WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 9 Determination Of Mean Proportional

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 9 Determination Of Mean Proportional

WBBSE Class 10 Mean Proportional Overview

Mean Proportional

Definition of Mean Proportional: If three homogeneous quantities a, b, c be such that \(\frac{a}{b}=\frac{b}{c}\) or, b² = ac

or, b = ± √ac, then b is said to be the mean-proportional of a and c.

For example, the mean-proportional of 1 and 4 is \(\sqrt{1 \times 4}\) = 2.

Similarly, if x, y and z be in continued proportion, then \(\sqrt{1 \times 4}\) or, y = zx or, y² = ± √zx, i.e., y is the mean-proportional of x and z.

Application of mean-proportional:

Since \(\frac{a}{b}=\frac{b}{c}\) or, b² = ac, ∴ the area of a rectangle of sides a and c is equal to the area of a square of side b.

So, by applying the principle of mean-proportional we can construct a square of area equal to the area of a rectangle.

Applications of Mean Proportional in Geometry

In the following, we have discussed how the mean-proportional of two given line segments is drawn in the geometric method by applying the principle of mean-proportional.

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Construction of mean-proportional of two given line segments.

Let AB and BC be two given line segments, where AB = a cm and BC = b cm. We have to draw the mean-proportional of them.

Method of construction:

1. Let us draw a ray AX of a length greater than (a + b) cm.
2. Let us cut off the part AB from AX equal to a cm and the part BC from BX equal to b cm.
3. Let us draw the perpendicular bisector PQ of AC. Let PQ intersects AC at O.
4. Let us draw a semi-circle by taking centre at O and radius equal to OA or OC.
5. Let us now draw a perpendicular on OC at B, which intersects the semi-circle at the point D.
6. Let us join B and D.

Hence BD is the required mean-proportional.

Proof: Let us join A, D and C, D.

∵ ∠ADC is a semi-circular angle, ∴ ∠ADC = 1 right-angle.

BD is the perpendicular drawn on the hypotenuse AC from the right-angular point D.

∴ ΔABD ~ ΔBCD.

∴ \(\frac{A B}{B D}=\frac{B D}{B C}\) [by Thales’ theorem]

or, BD² = AB x BC = a x b or, BD = √ab

‍ ∴ BD = √ab

Hence BD is the required mean-proportional. [Proved]

From the definition of mean-proportional, we have seen that if the length and breadth of a rectangle be a cm and c cm respectively, then the side of the square of equal area of the rectangle will be b cm, where b² = ac.

We shall now draw the following construction in the geometric method according to this principle.

Mean Proportional Examples with Solutions

Construction of a square of area equal to the area of a given rectangle.

Let the length of the given rectangle = a cm and breadth = b cm.

We have to construct a square of area equal to the area of this rectangle.

Since the length of the rectangle = a cm and breadth = b cm.

∴ the area of the rectangle = ab sq-cm.

Rule of construction:

1. Let us draw a rectangle ABCD at first, the length of which is AB = a cm and the breadth BC = b cm.
2. DC is extended to Y. The part CR is cut off from CY equal to b cm such that CR – b cm.
3. Let us draw the perpendicular bisector PQ of DE. Let PQ intersects DC at M.
4. Let us now draw a semicircle with centre at M and radius equal to DM or MR.
5. Let us draw a perpendicular CG on CR at the point C. Let CG intersects the semi-circle at G.
6. Let us now draw two arcs simultaneously in the same side of GC by taking centres at C and G respectively and with radius equal to CG.
7. Let us draw another arc with centre at E and radius equal to CG, which intersects the previous arc at the point F.
8. Let us join E, F and G, F.

Hence CEFG is the required square to be drawn.

Proof: In the rectangle AB = a cm and BC = b cm.

Again by construction, DC = AB = a cm and AD = BC = b cm and ∠BAD = 90°.

∴ ABCD is the required rectangle to be drawn.

Now, area of ABCD = ab sq-cm.

Again, in the quadrilateral CEFG, CE = EF = FG = CG [ by construction ] and ∠ECG = 90° [ by construction ]

∴ CEFG is a square.

Area of the square CEFG = (side)² = (CG)² sq-cm.

But by construction, CG is the mean-proportional of a cm and b cm.

∴ CG² = ab sq-cm.

∴ ab sq-cm = CG² = Area of the square.

or, area of the rectangle = area of the square.

Hence CEFG is the required square to be drawn. [Proved]

We shall now discuss about whether it is possible or not to construct a square of area equal to the area of a given triangle.

Construction of a square of area equal to the area of a given triangle.

Let ΔABC be a given triangle. We have to construct a square of area equal to the area of the given triangle ΔABC.

Rule of construction:

1. Let us draw a rectangle CDEF of area equal to the area of ΔABC.
2. Let us extend EF to N. Let us cut-off the part FG from extended FN equal to CF.
3. Let us draw perpendicular bisector of EG which intersects EG at the point O.
4. Let us now draw a semi-circle with centre at O and radius equal to OE or OG.
5. Let us extend CF in the upward direction, which intersects the semi-circle at R.
6. Now, let us draw an arc with centre F and radius equal to FR along FN, which intersects FN at R
7. Let us now draw two arcs, one with centre at R and radius equal to RF and another with centre at P and radius equal to RF. Let these two arcs intersect each other at the point Q.
8. Let us join P, Q and R, Q.

Hence FPQR is the required square to be drawn.

Proof: Let us join A, D.

∵ D is the mid-point of BC, ∵ AD is the median.

∴ area of ΔACD = \(\frac{1}{2}\) ΔABC …..(1)

∵ the median of a triangle bisects the triangle into two parts of equal areas.

Again, ΔACD and rectangle CDEF stand up on the same base CD and within the same parallels BC and MN,

∴ ΔACD = \(\frac{1}{2}\) (rectangle CDEF)

or, \(\frac{1}{2}\) ΔABC = \(\frac{1}{2}\)(rectangle CDEF)

or, ΔABC = rectangle CDEF……..(2)

Again, ∠CFE = 1 right angle [ by construction ]

∴ CF ⊥ EG.

By construction, RF is the mean-proportional of EF and CF.

∴ RF² = EF x CF = area of the rectangle CDEF…….(3)

Again, in quadrilateral PQRF

PQ = QR = RF = FP [ by construction ]

and ∠PFR = ∠CFE = 1 right angle, i.e., each and every angle is a right angle.

∴ PQRF is a square.

∴ the area of the square PQRF = PF² ……..(4)

Now from (3) and (4) we get,

area of the rectangle CDEF = area of the square PQRF……..(5)

Again, from (2) and (5) we get,

area of ΔABC = area of the square PQRF. [Proved]

In the following examples different applications of the above constructions are discussed thoroughly.

Solid Geometry Chapter 9 Determination Of Mean Proportional Examples

Example 1. Construct the mean-proportional of each of the following cases and also find the value of the mean-proportional:

1. 5 cm, 2.5 cm
2. 4 cm, 3 cm
3. 7.5 cm, 4 cm
4. 10 cm, 4 cm
5. 9 cm, 5 cm
6. 12 cm, 3 cm.

Solution:

Here BD is the required mean-proportional and by scale, BD = 3.53 cm (approx.)

2.

Here BD is the required mean-proportional and by scale, BD = 3.46 cm (approx.)

3.

Here, BD is the required mean-proportional, the length of which is by scale 5.47 cm (approx.)

4.

Here, the required mean-proportional = BD and by scale, BD = 6.22 cm (approx.)

5.

Here, the required mean-proportion = BD and by scale, BD = 6.7 cm (approx.)

Visual Representation of Mean Proportional

6.

Here, the required mean-proportional * BD and by scale, BD = 6 cm (approx.)

Example 2. Determine the square root of the following numbers in geometric method.

1. 7
2. 28
3. 13
4. 29

Solution:

1. We know that 7 = 7×1.

So, we can construct the mean-proportional of two line segments of length 7 cm and 1 cm, which will be the required square root of 7.

Here AB = 7 cm and BC = 1 cm and BD is the required mean-proportional of AB and BC. By scale, BD = 2.64 cm (approx.)

∴ √7 = 2.64 (approx.)

2. We know that 28 = 7 x 4

So, the required square root of 28 will be the mean-proportional of two line segments of length 7 cm and 4 cm.

 

Here, the required mean-proportional = BD and by scale, BD = 5.3 cm (approx.)

∴ √28 =5.3 (approx.)

3. We know that 13 = 13 x 1

Geometric Mean vs. Arithmetic Mean

Here, AB = 13 cm, BC = 1 cm and the required mean-proportional = BD and BD = 3.6 cm (approx.)

∴ √13 = 3.6 (approx.)

4. We know that 29 = 5.8 x 5

Here, AB = 5.8 cm, BC = 5 cm,

The required mean-proportional = BD and by scale

BD = 5.39 cm (approx.)

Hence √29=5.39 (approx.)

Construction Steps for Mean Proportional

Example 3. Construct a rectangle of each of the following cases by taking the given lengths as its two sides and also construct a square of area equal to this constructed rectangles.

1. 6 cm, 4 cm
2. 7.25 cm, 3.75 cm

Solution:

1.

Here, BPQR is the required square and ABCD is the required rectangle.

By construction, area of rectangle ABCD = area of the square BPQR.

2.

Here, ABEF is the required rectangle and BPQR is the required square.

Example 4. Construct a triangle at first by taking the given lengths as the sides of the triangle, then construct a square of area equal to the area of this drawn triangle.

1. The lengths of the three sides are 8.4 cm, 6.15 cm and 3.75 cm respectively.
2. An isosceles triangle, the base of which is 7 cm and the length of each of the equal sides is 5 cm.
3. An equilateral triangle the sides of which is 4.7 cm.

Solution:

1.

Here PQRF is the required square.

2.

3.

Here PQRF is the required square.

4.

Here PQRF is the required square.