Chapter 34 - Geometry and Algebra

In the previous section we completed a discussion on surface area and volume of some solids. In this chapter we will see Geometry and Algebra.

• This chapter is in fact a continuation of chapter 31 - Coordinate Geometry. 
• The last section of that chapter can be seen here. In that chapter 31, we saw an interesting case:
1. We were given the coordinates of two points
• Those were the end points of a diagonal of a rectangle or a square
2. We were asked to find the coordinates of the other two corners of the rectangle
• We solved such problems very easily. (Details here)

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• Now we will see a similar case. 
    ♦ This time, instead of rectangles or squares, we look at parallelograms. 
• Coordinates of any three of the four vertices of a parallelogram will be given. 
• We will be asked to find the coordinates of the fourth vertex. 
Let us see how it is done:
1. Consider the parallelogram OABC in fig.34.1(a) below

Fig.34.1

• Coordinates of three vertices O, A and C are known. We have to find the coordinates of the vertex B
2. In fig.b, green lines are drawn. 
• These green lines have a special property: They are all parallel to the axes
    ♦ The horizontal green lines are parallel to the x axis
    ♦ The vertical green lines are parallel to the y axis
3. Consider the two green lines attached to the side CB
• They intersect at B'. 
• We know that the angle between the axes will always be 90^o
• Since the green lines are parallel to the axes, the angle at B' will also be 90^o
• So triangle BB'C is a right triangle
4. In the same way, the bottom triangle AA'O is also a right triangle
5. Now we will see the relation between the two right triangles:
• OA and CB are the opposite sides of a parallelogram. 
    ♦ So OA and CB are parallel to each other 
    ♦ and they will be inclined at the same angle with the x axis
• CB' and OA' are both parallel to the x axis. So we get: ∠BCB' = ∠AOA'
• These two equal angles are denoted as α in fig.c
6. Again, OA and CB are parallel to each other and they will be inclined at the same angle with the y axis also
• BB' and AA' are both parallel to the y axis. So we get: ∠CBB' = ∠OAA'
• These two equal angles are denoted as β in fig.c
7. We have a situation:
(i) Angles at the ends of the hypotenuse BC are α and β
(ii) Angles at the ends of the hypotenuse OA are also α and β
(iii) The two triangles are right angled. 
(iv) The two hypotenuses, being opposite sides of a parallelogram, have the same lengths
■ In this situation, we can apply the RHS criterion. The two triangles are equal. (Details here)
Let us write the correspondence:
• The 90^o angle is at A' and B'. So A'↔B'
• The α^o angle is at O and C. So O↔C
• The β^o angle is at A and B. So A↔B
• So the correspondence is A'↔B', O↔C and A↔B. This is same as A'OA↔B'CB
• So we can write: 
The two triangles are congruent to one another under the correspondence: A'OA↔B'CB
• This is same as ΔA'OA ≅ ΔB'CB
Now we can write the corresponding sides. We get: 
    ♦ A'O↔B'C
    ♦ OA↔CB
    ♦ A'A↔B'B
8. Among the above three correspondences, the middle one is already known to us. 
• Because, they are the opposite sides of a parallelogram. 
• The first and the last are of great use to us.
■ The first one indicates this:
The bases of the two triangles are of equal length
■ The last one indicates this:
The altitudes of the two triangles are of equal length
9. Now consider fig.34.2(a) below:

Fig.34.2

• AA' is parallel to the y axis. So coordinates of A' are (6,0)
• Thus length of OA' is 6 units. This is shown in fig.34.2(b)
10. We have seen that the bases are equal. 
• So CB' must also be 6 units in length
• Then coordinates of B' will be (3+6, 5). That is., (9,5)
11. Now consider the coordinates of A. They are (6,2)
• So AA' will be 2 units in length
12. We have seen that the altitudes are equal
• So BB' will also be 2 units in length
13. We already wrote the coordinates of B' as (9,5). 
• So the coordinates of B will be (9,5+2). That is., (9,7)
• This is shown in fig.c. Thus we determined the unknown coordinates at B. The problem is solved.

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• We were able to solve it because, the two bases as well as the two altitudes are equal. We proved it using the RHS criterion.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.2(a) above, consider the right triangle BCB'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^BB'⁄_CB.
cos α = ^{adjacent side}⁄_hypotenuse = ^B'C⁄_CB.
2. Again in fig.34.2(a) above, consider the right triangle AOA'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^AA'⁄_OA.
cos α = ^{adjacent side}⁄_hypotenuse = ^A'O⁄_OA.
3. Now, the sine in ⊿BCB' can be equated to the sine in ⊿AOA'. Because, both are taken for the same angle α.
• For example, suppose α be 36^o. Then from the tables, we have: sin 36 = 0.5878
• So in ⊿BCB', sin α = sin 36 = ^BB'⁄_CB = 0.5878
• In ⊿AOA', ^AA'⁄_OA will also be equal to 0.5878. Because:
sin α = sin 36 = ^AA'⁄_OA= 0.5878
4. So we can write: ^BB'⁄_CB = ^AA'⁄_OA.
• But CB = OA. So we get: BB' = AA'
■ That is., altitudes are equal
5. The same is applicable to cosine also:
• Equating the cosines, we get:
^B'C⁄_CB = ^A'O⁄_OA.
• But CB = OA. So we get: B'C = A'O
■ That is., bases are equal

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• So now we are in a position to find the unknown vertex of any parallelogram. 
• The method can be applied very quickly just by doing some metal calculations. 
Let us analyse:
1. In fig.34.2(b), imagine a person standing at O
• To reach A, he must first travel 6 units horizontally to the right
    ♦ This '6 units' is determined from the x coordinate of A 
• Then he must travel 2 units vertically upwards
    ♦ This '2 units' is determined from the y  coordinate of A 
2. Another person at C must follow the same procedure to reach B. That is:
• He must first travel 6 units horizontally to the right
• Then he must travel 2 units vertically upwards
■ The direction of each travel must be clearly specified:
• Horizontal - Towards left OR Towards right
• Vertical - Upwards OR Downwards
3. We are able to use this easy method because:
    ♦ bases are equal
    ♦ altitudes are also equal
• This method will be more clear when we see a solved example

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Solved example 34.1:
What are the coordinates of the vertex C of the parallelogram shown in the fig.34.3 below:

Fig.34.3

Solution:
1. Group the vertices into two: [A,B] and [C,D]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach B from A:
• First travel 4 units horizontally to the right [∵ (5-1) = 4]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
3. The same procedure of travel must be followed for the travel from D to C
• First travel 4 units horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (2+4) = 6]
• Then travel 1 unit vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (5+1) = 6]
So the coordinates of C are (6,6)

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■ In the above steps, we chose [A,B] and [C,D]
Let us choose [A,D] and [B,C]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach A from D:
• First travel 1 unit horizontally to the right [∵ (2-1) = 1]
• Then travel 2 units vertically upwards [∵ (5-3) = 2]
3. The same procedure of travel must be followed for the travel from B to C
• First travel 1 unit horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,4) [∵ (5+1) = 6]
• Then travel 2 units vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (4+2) = 6]
So the coordinates of C are (6,6)
• We get the same answer. So we can choose groups in two different ways
■ But it is important to note that, the members of a group should not be diagonally opposite. Then the method will not work

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In the next section, we will see a few more solved examples.

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Chapter 11.3 - Construction of Right angled Triangle when Hypotenuse and One side are given

In the previous section we saw the method of construction of a triangle when any two angles and their included side are given. In this section we will learn another method.

Construction of a Right angled Triangle when Hypotenuse and one side are given

Mr. A now puts forward one more challenge. He has a drawing of a triangle ABC with him. As before, all the details (lengths of all 3 sides, and angles at all 3 corners) of the triangle is given in the drawing. He does not want to show us that drawing. But he wants us to draw an exact replica of the triangle. He will give us one information: The triangle is a right angled one. He also gives the lengths of the hypotenuse and one side. With this information, can we draw an exact replica? Let us try:

• The length of the hypotenuse is 5 cm. One side is 3 cm. The triangle is right angled
• We must draw a rough sketch with this given data. This is shown in the fig.11.14. Such a rough sketch will give us an idea about how to proceed. It is important to mark the hypotenuse as the side which is opposite to the 90^o angle.

Fig.11.14

• Let us begin the construction. The steps are shown in the fig.11.15
• First we draw a horizontal line 3.0 cm in length, and name it as AB. This is one side of the required triangle. It also fixes two corners A and B. 
• Now, if we can locate the correct position of ‘C’, the problem is solved. So our next aim is to locate ‘C’.

Fig.11.15

• From the rough fig., it is clear that, C lies some where on a line which is at right angle to AB. 
• So we draw a line AC' (of any convenient length) at an angle of 90^o to AB
• From the rough fig., it is also clear that, C is at a distance of 5.0 cm from B
• So with B as center, we draw an arc of radius 5.0 cm
• This arc cuts the line AC' at C
• ΔABC is the required triangle

It may be noted that, this method of constructing a triangle is related to the RHS criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The ΔABC that we have constructed, and the ΔABC which Mr. A is holding, are both right angled. Also, they have the hypotenuse and one side in common. If two right angled triangles have the hypotenuse and one side the same, they are congruent to one another. In other words, one is the exact replica of the other. So next time some one gives us a hypotenuse and one side, we can easily do the construction.

Solved example 11.7
Construct ΔPQR in which ∠P = 90^o, RQ = 8.1 cm, and one side = 3.9 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig. 11.16(a)

Fig.11.16

• Based on the rough sketch, we can proceed to do the construction:
• First draw a horizontal line PQ of length 3.9 cm
• Draw a line PR' (of any convenient length) at an angle of 90^o to PQ
• With Q as center,  draw an arc of radius 8.1 cm
• This arc cuts the line PR' at R
• ΔPQR is the required triangle

Solved example 11.8
ΔABC is an isosceles right angled triangle. Construct ΔABC, if AC = 7.5 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig.11.17(a)

Fig.11.17

• We are given one side: AB = 7.5 cm.
• In an isosceles triangle, two sides are equal. None of them can be the hypotenuse. This is because, the hypotenuse is the longest side, and it is unique. There cannot be two sides with the length of the hypotenuse
• So the rough fig. will be as shown in Fig.11.17(a)
• Now we can proceed to do the construction The steps are shown in fig.(b)
• Draw a horizontal line CA' of any convenient length
• At C draw a perpendicular CB' to CA'
• With C as center, draw two arcs, each of radius 7.5 cm
• These arcs cut CA' at A, and CB' at B
• Join A and B. ΔABC is the required triangle

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So we have learned the method to construct a right angled triangle when it's hypotenuse and one side are given. In the next chapter, we will learn the construction of Quadrilaterals.

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Chapter 10.5 - RHS Criterion for the Congruence of Triangles

In the previous section we have learned the ASA Criterion for the congruence of Triangles. In this section we will discuss about another criterion.

Fig.10.28(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, UVY, PQR, DEF, and MNO.

Fig.10.28

Our problem is this: Is there any triangle in fig.(b), which is congruent to ΔABC? If yes, which one?
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete. 
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence 
• If two sides and their included angle is given, we could use the SAS criterion
• If two angles and their included side are given, we could use the ASA criterion
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison, SAS comparison  or ASA comparison. In such situations, we must try to use other criteria.

We are going to use a rule known as the RHS criterion. First we will see the details about this criterion. Based on that, we will be able to understand how this criterion got it's name.

In the fig.10.28(a), we are given a right angled triangle. In fig.(b), all the triangles that are given for comparison are right angled triangles. Why is that so? 
The answer is that, a 'right angled triangle' will be congruent only with another 'right angled triangle'. There is no point in comparing other triangles.

Now, any given right angled triangle will have one hypotenuse. Any other triangle which is congruent to the given triangle should have the same hypotenuse. This is compulsory for congruence. If the triangle that we are comparing does not have the same hypotenuse, we can straight away discard it from our list.

If it does have the same hypotenuse, it could be a congruent triangle. But we need one more detail. Let us see how this is obtained:
Once we find that the hypotenuse of both the triangles are same, there remains two sides. Those are the ‘legs’. Any one of those legs in the given triangle should have it’s exact replica on the other triangle.

That means, for two right angled triangles to be congruent:
• The hypotenuse must be the same
• Any one leg should be the same
■ There is one more condition: The 90^o angle. But, as we are comparing only 'right angled triangles', the 90^o angle condition will be already satisfied. So we do not have to worry about it.

This RHS criterion gets it’s name thus:
• ‘R’ stands for Right angle
• ‘H’ stands for hypotenuse
• ‘S’ stands for side

Let us apply this rule to our problem:
• Consider fig.10.28(a). We know that, hypotenuse is the side opposite to the 90^o angle. 
• So in ΔABC, AB with a length of 8 cm is the hypotenuse. 
• In fig.(b), we find that, ΔMNO is a right angled triangle, and it’s hypotenuse is also 8 cm. 
• We need one more detail: One of the sides. 
• In ABC we have a side AC = 6.4 cm. This same length of 6.4 cm is possessed by the side MN in ΔMNO also. 
• So ΔABC and ΔMNO are congruent.
■ We are able to establish congruence just by using the hypotenuse and one side. And of course, the 90^o angle. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.

But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence:
• Consider ΔABC and ΔMNO. In ΔABC, the 90^o is at C. In ΔMNO, the 90^o is at N. So we get C↔N
• We have considered the hypotenuse of 8 cm length, and a side of 6.4 cm length
• In ΔABC, these two sides intersect at A. In ΔMNO, these two sides intersect at M. So we get A↔M
• The only remaining corners are B and O. So we get B↔O
• So the correspondence is C↔N, A↔M, and B↔O. This can be written as ABC ↔ MON
• So ΔABC and ΔMNO are congruent to one another under the correspondence: ABC ↔ MON
• This is same as writing: ΔABC ≅ ΔMON

So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔMON over ΔABC:

Fig.10.29

Based on the above discussion, we can write down the criterion:
■ RHS Congruence criterion:
If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.

Solved example 10.13
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 90^o,  AB = 3 cm, BC = 4.8 cm. │ΔPQR: ∠P = 90^o,  PR = 3 cm, RQ = 4.8 cm.
(ii) ΔXYZ: ∠X = 90^o, XY = 4.7 cm, XZ = 2.8 cm, YZ = 5.5 cm. │ΔMNO: ∠M = 90^o, OM = 2.8 cm, ON = 5.0 cm
Solution:
(i) A rough sketch is shown in the fig.10.30(i) below:

Fig.10.30

• Both are right angled triangles
■ From the rough sketch, it is clear that, both the triangles have the same 4.8 cm hypotenuse.
■ A side of 3 cm length is present in both the triangles
■ So the triangles are congruent to one another, based on the RHS criterion
• Now we have to write the correspondence:
• The 90^o angle is at A and P. So A↔P
• In ΔABC, the 4.8 cm side and 3.0 cm side intersect at B
• In ΔPQR, the 4.8 cm side and 3.0 cm side intersect at R. So we get B↔R
• The only remaining corners are C and Q. So we get C↔Q
• So the correspondence is A↔P, B↔R and C↔Q. This is same as ABC↔PRQ
• So we can write: ΔABC and ΔPQR are congruent to one another under the correspondence: ABC↔PRQ
• This is same as ΔABC ≅ ΔPRQ
(ii) A rough sketch is shown in the fig.10.30(ii)
• Both are right angled triangles
• From the rough sketch, it is clear that, ΔXYZ has a hypotenuse of 5.5 cm, while ΔMNO has a hypotenuse of 5.0 cm
• So the two triangles can never be congruent.
• A side of 2.8 cm length is present in both the triangles. But it does not alter the result.
• If the hypotenuse are different, the triangles can not be congruent
Solved example 10.14
Check whether ΔABC and ΔABD shown in fig.10.31 are congruent or not. Given that AD = BC. If congruent, write the correspondence, and all the corresponding parts.

Fig.10.31

Solution:
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is AD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side AD. These two intersect at A. So we get A↔B
• In ΔABC, the 90^o is at C. In ΔABD, the 90^o is at D. So we get C↔D
• So the correspondence is: A↔B and C↔D. This is same as ABC↔BAD - - - (1)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔BAD
• This is same as: ΔABC ≅ ΔBAD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (1). From that, we will get the corresponding sides also:
■ AB = BA, BC = AD and AC = BD
Solved example 10.15
Check whether ΔABC and ΔABD shown in fig.10.32 are congruent or not. Given that BD = BC. If congruent, write the correspondence, and all the corresponding parts.

Fig.10.32

Solution:
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is BD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side BD. These two intersect at B. So we get B↔B
• In ΔABC, the 90^o is at C. In ΔABD, the 90^o is at D. So we get C↔D
• The only remaining corner is A in both the triangles. So we get A↔A
• So the correspondence is: A↔A, B↔B and C↔D. This is same as ABC↔ABD - - - (2)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔ABD
• This is same as: ΔABC ≅ ΔABD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (2). From that, we will get the corresponding sides also:
■ AB = AB, BC = BD and AC = AD

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So we have completed the discussion on congruence of Triangles. In the next chapter, we will learn the construction of Triangles.

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