By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Carry out mensuration is a fundamental skill in Biomedical Engineering, especially when dealing with the design and analysis of medical devices and equipment. Accurate measurement of lengths, areas, and volumes is crucial for ensuring that biomedical instruments fit anatomical requirements or operate within specified parameters. In Kenya, where biomedical engineers often work in hospitals and medical research institutions, proficiency in mensuration supports effective problem-solving and innovation.
Square roots are essential in many biomedical engineering calculations, such as determining dimensions from areas or solving equations involving quadratic terms. Using a calculator to find square roots accurately saves time and reduces errors in calculations related to device design, signal processing, or biomechanical analysis. This topic explains how to correctly use calculators to extract square roots and applies this skill to typical biomedical engineering problems.
The square root of a number \(x\) is a value \(y\) such that \(y^2 = x\). It is denoted as \(\sqrt{x}\). In biomedical engineering, square roots appear in formulas calculating distances, standard deviations in data analysis, and physical properties like root mean square voltage. The principal square root is always non-negative.
The fundamental formula is:
$$ y = \sqrt{x} \quad \text{where} \quad y^2 = x $$
Example 1: Find the square root of 25.
Given: \(x = 25\)
$$ y = \sqrt{25} $$
$$ y = 5 $$
Answer: 5
Example 2: Find the square root of 0.81.
Given: \(x = 0.81\)
$$ y = \sqrt{0.81} $$
$$ y = 0.9 $$
Answer: 0.9
Example 3: Find the square root of 144.
Given: \(x = 144\)
$$ y = \sqrt{144} $$
$$ y = 12 $$
Answer: 12
Example 4: Find the square root of 2.25.
Given: \(x = 2.25\)
$$ y = \sqrt{2.25} $$
$$ y = 1.5 $$
Answer: 1.5
Example 5: Find the square root of 0.04.
Given: \(x = 0.04\)
$$ y = \sqrt{0.04} $$
$$ y = 0.2 $$
Answer: 0.2
Modern scientific calculators have a dedicated square root function, typically marked \(\sqrt{}\) or a similar symbol. To find \(\sqrt{x}\), enter the number \(x\) then press the square root key. It is important to ensure the number is positive, as the calculator will not compute real roots for negative numbers without complex number capability.
The governing formula remains:
$$ y = \sqrt{x} $$
Example 1: Use a calculator to find \(\sqrt{36}\).
Given: \(x = 36\)
$$ y = \sqrt{36} $$
$$ y = 6 $$
Answer: 6
Example 2: Use a calculator to find \(\sqrt{0.0025}\).
Given: \(x = 0.0025\)
$$ y = \sqrt{0.0025} $$
$$ y = 0.05 $$
Answer: 0.05
Example 3: Use a calculator to find \(\sqrt{81}\).
Given: \(x = 81\)
$$ y = \sqrt{81} $$
$$ y = 9 $$
Answer: 9
Example 4: Use a calculator to find \(\sqrt{0.16}\).
Given: \(x = 0.16\)
$$ y = \sqrt{0.16} $$
$$ y = 0.4 $$
Answer: 0.4
Example 5: Use a calculator to find \(\sqrt{225}\).
Given: \(x = 225\)
$$ y = \sqrt{225} $$
$$ y = 15 $$
Answer: 15
Square roots are used in biomedical engineering to calculate parameters such as the root mean square (RMS) voltage of bio-signals, the radius of circular cross-sections from their areas, and the calculation of standard deviation in data analysis.
The formula for radius \(r\) from area \(A\) of a circle is:
$$ r = \sqrt{\frac{A}{\pi}} $$
Example 1: Calculate the radius of a circular electrode with an area of 3.14 cm\(^2\).
Given: \(A = 3.14 \, \text{cm}^2\), \(\pi = 3.14\)
$$ r = \sqrt{\frac{3.14}{3.14}} $$
$$ r = \sqrt{1} $$
$$ r = 1 \, \text{cm} $$
Answer: 1 cm
Example 2: Find the radius of a circular cross-section with an area of 12.57 cm\(^2\).
Given: \(A = 12.57 \, \text{cm}^2\), \(\pi = 3.14\)
$$ r = \sqrt{\frac{12.57}{3.14}} $$
$$ r = \sqrt{4} $$
$$ r = 2 \, \text{cm} $$
Answer: 2 cm
Example 3: Calculate the radius of a circular bio-sensor area of 50.24 cm\(^2\).
Given: \(A = 50.24 \, \text{cm}^2\), \(\pi = 3.14\)
$$ r = \sqrt{\frac{50.24}{3.14}} $$
$$ r = \sqrt{16} $$
$$ r = 4 \, \text{cm} $$
Answer: 4 cm
Example 4: A circular patch electrode has an area of 7.07 cm\(^2\). Find its radius.
Given: \(A = 7.07 \, \text{cm}^2\), \(\pi = 3.14\)
$$ r = \sqrt{\frac{7.07}{3.14}} $$
$$ r = \sqrt{2.25} $$
$$ r = 1.5 \, \text{cm} $$
Answer: 1.5 cm
Example 5: Determine the radius of a circular membrane with area 28.26 cm\(^2\).
Given: \(A = 28.26 \, \text{cm}^2\), \(\pi = 3.14\)
$$ r = \sqrt{\frac{28.26}{3.14}} $$
$$ r = \sqrt{9} $$
$$ r = 3 \, \text{cm} $$
Answer: 3 cm
Biomedical engineers sometimes solve quadratic equations to determine system parameters, such as blood flow velocity or electrical circuit characteristics. Quadratic equations of the form \(ax^2 + bx + c = 0\) can be solved by isolating the squared term and taking the square root.
The formula for solving by square root method when \(bx = 0\) is:
$$ ax^2 + c = 0 \Rightarrow x^2 = -\frac{c}{a} \Rightarrow x = \pm \sqrt{-\frac{c}{a}} $$
Example 1: Solve \(4x^2 - 36 = 0\).
Given: \(a = 4\), \(c = -36\)
$$ 4x^2 - 36 = 0 $$
$$ 4x^2 = 36 $$
$$ x^2 = \frac{36}{4} = 9 $$
$$ x = \pm \sqrt{9} $$
$$ x = \pm 3 $$
Answer: x = 3 or x = -3
Example 2: Solve \(9x^2 - 81 = 0\).
Given: \(a = 9\), \(c = -81\)
$$ 9x^2 = 81 $$
$$ x^2 = \frac{81}{9} = 9 $$
$$ x = \pm \sqrt{9} $$
$$ x = \pm 3 $$
Answer: x = 3 or x = -3
Example 3: Solve \(25x^2 - 100 = 0\).
Given: \(a = 25\), \(c = -100\)
$$ 25x^2 = 100 $$
$$ x^2 = \frac{100}{25} = 4 $$
$$ x = \pm \sqrt{4} $$
$$ x = \pm 2 $$
Answer: x = 2 or x = -2
Example 4: Solve \(16x^2 - 64 = 0\).
Given: \(a = 16\), \(c = -64\)
$$ 16x^2 = 64 $$
$$ x^2 = \frac{64}{16} = 4 $$
$$ x = \pm \sqrt{4} $$
$$ x = \pm 2 $$
Answer: x = 2 or x = -2
Example 5: Solve \(36x^2 - 144 = 0\).
Given: \(a = 36\), \(c = -144\)
$$ 36x^2 = 144 $$
$$ x^2 = \frac{144}{36} = 4 $$
$$ x = \pm \sqrt{4} $$
$$ x = \pm 2 $$
Answer: x = 2 or x = -2
This chapter focused on carrying out mensuration with an emphasis on extracting square roots using calculators. It began by introducing the process of finding square roots efficiently through calculator functions, which is essential for solving various mensuration problems. The chapter then progressed to applying these square root values in practical civil engineering contexts, such as calculating areas and volumes of geometric shapes. Techniques for handling complex mensuration calculations involving roots were demonstrated with step-by-step guidance. Attention was given to accuracy and precision when using calculators to ensure reliable results in measurements. The chapter also included progressively challenging worked examples to build confidence in using square roots for mensuration calculations. Finally, practice questions provided opportunities to reinforce the skills acquired and prepare for real-world engineering applications.
Calculate the square root of 256 using a calculator. (1 mark)
Find \(\sqrt{0.81}\) as it might be used to determine the radius of a circular electrode with an area of 0.81 cm². (1 mark)
A biomedical sensor has a square footprint of area 144 mm². Calculate the length of one side. (2 marks)
Determine the square root of 62500 to find the magnitude of a signal voltage in microvolts, given \(V^2 = 62500\) \(\mu V^2\). (2 marks)
Calculate \(\sqrt{2.56 \times 10^{-4}}\) which represents the resistivity in ohm-meters of a biological tissue sample. (2 marks)
A circular catheter cross-section has an area of 3.14 mm². Use the square root to estimate its radius, knowing that area \(A = \pi r^2\). (3 marks)
The power of a biomedical amplifier is given as \(P^2 = 1.44\) watts squared. Calculate the power \(P\). (2 marks)
A rectangular biosensor has an area of 0.0025 m². Calculate the length of one side if it is square-shaped. (2 marks)
The signal intensity \(I\) is related to the square of the amplitude \(A\) by \(I = A^2\). If \(I = 0.0009\) W/m², find \(A\). (3 marks)
A microfluidic channel has a square cross-section with an area of \(9 \times 10^{-8}\) m². Calculate the side length in micrometers. (3 marks)
$$\sqrt{256} = 16$$
Answer: \(\boxed{16}\)
$$\sqrt{0.81} = 0.9$$
Answer: \(\boxed{0.9}\) cm
Side length \(s = \sqrt{A} = \sqrt{144} = 12\) mm
Answer: \(\boxed{12}\) mm
$$V = \sqrt{62500} = 250 \, \mu V$$
Answer: \(\boxed{250}\) microvolts
$$\rho = \sqrt{2.56 \times 10^{-4}} = \sqrt{2.56} \times \sqrt{10^{-4}} = 1.6 \times 10^{-2} = 0.016 \, \Omega \cdot m$$
Answer: \(\boxed{0.016 \, \Omega \cdot m}\)
$$r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{3.14}{3.14}} = \sqrt{1} = 1 \text{ mm}$$
Answer: \(\boxed{1}\) mm
$$P = \sqrt{1.44} = 1.2 \text{ W}$$
Answer: \(\boxed{1.2}\) watts
$$s = \sqrt{0.0025} = 0.05 \text{ m} = 5 \text{ cm}$$
Answer: \(\boxed{0.05}\) m or \(\boxed{5}\) cm
$$A = \sqrt{0.0009} = 0.03 \text{ (units of amplitude)}$$
Answer: \(\boxed{0.03}\)
Given square cross-sectional area \(A = 9 \times 10^{-8}\) m²:
$$s = \sqrt{9 \times 10^{-8}} = \sqrt{9} \times \sqrt{10^{-8}} = 3 \times 10^{-4} \text{ m} = 300 \, \mu m$$
Answer: \(\boxed{300}\) micrometers
Question 11 (Compulsory - 20 marks)
The biomedical engineering team at Nairobi Hospital is designing a square-shaped biosensor patch for glucose monitoring. The patch area is specified as 10,000 mm².
a) Calculate the length of one side of the biosensor patch using a calculator. (10 marks)
b) If the patch is to be enclosed in a circular protective casing with the diameter equal to the diagonal of the square patch, calculate the diameter of the casing. (10 marks)
Question 12 (20 marks)
A spherical implant used in neurostimulation has a volume of 36,000 mm³. Using a calculator:
a) Calculate the radius of the sphere.
b) Calculate the surface area of the spherical implant.
Question 13 (20 marks)
At Kenyatta University, a rectangular biomedical sensor has a length of 80 mm and a diagonal of 100 mm. Use a calculator to:
a) Calculate the width of the sensor.
b) Calculate the area of the sensor.
Question 14 (20 marks)
A cylindrical blood oxygen sensor has a height of 120 mm and a circular base with radius 15 mm. Using a calculator:
a) Calculate the surface area of the cylinder (including top and bottom).
b) Calculate the volume of the cylinder.
Question 11
a) Side length \(s = \sqrt{10,000} = 100 \text{ mm}\)
b) Diagonal \(d = \sqrt{2} \times 100 = 141.42 \text{ mm}\)
Diameter of casing = 141.42 mm
Question 12
Volume of sphere \(V = \frac{4}{3} \pi r^3\)
a) \(r = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3 \times 36,000}{4 \times 3.1416}} = \sqrt[3]{8,591} = 20.5 \text{ mm}\)
b) Surface area \(S = 4 \pi r^2 = 4 \times 3.1416 \times (20.5)^2 = 4 \times 3.1416 \times 420.25 = 5,277 \text{ mm}^2\)
Question 13
a) Using Pythagoras theorem:
Width \(w = \sqrt{100^2 - 80^2} = \sqrt{10,000 - 6,400} = \sqrt{3,600} = 60 \text{ mm}\)
b) Area \(A = 80 \times 60 = 4,800 \text{ mm}^2\)
Question 14
a) Surface area \(S = 2 \pi r h + 2 \pi r^2 = 2 \times 3.1416 \times 15 \times 120 + 2 \times 3.1416 \times 15^2 = 11,310 + 1,413 = 12,723 \text{ mm}^2\)
b) Volume \(V = \pi r^2 h = 3.1416 \times 15^2 \times 120 = 3.1416 \times 225 \times 120 = 84,823 \text{ mm}^3\)
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