Name: Simplifying Surds: Expanding Brackets
Uploaded: 2019-07-19
Duration: 7 min 48 s
Description: Simplifying surds is where we rewrite a surd in its simplest form by ensuring the number underneath the square root has no square numbers as factors. Let us lea

Question

Any odd number can be written as the difference of two perfect squares. Use this to draw squares of area 7 square centimetres and 11 square centimetres. What are the lengths of the sides of these squares?

Accepted Answer

To find the lengths of the sides of squares with areas of $7$ square centimetres and $11$ square centimetres, we need to use the fact that an odd number can be expressed as the difference of two perfect squares.

For the area of $7$ square centimetres:

Let the sides of the squares be $a$ and $b$ , where $a^{2} - b^{2} = 7$ .

This can be factored as: $(a + b) (a - b) = 7$

Since $7$ is a prime number, the only pairs $(a + b)$ and $(a - b)$ that multiply to $7$ are $(a + b) = 7$ and $(a - b) = 1$

Solving these $(a + b) = 7$ and $(a - b) = 1$

Add these two equations:

$2 a = 8 \Rightarrow a = 4$

Subtract them:

$2 b = 6 \Rightarrow b = 3$

So, the value of $a$ is $4$ and $b$ is $3$

Using the Pythagoras theorem,

$c = \sqrt{4^{2} - 3^{2}} = \sqrt{7} cm$

Length of the side of the square with area $7 {cm}^{2}$ is $\sqrt{7} cm$

For the area of $11$ square centimetres:

Let the sides of the squares be $a$ and $b$ , where $a^{2} - b^{2} = 11$ .

This can be factored as: $(a + b) (a - b) = 11$

Since $11$ is a prime number, the only pairs $(a + b)$ and $(a - b)$ that multiply to $11$ are $(a + b) = 11$ and $(a - b) = 1$

Solving these $(a + b) = 11$ and $(a - b) = 1$

Add these two equations:

$2 a = 12 \Rightarrow a = 6$

Subtract them:

$2 b = 10 \Rightarrow b = 5$

So, the value of $a$ is $6$ and $b$ is $5$

Using the Pythagoras theorem,

$c = \sqrt{6^{2} - 5^{2}} = \sqrt{11} cm$

Length of the side of the square with area $11 {cm}^{2}$ is $\sqrt{11} cm$

Exercise 1

Kerala Board Mathematics Solutions for Exercise 1

Questions from Exercise 1 with Hints & Solutions

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