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1.04 Estimating square roots

Lesson

Previously we have looked at evaluating square roots of perfect squares such as $\sqrt{25}$25 or $\sqrt{361}$361. Most of the time when we take square roots it will not be of a perfect square. We can use calculators or estimate in this case.

This applet can help you to learn the first $20$20 perfect squares from $1$1 to $400$400.

There are also some ways to approximate square roots without using a calculator directly. One way is to consider the nearest integer value as a way to estimate or check our work. 

Estimating square roots

If $aa<b,

then $\sqrt{a}<\sqrt{b}$a<b 

Let's see how it can help us to approximate. Let's say we have a square root, $\sqrt{40}$40. If we ask ourselves what are the closest square numbers that are bigger and smaller than $40$40, then we'll find that they're $36$36 and $49$49. So then we have $36<40<49$36<40<49, which leads us to say that $\sqrt{36}$36$<$<$\sqrt{40}$40$<$<$\sqrt{49}$49. And if we evaluate that further we get $6$6$<$<$\sqrt{40}$40$<$<$7$7, so we've managed to narrow this square root down to somewhere between $6$6 and $7$7!

What if we wanted to approximate it further? There's a method for that as well! Once you know what integers the square root lies between, you can find the decimal part by using the following formula:

Decimal approximation

decimal part of approximation = $\frac{\text{number inside square root - closest smaller square}}{\text{closest bigger square - closest smaller square}}$number inside square root - closest smaller squareclosest bigger square - closest smaller square

This means that, still using our $\sqrt{40}$40 example, the decimal part would be equal to $\frac{40-36}{49-36}$40364936$0.3$0.3 so $\sqrt{40}$40 can be approximated to $6.3$6.3. If you plug this square root into a calculator, you'll see that it is indeed rounded to $6.3$6.3! However this method only works well on larger numbers, and bigger they are the better they'll work! Try and see the difference between using this on say, $\sqrt{2}$2 and $\sqrt{300}$300

 

Worked examples

question 1

Find the largest value out of the following

A) $2\pi$2π   B) $\sqrt{50}$50    C) $4.21$4.21    D) $\sqrt{49}$49

Think: Let's convert all the numbers to decimals or approximate them using decimals, so that we can compare them. Remember that $\pi$π is approximately $3.14$3.14.

Do:

$2\pi$2π $2\times3.14$2×3.14
  $6.28$6.28

$\sqrt{50}$50 is bigger than $\sqrt{49}$49 but smaller than $\sqrt{64}$64 so we can say it's between $7$7 and $8$8.

$\sqrt{49}$49 can be evaluated to $7$7 exactly.

Therefore the biggest value is $\sqrt{50}$50.

 

Question 2

Approximate $\sqrt{95}$95 to the nearest hundredth without using a calculator.

Think: Hundredths are represented by the second decimal place, so we need two decimal places. Since $95$95 is a large number, we can use the formula above to approximate its value.

Do:

$81<95$81<95$<$<$100$100

$\sqrt{81}<\sqrt{95}$81<95$<$<$\sqrt{100}$100

$9<\sqrt{95}$9<95$<$<$10$10

So now we know the square root is equal to $9$9 point something.

For the decimal part:

$\frac{\text{number inside square root - closest smaller square}}{\text{closest bigger square - closest smaller square}}$number inside square root - closest smaller squareclosest bigger square - closest smaller square $=$= $\frac{95-81}{100-81}$958110081
  $=$= $\frac{14}{19}$1419
  $0.74$0.74

Therefore $\sqrt{95}$95 is approximately $9.74$9.74.

 

question 3

Andrea needs fences for all $4$4 sides for each of her $3$3 fields. If each field is square and has an area of $27$27 m2, how many meters of fencing would she need (to the nearest whole number)?

Think: We need to figure out what the perimeter is for one paddock and then multiply it by the number of paddocks.

Do:

The area of one square paddock is $27$27 m2 so $x^2=27$x2=27 where $x$x is the length of one of its sides. That means $x=\sqrt{27}$x=27.

Using our calculator, this square root is approximately $5.2$5.2.

That means one paddock needs $4\times5.2=20.8$4×5.2=20.8 m of fencing.

Therefore she needs $3\times20.8=62$3×20.8=62 m (nearest whole number) of fencing in total.

 

Practice questions

Question 4

Between which two consecutive integers does $\sqrt{74}$74 lie?

  1. Complete the inequality.

    $\editable{}<\sqrt{74}<\editable{}$<74<

Question 5

Between which two consecutive integers does $\sqrt{38}$38 lie?

  1. $\editable{}<\sqrt{38}<\editable{}$<38<

Question 6

Bob has a square-shaped pool with an area of $59$59 m2. What is the approximate length of each side of his pool to the nearest meter?

 

Outcomes

8.NS.2

Use rational approximations of irrational numbers to compare the size of irrational numbers, plot them approximately on a number line, and estimate the value of expressions involving irrational numbers

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