Getting confused while solving the mean proportional problems then practice more with this Mean Proportional Worksheet pdf and learn the concept of mean proportional efficiently. Mean proportional is also known as Geometric Mean and it is not similar to the arithmetic mean. The Mean Proportion is calculated between two terms of a ratio by taking the square root of the product of those two quantities in terms of ratio.

Let us understand more about mean proportional by taking help from this free printable mean proportional of two numbers worksheet. For example, if a, b, and c are in continued proportion then b is called the mean proportional of a and c. The mean proportion is expressed as b = âˆšac. This free printable Worksheet on Mean Proportional with Answers PDF helps you understand the problem-solving techniques and feels fun to practice.

Also Read:

- Worked out Problems on Ratio and Proportion
- Worksheet on Proportions
- Worksheet on Proportion and Continued Proportion

## Mean Proportional Worksheet PDF with Solutions

**Example 1:Â **

Find the mean proportional of the following sets of positive integers:

(i) x- y, xÂ³- xÂ²y

(ii) xÂ³y, xyÂ³

**Solution:Â **

(i) Given x- y, xÂ³- xÂ²y

Now, to find the mean proportion.

Let p be the mean proportional between x- y, xÂ³- xÂ²y.

So, x- y: p:: p: xÂ³- xÂ²y

Product of extremes = Product of means.

Now,

pÂ² = (x- y)(xÂ³- xÂ²y)

â‡’ pÂ² = xÂ²(x- y)(x- y)

â‡’ pÂ² = xÂ²(x- y)Â²

â‡’ p = âˆš(xÂ²(x- y)Â²)

â‡’ p = x(x- y)

Thus, the mean proportion of x- y, xÂ³- xÂ²y is **x(x- y)**.

(ii) Given xÂ³y, xyÂ³

Let m be the mean proportion of xÂ³y, xyÂ³.

So, xÂ³y: m:: m: xyÂ³.

Product of extremes = Product of means.

Product of extremes = xÂ³y Ã— xyÂ³

Product of means = m Ã— m = mÂ²

mÂ² = xÂ³y Ã— xyÂ³

â‡’ mÂ² = (xÂ³ Ã— x) Ã— (yÂ³ Ã— y)

â‡’ mÂ² = x^{4} Ã— y^{4}

â‡’ m = âˆš(x^{4} Ã— y^{4})

â‡’ m = âˆš(xÂ²yÂ²)Â²

â‡’ m = xÂ²yÂ²

Hence, the mean proportion of xÂ³y, xyÂ³ is **xÂ²yÂ²**.

**Example 2:Â Â **

Find the mean proportional of the following:

(i) 8 and 32

(ii) 0.04 and 0.56

(iii) 4 and 25

**Solution:Â **

(i) Given 8 and 32

Let the mean proportion between 8 and 32 is a.

Now, 8: a:: a: 32

We know that, Product of extremes = Product of means.

Here, extremes are 8 and 32 and means are a and a.

So, 8 Ã— 32 = a Ã— a

â‡’ aÂ² = 256

â‡’ a = âˆš256

â‡’ a = 16

Thus, the value of mean proportion ‘a’ is **16**.

(ii) Given 0.04 and 0.56

Let the mean proportion be p.

Now, 0.04: p:: p: 0.56

We know that, Product of extremes = Product of means.

The extremes are 0.04 and 0.56 and the means are p and p.

So, 0.04 Ã— 0.56 = p Ã— p

â‡’ pÂ² = 0.02

â‡’ p = âˆš0.02

â‡’ p = 0.14

Therefore, the value of the mean proportion p is **0.14**.

(iii) Given 4 and 25

Let the mean proportion between 4 and 25 be x.

Thus, 4: x:: x: 25

We know that, Product of extremes = Product of means.

Here, the extremes are 4 and 25 and the means are x and x.

So, 4 Ã—25 = x Ã— x

â‡’ xÂ² = 100

â‡’ x = âˆš100

â‡’ x = 10

Hence, the mean proportion of 4 and 25 is **10**.

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**Example 3:Â **

If b is the mean proportion between a and c, show that a^{4}+ a^{2}b^{2}+ b^{4}/b^{4}+ b^{2}c^{2}+ c^{4}Â = \(\frac{aÂ²}{cÂ²}\).

**Solution:Â **

Given b is the mean proportion between a and c, then we have bÂ² = ac.

Now, we have to prove that a^{4}+ a^{2}b^{2}+ b^{4 }/ b^{4}+ b^{2}c^{2}+ c^{4} = \(\frac{aÂ²}{cÂ²}\) i.e., L.H.S = R.H.S.

LHS = a^{4}+ a^{2}b^{2}+ b^{4}/ b^{4}+ b^{2}c^{2}+ c^{4}

Let us substitute bÂ² = ac in LHS.

LHS = a^{4}+ aÂ²(ac)+ (ac)Â²/(ac)Â²+ (ac)cÂ²+ c^{4}

â‡’ LHS = \(\frac{aÂ²( aÂ²+ ac+ cÂ² )}{cÂ²( aÂ²+ ac+ cÂ² )}\)

â‡’ LHS = \(\frac{aÂ²}{cÂ²}\) = RHS

â‡’ LHS = RHS

Therefore, a^{4}+ a^{2}b^{2}+ b^{4}/ b^{4}+ b^{2}c^{2}+ c^{4}= \(\frac{aÂ²}{cÂ²}\).

**Example 4:Â **

Find the mean proportion of the following

(i) 4\(\frac{4}{5}\), 2\(\frac{1}{2}\)

(ii) aÂ²b, abÂ²

**Solution:Â **

(i) Given 4\(\frac{4}{5}\), 2\(\frac{1}{2}\)

Now, change the mixed fraction into proper fraction

4\(\frac{4}{5}\) = \(\frac{24}{5}\)

2\(\frac{1}{2}\) = \(\frac{5}{2}\)

Let m be the mean proportion of \(\frac{24}{5}\), \(\frac{5}{2}\).

Product of extremes = Prouct of means

Here, the extremes are \(\frac{24}{5}\) and \(\frac{5}{2}\), the means are m and m.

\(\frac{24}{5}\) Ã— \(\frac{5}{2}\) = m Ã— m

â‡’ mÂ² = \(\frac{24}{2}\)

â‡’ mÂ² = 12

â‡’ m = âˆš12

â‡’ m = 3.46

Thus, the value of mean proportion is **3.46**.

(ii) Given aÂ²b, abÂ²

Let k be the mean proportion of aÂ²b and abÂ².

So, aÂ²b: k:: k: abÂ².

In mean proportion, Product of extremes = Product of means.

Now, aÂ²b Ã— abÂ² = kÂ²

â‡’ kÂ² = aÂ³bÂ³

â‡’ kÂ² = (ab)Â¹ Ã— (ab)Â²

â‡’ k = âˆš((ab)Â¹)Â²

â‡’ k = ab

Therefore, **ab** is the mean proportion of aÂ²b, abÂ².