In the previous section we completed the discussion on the division of triangles. We also saw some solved examples. In this section we will see some more solved examples that demonstrate the topics that we have discussed in this chapter as a whole.
Solved example 14.20
ABCD is a trapezium. It’s diagonals AC and BD meet at O. Prove that the magenta coloured ΔAOD and the red coloured ΔBOC have the same area.
 |
| Fig.14.33 |
Solution:
1. In any trapezium, two opposite sides are parallel. In the fig., the parallel sides are AB and CD
2. In the fig., ΔABD and ΔABC have the same area. [since they are triangles between the same parallels, and they have the same base AB]
3. So we can write: ar (ABD) = ar (ABC)
4. But [ar (ABD) = ar (ABO) + ar (AOD)] and [ar (ABC) = ar (ABO) + ar (BOC)]
5. Substituting these values in 3, we get:
[ar (ABO) + ar (AOD)] = [ar (ABO) + ar (BOC)] That is., the yellow ΔABO is common to both
6. ar (ABO) present on both sides will cancel out. So we get:
ar (AOD) = ar (BOC)
Solved example 14.21
In the previous example, what is the total area of the trapezium ABCD, if the area of the blue triangle is 4 cm2 and yellow triangle is 9 cm2
Solution:
1. In the previous example, we have already proved that ar (AOD) = ar (BOC). Let each be equal to x. That is.,
2. Let ar (AOD) = ar (BOC) = x cm2
3. From fig.14.33(b) we get • ar (AOD) = 1⁄2 × AO × h1 and • ar (COD) = 1⁄2 × CO × h1
4. Also we get • ar (AOB) = 1⁄2 × AO × h2 and • ar (COB) = 1⁄2 × CO × h2
5. From (3) and (4) we get
6. So we get, total area of the trapezium = 9 + 4 + x + x = 9 + 4 + 6 + 6 = 25 cm2
Solved example 14.22
In fig.14.34(a), CD is a median of the ABC. This median CD is divided at E in such a way that CE : DE = 2:1. Prove that area of each triangle in fig.13.34(b) is one third of the whole area of ΔABC
 |
| Fig.14.34 |
Solution:
1. Consider ΔADC. It is split into two triangles: ΔADE and ΔACE.
2. Given that CE : DE = 2 :1. So ar (ACE) : ar (ADE) = 2 :1
3. That means ar (ACE) = 2 × ar (ADE)
4. In a similar way, ar (BCE) = 2 × ar (BDE)
5. Consider ΔABE. It is split into two triangles: ΔADE and ΔBDE
6. CD is a median. So AD : BD = 1 : 1. So ar (ADE) = ar (BDE)
7. So we can put ar (ADE) in the place of ar (BDE) in (4)
8. We get ar (BCE) = 2 × ar (ADE)
9. Compare (3) and (8). The right sides are the same. So left sides also must be equal
10. We get ar (ACE) = ar (BCE)
11. Take the sum of two triangles: ΔADE and ΔBDE:
12. We get ar (ADE) + ar (BDE) = ar (ABE). But from (6), ar (BDE) = ar (ADE)
13. ∴ 2 × ar (ADE) = ar (ABE) - - -(6)
14. Comparing the above with (3) we get ar (ACE) = ar (ABE)
15. Comparing (14) and (10) we get ar (ACE) = ar (ABE) = ar (BCE)
16. The sum of the 3 triangles in (15) is the total area ar (ABC). Each of the three are equal. That means each triangle is equal to one third of the total area
Solved example 14.23
In fig.14.35(a), ABCD is a parallelogram. AB is extended to any point P. Line AQ is drawn through A, parallel to PC. AQ meets CB produced at Q. Parallelogram BQRP is completed by drawing QR and RP. Prove that ar (ABCD) = ar (BQRP). [Hint: Draw the diagonals of the parallelograms]
 |
| Fig.14.35 |
Solution:
1. The diagonals AC and PQ are added to the fig. in 14.25(a). The modified fig. is shown in (b)
2. Given that PC is parallel to AQ
3. So AQC and AQP are two triangles with the same base, and between same parallels.
4. Thus, they have the same area. That is., ar (AQC) = ar (AQP)
5. Let us split the above two areas:
• ar (AQC) = ar (ABC) + ar (AQB)
• ar (AQP) = ar (BQP) + ar (AQB)
6. Let us equate as in (4): [ar (ABC) + ar (AQB)] = [ar (BQP) + ar (AQB)]
7. ar (AQB) is common. It will cancel out.
8. So we get ar (ABC) = ar (BQP)
9. In (8) above, each is half of the corresponding parallelogram. (∵ AC and PQ are diagonals)
10. Doubling each will give the corresponding parallelogram. So we get ar (ABCD) = ar (BQRP)
Solved example 14.24
Prove that the perpendiculars drawn from any point on the angle bisector to the sides are equal
Solution:
1. The diagram for this example is given in fig.14.36(a) below:
 |
| Fig.14.36 |
We have:
• an ∠BCA, • it's bisector BG, • 'any point' D on the bisector, • perpendiculars DE and DF from D, to the sides
2. We have to prove that DF and DE are equal
3. We have proved the above equality when we discussed Theorem 14.7
4. In fact, the fig. 14.36(a) given above is the same fig.14.28(b) that we saw when we discussed the theorem
5. There we proved that DF and DE are equal, and so, indicated each of them as 'h'. The same steps can be followed in this example also
Solved example 14.25
In the fig.14.36(b), ABCD is a rectangle of length 14 cm, and width 6 cm. E is the midpoint of BC. F is the mid point of AE. Find the areas of ΔABF and ΔEBF
Solution:
1. Consider ΔABE. Base AB = 14 cm. Height BE = 6/2 = 3 cm (∵ E is the midpoint of BC)
2. So ar (ABE) = 1⁄2 × 14 × 3 = 21 cm2
3. F is the midpoint of AE. So BF is a median of ΔABE
4. So ar (ABF) = ar (EBF) [By Theorem 14.5]
5. If the two areas are equal, each must be exact half of the total ar (ABE)
6. But ar (ABE) = 21 cm2
7. So we get ar (ABF) = ar (EBF) = 21/2 = 10.5 cm2
This completes the discussion on Triangles. In the next section we will see 'Pairs of Equations'.
In the previous section we saw how a quadrilateral, or even a polygon, can be transformed into a triangle having the same area. In this section, we will see some solved examples.
Solved example 14.8
Draw the two quadrilaterals shown in fig.14.15(a) and 14.16(a) below. For each of them, draw a triangle of same area, and calculate the area. Lengths required for the area calculations may be measured
 |
| Fig.14.15 |
Solution:
■ Fig.14.15(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of ΔAED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 10.6 cm and Height DF = 2.9 cm
• So Area = 1⁄2 × b × h = 1⁄2 × 10.6 × 2.9 = 15.66 cm2
 |
| Fig.14.16 |
■ Fig. 14.16(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of AED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 12.6 cm and Height DF = 5.2 cm
• So Area = 1⁄2 × b × h = 1⁄2 × 12.6 × 5.2 = 32.76 cm2
Solved example 14.9
Draw a rhombus of sides 6 cm, and one angle 60o. Then draw a right triangle of the same area
Solution:
 |
| Fig.14.17 |
• We have learned to construct a rhombus in lower classes. In fig.14.17, ABCD is the required rhombus. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• But we want a right triangle of the same area as ABCD
• So we need to transform ΔAED into a right triangle
• For that, extend CD to the left. Draw a perpendicular AF to this extended line
• Join A and E to F. ΔAEF is the required right triangle
Solved example 14.10
Draw a regular pentagon, and then a triangle of the same area. Calculate the area of the triangle.
Solution:
 |
| Fig.14.18 |
• We have learned to construct a regular pentagon in lower classes. In this problem, we can draw a regular pentagon of any side. In fig.14.18, ABCDE is the required regular pentagon with sides 4 cm. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark F, in line with AB
Join B and D to F. The pentagon ABCDE has now become a quadrilateral AFDE. Now work on the left side:
• Draw diagonal AD
• Draw M'N' parallel to AD through E
• On M'N', mark G, in line with AB
Join A and D to G. GFD is the required triangle. Now we want it's area
• From the actual construction, the required lengths can be measured. Base GF is already drawn. To obtain height, draw a perpendicular DH to AB
• By measurement, base AE = 8 cm and Height DF = 6.5 cm
• So Area = 1⁄2 × b × h = 1⁄2 × 8 × 6.5 = 26 cm2
Solved example 14.11
Fig.14.19(a) shows a rectangle ABCD. It is split into an yellow part and a green part. The lines EF and FG does this separation. Instead of the two lines, draw a single line to make the separation in such a way that the areas of the two parts remain the same. Calculate the area of the two parts
 |
| Fig.14.19 |
Solution:
• We are given the fig.14.19(a). It has all the required details for us to draw it in our note book:
• The rectangle has length 5 cm, and width 3 cm. So the rectangle can be easily drawn
• Mark a point G on CD, 1 cm from D. Mark a point E on AB 2 cm from A. So we get G and E. The only remaining point is F. It's details are also given
• Point F satisfies two conditions: It is 3 cm from side AD and 2 cm from side AB
• So draw a line UV (see fig.b), parallel to AD, at a distance of 3 cm from AD
• Draw line XY, parallel to AB, at a distance of 2 cm from AB
• The point of intersection of UV and XY is the point F
• Now we can proceed to solve the actual problem
• AEFGD is a pentagon. Two of it's sides, EF and FG is separating the rectangle into two parts.
• We want the two lines to become one. Then the pentagon will become a quadrilateral.
• The area of the new quadrilateral should be same as the original pentagon
• For that, draw the diagonal EG. Draw MN parallel to EG through F
• MN will intersect AB at H. Join H to G. AHGD is the required quadrilateral.
• The final two areas are shown in fig.(d). We can now proceed to find the areas:
• AHGD is a quadrilateral. Not just a quadrilateral, it is a non isosceles trapezium. We have seen how to find it's area
• We have, Area = 1⁄2 × (a + b)d. In this problem, a = 1, b = 4 (by measurement), and d = 3 cm
• Area = 1⁄2 × (1 + 4) × 3 = 7.5 cm2
• Consider the green portion HBCG. It is an inverted shape of the yellow portion (∵ it has the parallel sides equal to 1 cm and 4 cm, and the height 3 cm)
• So the area of the green portion is also 7.5
In the next section we will see unchanging areas of parallelograms.
In the previous section we completed the discussion on 'unchanging areas of triangles between parallel lines'. In this section, we will see the relation between the areas of quadrilaterals and triangles.
• Fig.14.11(a) below, shows a quadrilateral ABCD
 |
| Fig.14.11 |
• In the fig.b, a diagonal BD is drawn. This diagonal splits the quadrilateral into two triangles: ΔABD and ΔCBD. Through C, draw a line MN, parallel to the diagonal BD. This is shown in fig.b.
• Extend the side AB towards the right. It will meet MN at E. This is shown in the fig.c
• Join B and D to E. We will get a new ΔEBD as shown in fig.d
• What is the peculiarity of this new ΔEBD ?
ΔEBD and ΔCBD has the same base BD. Their third vertices E and C lies on the same line MN parallel to the base BD, and passing through C. So ΔEBD has the same area as ΔCBD
• Now ignore the original ΔCBD. Because, we have it's area in another shape, which is ΔEBD.
We split the original quadrilateral into two triangles, and now, one of them has transformed into another shape with the same area. So the total area of the quadrilateral remains unchanged
• Besides 'keeping the area the same', we achieve one more thing in this transformation:
Look at the final shape in fig.e: AED is a triangle. This was achieved by marking 'E' in line with the side AB. When one vertex comes in line with another two vertices, one side will disappear. This is shown in the animation in fig.14.12 below:
 |
| Fig.14.12 |
In fig.b, the vertex 'C' comes in line with side AB. So two sides: AB and BC becomes one side AE
• So we transformed a '4 sided quadrilateral' into a '3 sided triangle' with the same area.
Figs. (a) to (e) in 14.11 were drawn to show each step of the transformation. In an actual construction, We can do all the steps in a single fig. This is shown in the fig. 14.13 below:
 |
| Fig.14.13 |
The fig.14.13 was drawn quickly by following the steps below:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
So now we know how to transform a quadrilateral into a triangle of same area. The same method can be used to transform any polygon into a triangle of same area. But we may have to do the above process many number of times. The 'number of times' will depend on the number of sides of the polygon. For example, if we are given a pentagon, we will apply the method once to get a quadrilateral. So the five sides of the pentagon will become four sides of a quadrilateral. Then we apply the method a second time to change the quadrilateral into a triangle. Like wise, if we are given a hexagon, we will have to apply the method 3 times. Fig.14.14 below shows the transformation of a pentagon.
• Fig.14.14(a) shows the original pentagon
• In fig.(b), the pentagon is split into a quadrilateral and a triangle, by drawing the diagonal BD
• In fig(c), one side of the pentagon is reduced, by bringing the vertex C to F. Thus we get a quadrilateral AFDE. This quadrilateral has the same area as the original pentagon ABCDE
• In fig(d), the method is applied again. This time, on the left side of the quadrilateral. (This is because, if we continue working on the right side, the resulting triangle will be too elongated towards the right. The reader may try that too). In fig.(d), the quadrilateral is split into two triangles.
• In fig(e), the vertex E is brought down to G. We get the final ΔGFD
■ While doing the actual construction, the above steps can be done in a single fig. This is shown in fig(f). The steps involved are:
• Draw the diagonal BD. Draw MN through C, parallel to BD
• Extend AB towards the right, to meet MN at F. Join B and D to F
• Now work on the left side of the pentagon. Draw diagonal AD. Draw M'N' through E, parallel to AD
• Extend AB towards the left, to meet M'N' at G. Join A and D to G. ΔGFD is the required triangle
In the next section we will see some solved examples.
In the previous section we completed the discussion on the area of Non Isosceles Trapezium. In this section, we will discuss about Area of Quadrilaterals in general.
Area of Quadrilaterals
Fig.13.23(a) shows a quadrilateral ABCD with no specified shape. To find it's area, we can split it into two triangles as shown in fig.b.
 |
| Fig.13.23 |
The splitting is done by drawing a diagonal BD. The length of this diagonal is denoted as 'd'. We get two triangles ΔABD and ΔBCD. We can find the areas of these triangles. The total area of the two triangles will be the area of the quadrilateral ABCD.
But to find the area of a triangle, we need it's base and height. For both the triangles, base is equal to d. To get the height, perpendicular lines AE and CF are dropped from A and C. The length of AE is h1 and length of CF is h2. Now we can calculate the areas of each triangle:
• Area of ΔABD = 1⁄2 × dh1
• Area of ΔBCD = 1⁄2 × dh2
• Total area of the two triangles = Area of the trapezium = 1⁄2 × dh1 + 1⁄2 × dh2 = 1⁄2 × d(h1 + h2)
So we can write:
■ Area of a quadrilateral is half of the product: [Length of a diagonal] × [Sum of the perpendicular distances from opposite vertices to the diagonal]
Solved example 13.17
Compute the area of the quadrilateral ABCD shown in the fig.13.24 below
 |
| Fig.13.24 |
Solution:
• We have, Area = 1⁄2 × d(h1 + h2)
• Given; d = 8 cm, h1 = 4 cm and h2 = 6 cm
• Substituting the values, we get Area = 1⁄2 × 8(4 + 6) = 1⁄2 × 80 = 40 cm2
Solved example 13.18
Compute the area of the quadrilateral ABCD shown in the fig.13.25 below
 |
| Fig.13.25 |
Solution:
In the fig., only the lengths of four sides are given. We need one diagonal and the perpendicular distances. We must calculate them ourselves. There are two clues:
■ Angle at C is 90o
■ Sides AB and AD are equal in length
We can use these clues to find the area:
• Draw the diagonal BD. (shown in fig.b) The quadrilateral is then split into ΔABD and ΔBCD
• ΔBCD is a right triangle. We can find it's area easily because base BC and height CD are known. Thus:
■ Area of ΔBCD = 1⁄2 × b × d = 1⁄2 × 16 × 12 = 96 cm2
• While we are working on this ΔBCD, let us calculate the hypotenuse BD also. Because, this hypotenuse will be required at a later stage in this problem.
• We can apply the Pythagoras theorem:
• BC2 + CD2 = BD2
• So 162 + 122 = BD2 ⇒ 256 + 144 = BD2 ⇒ 400 = BD2 ∴ BD = √400 = 20
• Now we have to calculate the area of ΔABD. We have the base BD. We want the height to calculate the area. We can determine the height as follows:
• ΔABD is an isosceles triangle, with equal sides AB and AD, and the base BD. So, if we drop a perpendicular AE (as shown in fig.c) from the vertex A to the base BD, that perpendicular will bisect the base. That means, BE = DE
• We get two right triangles ΔAEB and ΔAED. We need only one of them for our calculations, because, they are equal.
• Consider ΔAEB: AB = 26 cm, BE = BD/2 = 20/2 = 10
Apply Pythagoras theorem:
• AB2 - BE2 = AE2
• So 262 - 102 = AE2 ⇒ 676 - 100 = AE2 ⇒ AE2 = 576 ∴ AE = √576 = 24
• Now ΔABD is complete. We have the base and height. So area = 1⁄2 × BD × AE = 1⁄2 × 20 × 24 = 240 cm2
• So area of quadrilateral ABCD = area of ΔBCD + area of ΔABD = 96 + 240 = 336 cm2
Solved example 13.19
The three blue lines in the fig.13.26(a) are parallel.
(i) Prove that: [Area of ABCD ÷ Area of PQRS] = [Length of diagonal AC ÷ Length of diagonal PR]
(ii) How should the diagonals be related, if ABCD and PQRS is to have the same area?
(iii) Draw two quadrilaterals, neither parallelograms, nor trapeziums, each with area 15cm2
 |
| Fig.13.26 |
Solution:
• We are given two quadrilaterals and three parallel lines in fig.a. The diagonals AC and PR of the quadrilaterals are shown in yellow colour in fig.b.
• These diagonals split each of the quadrilaterals into two triangles. The heights of the triangles are also marked: h1 and h2
• Area of ΔACD = 1⁄2 × AC × h1 • Area of ΔACB = 1⁄2 × AC × h2
■ Area of quadrilateral ABCD = sum of above two areas = [1⁄2 × AC × h1 + 1⁄2 × AC × h2] = 1⁄2 × AC × (h1 + h2)
• Area of ΔPRS = 1⁄2 × PR × h1 • Area of ΔPQR = 1⁄2 × PR × h2
■ Area of quadrilateral PQRS =[1⁄2 × PR × h1 + 1⁄2 × PR × h2] = 1⁄2 × PR × (h1 + h2)
∴ [Area of ABCD ÷ Area of PQRS] = {[1⁄2 × AC × (h1 + h2)] ÷ [1⁄2 × PR × (h1 + h2)]} = AC ÷ PR - - - (1)
■ The above result can be written in another form:
The ratio of 'Area of ABCD' to 'Area of PQRS' is same as the ratio of 'Length of AC' to 'Length of PR'
(ii) • If ABCD and PQRS have the same area, [Area of ABCD ÷ Area of PQRS] will be equal to 1
• But from (1), [Area of ABCD ÷ Area of PQRS] = AC ÷ PR
• So, if the areas are equal, AC ÷ PR will also become equal to 1
• If AC ÷ PR is equal to 1, AC = PR
■ So we can write: If the areas are equal, their diagonals will also be equal
(iii) Consider the two quadrilaterals ABCD and PQRS in fig.13.27 below:
 |
| Fig.13.27 |
• The three blue lines are parallel
• Both the quadrilaterals must satisfy the following conditions:
♦ AC and PR must always be 6 cm
♦ AC and PR must always lie on the middle blue line
♦ Vertices 'S' and 'D' must always lie on the top most blue line
♦ Vertices 'B' and 'Q' must always lie on the bottom most blue line
• If the above conditions are satisfied, both ABCD and PQRS will have an area of 15 cm2, what ever be the shapes.
Proof:
• Area of ΔACD = 1⁄2 × 6 × 3 = 9.
• Δ PRS has the same base and height. So it also has an area of 9
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the top vertex lies on the top blue line
• Area of ΔABC = 1⁄2 × 6 × 2 = 6
• Δ PQR has the same base and height. So it also has an area of 6
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the bottom vertex lies on the bottom blue line
■ So the total area of each quadrilateral = 9 + 6 = 15 cm2
In the next section, we will discuss about 'Unchanging areas of triangles'.
In the previous section we completed the discussion on the calculation of area of a parallelogram. We also saw a number of solved examples. In this section, we will see the area of a Rhombus.
Area of a Rhombus
Consider a square shown in fig.13.10(a). Usually, a square is specified by it's side. But here, side is not given. Instead, the length of diagonal is given as 6 cm as in fig.b (for a square, both diagonals are equal).
 |
| Fig.13.10 |
When we are given the length of side of a square, we can easily calculate the area by using the formula:
Area of a square of side l = l2
But how do we calculate the area when diagonal is given? For that, we use a 'special property of diagonals of a square':
■ The diagonals of a square bisect each other, and are perpendicular to each other.
So the area consists of 4 equal triangles. This is shown in fig.b. They are: OAB, OBC, OCD and OAD. Not just triangles, they are isosceles triangles. Because two sides are equal (3 cm).
• In each triangle, the base is 3 cm, and height is 3 cm. So area of one triangle = 1⁄2 × b × h = 1⁄2 × 3 × 3 = 41⁄2
• ∴ Total area = 4 × 41⁄2 = 18 cm2
• Fig.c shows a square PQRS whose diagonal is 5 cm. Area of PQRS will be equal to 4 × 1⁄2 × 2.5 × 2.5 = 12.5 cm2
We can use algebra to get a general form:
From the above we can write:
■ Area of a square is half the square of it's diagonal
• Now consider a Rhombus ABCD in fig.13.11(a). Here also diagonals are given instead of sides. Diagonals of rhombus also have a special property: They bisect each other, and are perpendicular to each other.
 |
| Fig.13.11 |
• But unlike a square, the diagonals of a rhombus are not of equal length. So the four segments of the diagonals will not be of equal length.
• Consequently, the triangles formed by the diagonals are not isosceles.
• Even if they are not isosceles, we can calculate the area of each triangle easily. Because, we know the base and height. In fig.13.11(a), the area of each triangle is 1⁄2 × b × h = 1⁄2 × 3 × 2 = 3
• So the total area of the Rhombus ABCD = Total area of 4 squares = 4 × 3 = 12 cm2
• Similarly, the area of the rhombus PQRS in fig will be equal to 4 × 1⁄2 × 2.5 × 2 = 10 cm2
• We can use algebra to get a general form:
From the above result, we can write:
■ Area of a Rhombus is half the product of it's diagonals
Solved example 13.7
A square has an area of 4.5 cm2. What is the length of it’s diagonal?
Solution:
Solved example 13.8
The area of a non square rhombus is 216 cm2. The length of one of it’s diagonal is 24 cm. Compute the following:
(1) Length of the other diagonal (2) Length of a side (3) Perimeter (4) Distance between sides
Solution:
(1) Length of the other diagonal:
(2) Length of a side:
Let us draw a rough sketch (shown in fig.13.12.a), which shows all the available details:
 |
| Fig.13.12 |
• We have two diagonals BD = 24 cm and AC = 18 cm.
• The diagonals of a rhombus bisects each other. So AC will be split into AO and CO, each of 9 cm. Similarly, BD will be split into BO and DO, each of 12 cm.
• Consider any triangle, say ΔOBC. It is a right triangle, right angled at O. (∵ diagonals of a rhombus are perpendicular to each other)
• We can apply the Pythagoras theorem to this triangle: OB2 + OC2 = BC2
So 122 + 92 = BC2 ⇒ 144 + 81 = BC2 ⇒ 225 = BC2 ∴ BC = √225 = 15
(3) Perimeter: We have obtained the length of one side. For a rhombus, all four sides are equal. So Perimeter = 4 × 15 = 60 cm
(4) Distance between sides:
• Consider fig 13.12.b above. CE is drawn perpendicular to AB. So the length of CE is the distance between sides AB and CD. This will be the same distance between the other two sides AD and CB. So our aim is to find the length of CE.
• For that, we consider the rhombus as a parallelogram. A rhombus is actually a special parallelogram. So the method that we use to calculate the area of a parallelogram can be use to calculate the area of a rhombus also
• The method for parallelogram was explained here. We have to multiply the base by the height
• In our present problem, AB is the base, and CE is the height. So area of the rhombus ABCD = AB×CE
• The area is given to us as 216 cm2. earlier, we calculated AB (length of one side) as 15 cm. So, in the above equation, CE is the only unknown
• We can write: 216 = 15×CE. ∴ CE = 216/15 = 14.4 cm
Solved example 13.9
A 68 cm long rope is used to make a rhombus. The distance between two opposite corners is 16 cm. (1) What is the distance between the other two opposite corners? (2) What is the area of the rhombus?
Solution:
A rough sketch is shown in fig.13.13 below:
 |
| Fig.13.13 |
• The perimeter is given as 68 cm. We know that all the four sides of a rhombus are equal. So length of one side = 68/4 = 17 cm.
• Distance between one pair of opposite corners is given as 16 cm. This is the length d1 of one diagonal
• Let DB = 16 cm. Then DO = OB = 8 cm (∵ diagonals of a rhombus are perpendicular bisectors of each other)
• We can apply the Pythagoras theorem to the ΔOBC:
• OC2 + 82 = 172 ⇒ OC2 = 172 – 82 = 289 – 64 = 225 ∴ OC = √225 = 15
• Now, OA will also be 15 cm. So distance between the other pair of opposite corners = d2 = AC = 15 + 15 = 30 cm
• To find the area:
In the next section, we will learn about the Area of Isosceles Trapezium.
6. So we get, total area of the trapezium = 9 + 4 + x + x = 9 + 4 + 6 + 6 = 25 cm2 
