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Squares and Square Roots - Estimating Square Roots

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Perfect squares are numbers like 1,4,9,16,25...1, 4, 9, 16, 25... which are obtained by multiplying an integer by itself. For non-perfect squares, we must estimate their square root as it will not be a whole number.

To estimate a square root, identify the two consecutive perfect squares between which the given number lies. For example, to estimate 20\sqrt{20}, we note that it lies between 42=164^2 = 16 and 52=255^2 = 25.

On a horizontal number line, the estimated square root n\sqrt{n} is positioned between the integers representing the square roots of the bounding perfect squares. If nn is closer to the smaller perfect square, the estimate will be closer to the lower integer.

The 'Proximity Rule' helps refine the estimate. If a number is almost equal to a perfect square, its square root is almost equal to the root of that square. For instance, 99\sqrt{99} is very close to 100\sqrt{100}, so it is approximately 9.99.9.

Visually, if you imagine a square with an area of 3030 units, its side length 30\sqrt{30} would be slightly larger than a square of area 2525 (side 55) and smaller than a square of area 3636 (side 66).

You can further narrow down an estimate by checking the square of the midpoint. If we know n\sqrt{n} is between 66 and 77, we check 6.52=42.256.5^2 = 42.25. If n<42.25n < 42.25, then n\sqrt{n} is between 66 and 6.56.5.

Estimation is particularly useful in real-life geometry problems where exact measurements are not required, such as finding the approximate length of a diagonal or the side of a square plot.

📐Formulae

If a2<n<(a+1)2a^2 < n < (a+1)^2, then a<n<a+1a < \sqrt{n} < a+1

(Side of a square)=Area(Side \ of \ a \ square) = \sqrt{Area}

(Approximate value)Integer closest to the actual root(Approximate \ value) \approx \text{Integer closest to the actual root}

💡Examples

Problem 1:

Estimate the value of 80\sqrt{80} to the nearest whole number.

Solution:

Step 1: Identify perfect squares near 8080. We know that 82=648^2 = 64 and 92=819^2 = 81. Step 2: Compare 8080 with these squares: 64<80<8164 < 80 < 81. Step 3: This implies that 64<80<81\sqrt{64} < \sqrt{80} < \sqrt{81}, which simplifies to 8<80<98 < \sqrt{80} < 9. Step 4: Determine which square 8080 is closer to. Since 8180=181 - 80 = 1 and 8064=1680 - 64 = 16, 8080 is much closer to 8181. Step 5: Therefore, 80\sqrt{80} is approximately 99.

Explanation:

By finding the bounding perfect squares and checking for proximity, we determine that the root is extremely close to the upper bound.

Problem 2:

Estimate 250\sqrt{250} to one decimal place.

Solution:

Step 1: Find perfect squares near 250250. 152=22515^2 = 225 and 162=25616^2 = 256. Step 2: Since 225<250<256225 < 250 < 256, the root is between 1515 and 1616. Step 3: Observe that 250250 is much closer to 256256 (difference of 66) than to 225225 (difference of 2525). Step 4: Try a decimal close to 1616, such as 15.815.8. Step 5: Calculate 15.82=249.6415.8^2 = 249.64. This is very close to 250250. Step 6: Try 15.92=252.8115.9^2 = 252.81. Since 250250 is closer to 249.64249.64, the estimate is 15.815.8.

Explanation:

After identifying the integer range, we use trial and error with decimals near the closer bound to find a more precise estimate.