A + 4 \times \text{Area of one triangular face} The area of one triangular face (with base aa and height ll) is: 12al=12a×56a=512a2\frac{1}{2} a l = \frac{1}{2} a \times \frac{5}{6}a = \frac{5}{12} a^2 Thus, the total surface area is: A+4×512a2=a2+53a2=83a2A + 4 \times \frac{5}{12} a^2 = a^2 + \frac{5}{3} a^2 = \frac{8}{3} a^2 Given that the total surface area is 1536 cm²: 83a2=1536\frac{8}{3} a^2 = 1536 Solving for a2a^2: a2=1536×38=576a=576=24cma^2 = \frac{1536 \times 3}{8} = 576 \quad \Rightarrow \quad a = \sqrt{576} = 24 \, \text{cm}

Step 3: Solve for the Volume

Now, with a=24cma = 24 \, \text{cm} and l=56×24=20cml = \frac{5}{6} \times 24 = 20 \, \text{cm}, we can use the Pythagorean theorem to find hh: h2+(242)2=202h^2 + \left(\frac{24}{2}\right)^2 = 20^2 h2+122=202h^2 + 12^2 = 20^2 h2+144=400h2=256h=256=16cmh^2 + 144 = 400 \quad \Rightarrow \quad h^2 = 256 \quad \Rightarrow \quad h = \sqrt{256} = 16 \, \text{cm}

Finally, the volume of the pyramid is: V=13×242×16=13×576×16=92163=3072cm3V = \frac{1}{3} \times 24^2 \times 16 = \frac{1}{3} \times 576 \times 16 = \frac{9216}{3} = 3072 \, \text{cm}^3

Final Relations:

  1. Base Area: A=a2=576cm2A = a^2 = 576 \, \text{cm}^2.
  2. Height: h=16cmh = 16 \, \text{cm}.
  3. Volume: V=3072cm3V = 3072 \, \text{cm}^3.

Would you like to explore further details, or do you have any specific questions?

Additional Questions:

  1. What formula links the slant height to the height of a pyramid?
  2. How does the surface area of a square pyramid differ from other pyramids?
  3. Can you derive the formula for the volume of any pyramid?
  4. What are the units for volume and how do they relate to area and height?
  5. What other shapes could have similar surface area calculations?

Tip: Always use the Pythagorean theorem to relate slant height and height in pyramid problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometry
Pyramids
Surface Area
Volume

Formulas

Base Area (A) = a^2
Height (h): h^2 + (a/2)^2 = l^2
Volume (V) = (1/3) * A * h = (1/3) * a^2 * h

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 9-12