In the previous section we saw how to find the unknown coordinates of a vertex in a parallelogram. We also saw a solved example. In this section we will see a few more solved examples.
Solved example 34.2
In the fig.34.4(a) below, the midpoints of the sides of the large triangle are joined to make a small triangle inside.
Calculate the coordinates of the vertices of the large triangle.
Solution:
• Let the outer triangle be PQR and the inner triangle be ABC. This is shown in fig.b
♦ When the midpoints are marked and joined as in the fig.a, we get a triangle inside.
♦ Also we will get 3 inside parallelograms (Details here)
• So we need to find the unknown fourth vertex of those parallelograms. The steps are given below:
1. One of the 3 parallelograms is highlighted in fig.c. It is AQBC.
Q is the unknown vertex. When we find Q of the parallelogram, we get one of the vertices of the main triangle also
(i) Group the vertices into two: [A,Q] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach B from C:
• First travel 2 units horizontally to the right [∵ (5-3) = 2]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the right
♦ At the end of this travel, the coordinates will be (6,2) [∵ (4+2) = 6]
• Then travel 1 unit vertically upwards
♦ At the end of this final lap, the coordinates will be (6,3) [∵ (2+1) = 3]
So the coordinates of Q are (6,3)
2. The next parallelogram is highlighted in fig.34.4(d) below. It is ABRC.
R is the unknown vertex.
(i) Group the vertices into two: [A,B] and [C,R]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
(ii) To reach B from A:
• First travel 1 unit horizontally to the right [∵ (5-4) = 1]
• Then travel 2 units vertically upwards [∵ (4-2) = 2]
(iii) The same procedure of travel must be followed for the travel from C to R
• First travel 1 unit horizontally to the right
♦ At the end of this travel, the coordinates will be (4,3) [∵ (3+1) = 4]
• Then travel 2 units vertically upwards
♦ At the end of this final lap, the coordinates will be (4,5) [∵ (3+2) = 5]
So the coordinates of R are (4,5)
3. The third parallelogram is highlighted in fig.34.4(e) below. It is PABC.
P is the unknown vertex.
(i) Group the vertices into two: [P,A] and [B,C]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach C from B:
• First travel 2 units horizontally to the left [∵ (3-5) = -2]
• Then travel 1 unit vertically downwards [∵ (3-4) = -1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the left
♦ At the end of this travel, the coordinates will be (2,2) [∵ (4-2) = 2]
• Then travel 1 unit vertically downwards
♦ At the end of this final lap, the coordinates will be (2,1) [∵ (2-1) = 1]
So the coordinates of P are (2,1)
Solved example 34.3
In the parallelogram OABC shown in fig.34.5 below, the coordinates of A are (x2,y2). The coordinates of C are (x1,x2).
What are the coordinates of B?
Solution:
1. Group the vertices into two: [O,A] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
2. To reach A from O:
• First travel x2 units horizontally to the right [∵ (x2-0) = x2]
• Then travel y2 units vertically upwards [∵ (y2-0) = y2]
3. The same procedure of travel must be followed for the travel from C to B
• First travel x2 units horizontally to the right
♦ At the end of this travel, the coordinates will be [(x1+x2),y1]
• Then travel y2 units vertically upwards
♦ At the end of this final lap, the coordinates will be [(x1+x2),(y1+y2)]
So the coordinates of B are [(x1+x2),(y1+y2)]
Solved example 34.4
Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals
Solution:
• Consider any parallelogram such as the one shown in fig.34.6(a) below:
• Any parallelogram will have:
♦ An unique value for it's base length
♦ An unique value for it's height
♦ An unique value for the offset
• If we know these three values of a parallelogram, we can completely define it.
• For our parallelogram,
♦ Let the base length be a
♦ Let the height be y
♦ Let the offset be x
• These are shown in fig.b. We will use them to prove the result.
• We have see that, the origin of the axes can be placed any where.
• And also, the axes can have any orientation.
♦ But they must always be perpendicular to each other. (Details here)
• So let us place the origin at the bottom left vertex of the parallelogram
• Let the orientation be in such a way that, the base of the parallelogram coincides with one of the axes. This is shown in fig.c
• Now we can easily write the coordinates of the various points. Fig.c is self explanatory
• Once the coordinates are written, we can calculate the distances using the distance formula. The steps are given below.
(Note that, Pythagoras theorem is used in those cases where it can be used with more ease than the distance formula)
1. Lengths of sides and their squares:
(i) Length of OA = a
So OA2 = a2
(ii) Length of AB using Pythagoras theorem = √[x2+y2]
So AB2 = [x2+y2]
(iii) Length of BC = OA = a
So BC2 = a2
(ii) Length of OC = AB = √[x2+y2]
So OC2 = [x2+y2]
2. Sum of the above squares = 2x2+2y2+2a2 = 2(x2+y2+a2)
3. Length of diagonals and their squares:
(i) Length of OB using Pythagoras theorem = √[(a+x)2+y2] = √[a2+2ax+x2+y2]
So OB2 = [a2+2ax+x2+y2]
(ii) Length of AC using distance formula = √[(a-x)2+(0-y)2] = √[a2-2ax+x2+y2]
So AC2 = [a2-2ax+x2+y2]
4. Sum of the above squares = 2(x2+y2+a2)
Results in (2) and (4) are the same. Hence proved
In the next section, we will see a few more solved examples.
Solved example 34.2
In the fig.34.4(a) below, the midpoints of the sides of the large triangle are joined to make a small triangle inside.
 |
| Fig.34.4 |
Solution:
• Let the outer triangle be PQR and the inner triangle be ABC. This is shown in fig.b
♦ When the midpoints are marked and joined as in the fig.a, we get a triangle inside.
♦ Also we will get 3 inside parallelograms (Details here)
• So we need to find the unknown fourth vertex of those parallelograms. The steps are given below:
1. One of the 3 parallelograms is highlighted in fig.c. It is AQBC.
Q is the unknown vertex. When we find Q of the parallelogram, we get one of the vertices of the main triangle also
(i) Group the vertices into two: [A,Q] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach B from C:
• First travel 2 units horizontally to the right [∵ (5-3) = 2]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the right
♦ At the end of this travel, the coordinates will be (6,2) [∵ (4+2) = 6]
• Then travel 1 unit vertically upwards
♦ At the end of this final lap, the coordinates will be (6,3) [∵ (2+1) = 3]
So the coordinates of Q are (6,3)
2. The next parallelogram is highlighted in fig.34.4(d) below. It is ABRC.
 |
| Fig.34.4 |
(i) Group the vertices into two: [A,B] and [C,R]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
(ii) To reach B from A:
• First travel 1 unit horizontally to the right [∵ (5-4) = 1]
• Then travel 2 units vertically upwards [∵ (4-2) = 2]
(iii) The same procedure of travel must be followed for the travel from C to R
• First travel 1 unit horizontally to the right
♦ At the end of this travel, the coordinates will be (4,3) [∵ (3+1) = 4]
• Then travel 2 units vertically upwards
♦ At the end of this final lap, the coordinates will be (4,5) [∵ (3+2) = 5]
So the coordinates of R are (4,5)
3. The third parallelogram is highlighted in fig.34.4(e) below. It is PABC.
P is the unknown vertex.
(i) Group the vertices into two: [P,A] and [B,C]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach C from B:
• First travel 2 units horizontally to the left [∵ (3-5) = -2]
• Then travel 1 unit vertically downwards [∵ (3-4) = -1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the left
♦ At the end of this travel, the coordinates will be (2,2) [∵ (4-2) = 2]
• Then travel 1 unit vertically downwards
♦ At the end of this final lap, the coordinates will be (2,1) [∵ (2-1) = 1]
So the coordinates of P are (2,1)
Solved example 34.3
In the parallelogram OABC shown in fig.34.5 below, the coordinates of A are (x2,y2). The coordinates of C are (x1,x2).
 |
| Fig.34.5 |
Solution:
1. Group the vertices into two: [O,A] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
2. To reach A from O:
• First travel x2 units horizontally to the right [∵ (x2-0) = x2]
• Then travel y2 units vertically upwards [∵ (y2-0) = y2]
3. The same procedure of travel must be followed for the travel from C to B
• First travel x2 units horizontally to the right
♦ At the end of this travel, the coordinates will be [(x1+x2),y1]
• Then travel y2 units vertically upwards
♦ At the end of this final lap, the coordinates will be [(x1+x2),(y1+y2)]
So the coordinates of B are [(x1+x2),(y1+y2)]
Solved example 34.4
Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals
Solution:
• Consider any parallelogram such as the one shown in fig.34.6(a) below:
 |
| Fig.34.6 |
♦ An unique value for it's base length
♦ An unique value for it's height
♦ An unique value for the offset
• If we know these three values of a parallelogram, we can completely define it.
• For our parallelogram,
♦ Let the base length be a
♦ Let the height be y
♦ Let the offset be x
• These are shown in fig.b. We will use them to prove the result.
• We have see that, the origin of the axes can be placed any where.
• And also, the axes can have any orientation.
♦ But they must always be perpendicular to each other. (Details here)
• So let us place the origin at the bottom left vertex of the parallelogram
• Let the orientation be in such a way that, the base of the parallelogram coincides with one of the axes. This is shown in fig.c
• Now we can easily write the coordinates of the various points. Fig.c is self explanatory
• Once the coordinates are written, we can calculate the distances using the distance formula. The steps are given below.
(Note that, Pythagoras theorem is used in those cases where it can be used with more ease than the distance formula)
1. Lengths of sides and their squares:
(i) Length of OA = a
So OA2 = a2
(ii) Length of AB using Pythagoras theorem = √[x2+y2]
So AB2 = [x2+y2]
(iii) Length of BC = OA = a
So BC2 = a2
(ii) Length of OC = AB = √[x2+y2]
So OC2 = [x2+y2]
2. Sum of the above squares = 2x2+2y2+2a2 = 2(x2+y2+a2)
3. Length of diagonals and their squares:
(i) Length of OB using Pythagoras theorem = √[(a+x)2+y2] = √[a2+2ax+x2+y2]
So OB2 = [a2+2ax+x2+y2]
(ii) Length of AC using distance formula = √[(a-x)2+(0-y)2] = √[a2-2ax+x2+y2]
So AC2 = [a2-2ax+x2+y2]
4. Sum of the above squares = 2(x2+y2+a2)
Results in (2) and (4) are the same. Hence proved
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