1.04 Estimating square roots

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.

Created with Geogebra

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. 

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:

This means that, still using our $\sqrt{40}$√40 example, the decimal part would be equal to $\frac{40-36}{49-36}$40−3649−36​≈$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}$95−81100−81​
	$=$=	$\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

Question 5

Question 6