In the previous section we completed the discussion on parallelograms between parallel lines. In this section, we will see division of triangles. Consider the ΔABC in fig.14.25(a). It has a height h. In fig.b. a median CD is drawn to this triangle. So point D is the midpoint of side AB. That is., AD = BD. Since they are equal, we will denote both by ‘x’. So AD = BD = x.
The median CD splits the triangle ABC into two smaller triangles: ΔADC and ΔDBC. What about the areas of these two smaller triangles? Both have the same base ‘x’ and the same height ‘h’. So both have the same area . We can write:
• Area of ΔADC = 1⁄2 × b × h = 1⁄2 × x × h
• Area of ΔBDC = 1⁄2 × b × h = 1⁄2 × x × h
These two areas are the same. So we can write:
Theorem 14.5
A median splits the triangle into two smaller triangles, and the areas of these smaller triangles will be equal
Let us take a more advanced case. In this case, the line CD is not a median. This is shown in fig.14.26(a).
If it is not a median, the side AB will not be split equally. That is AD ≠ BD. So we will put AD = x and BD = y. This is shown in fig.a. We can easily calculate the areas of both ΔADC and ΔBDC, because, their base and height are known. Once we calculate the areas, some interesting properties can be determined as shown below:
So we find that the ratio of the areas is equal to the ratio of the bases. We will write the above findings in the form of a theorem. Usually theorems are written as a single sentence. Here we will write in a step by step manner:
Theorem 14.6:
• A line is drawn from a vertex to any point on the opposite side
• This line divides the triangle into two smaller triangles
• The point of intersection of this line with the opposite side divides the opposite side into two smaller line segments
• If the lengths of these smaller line segments are in the ratio x:y
• Then the areas of the two smaller triangles will also be in the ratio x:y
• While taking ratios, x and y should be assigned to the corresponding triangles
We did the above discussion in terms of x and y. Let us check with actual numbers:
Fig.14.26(b) shows a ΔPQR. A line RS is drawn from the vertex R to the base PQ. Point R divides the line PQ into PS and RS in such a way that PS = 6 cm and RS = 4 cm. The height of PQR is 5 cm. Let us calculate the areas:
• ar (PSR) = 1⁄2 × b × h = 1⁄2 × 6 × 5 = 15 cm2
• ar (QSR) = 1⁄2 × 4 × 5 = 10 cm2
• Ratio of areas = ar (PSR) : ar (QSR) = 15 :10 = 3:2
• Ratio of lengths = length of PS : Length of QS = 6 : 4 = 3 :2
■ Both ratios are the same
Solved example 14.16
A median AD is drawn in the triangle ABC. The midpoint of this median is marked as E. Show that (i) ar (AEC) = ar (BEC) (ii) ar (AED) = 1/4 ar (ABC)
Solution:
The rough sketch is shown in fig.14.27(a)
• ar (ADC) = ar (BDC) [∵ D is the midpoint of AB]
• Since the areas are the same, let us denote each area as X
• The sum of the two areas = Total area of ABC = ar (ABC) = X + X = 2X
■ From this we get X = ar (ABC)/2. = ar (ADC) = ar (BDC). That is., each area is half of the total area of ABC - - - (1)
• ar (AEC) = ar (AED) [∵ E is the midpoint of CD]
• Let us denote each area as Y
• The sum of the two areas = Total area of ADC = ar (ADC) = Y + Y = 2Y
■ From this we get Y = ar (ADC)/2. That is., each area is half of the total area of ΔADC. But from (1), ar (ADC) = X
So we get: Y = ar (AEC) = ar (AED) = ar (ADC)/2 = X/2 - - - (2)
• ar (BDE) = ar (BCE) [∵ E is the midpoint of CD]
• Let us denote each area as Z
• The sum of the two areas = Total area of BDC = ar (BDC) = Z + Z = 2Z
• From this we get Z = ar (BDC)/2. That is., each area is half of the total area of ΔBDC
• But from (1), ar (BDC) = X
■ So we get: Z = ar (BDE) = ar (BCE) = ar(BDC)/2 = X/2 - - - (3)
• From (2) we get ar (AEC) = X/2
• From (3) we get ar (BCE) = X/2
■ So we can write ar (AEC) = ar (BCE) - - - Answer 1
• From (2) we get AED = X/2. But from (1) X = ar (ABC)/2
■ So ar (AED) = [ar (ABC)/2]/2 = ar (ABC)/4 - - - Answer 2
Solved example 14.17
In the fig.14.27(b), CD is bisected by AB at O. Show that ar (ABC) = ar (ABD)
Solution:
• AB bisects CD at O. So we get OC = OD
• So in the ΔADC, AO is a median, and thus, ar (AOD) = ar (AOC)
• Since the areas are equal, we will denote both the areas as X
• That is., ar (AOD) = ar (AOC) = X
• In a similar way, we can write: ar (BOD) = ar (BOC) = Y
■ ar (ABC) = ar (AOC) + ar (BOC) = X + Y
■ ar (ABD) = ar (AOD) + ar (BOD) = X + Y
■ So they are equal
Solved example 14.18
In the ΔABC in fig(c), D is any point on AB, and E is any point on AC. If ar (CBD) = ar (BCE), prove that DC is parallel to BC
Solution:
• Given that ar (CBD) = ar (BCE)
• These two triangles have the same base BC
• So if they have the same area, it means that, they have the same height
• So the third vertex D of CBD and, the third vertex E of BCE are at the same height from the base BC
• That means the line joining D and E will be parallel to BC as shown in fig(d)
In the next section we continue our discussion on Division of Triangles.
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| Fig.14.25 |
• Area of ΔADC = 1⁄2 × b × h = 1⁄2 × x × h
• Area of ΔBDC = 1⁄2 × b × h = 1⁄2 × x × h
These two areas are the same. So we can write:
Theorem 14.5
A median splits the triangle into two smaller triangles, and the areas of these smaller triangles will be equal
Let us take a more advanced case. In this case, the line CD is not a median. This is shown in fig.14.26(a).
 |
| Fig.14.26 |
So we find that the ratio of the areas is equal to the ratio of the bases. We will write the above findings in the form of a theorem. Usually theorems are written as a single sentence. Here we will write in a step by step manner:
Theorem 14.6:
• A line is drawn from a vertex to any point on the opposite side
• This line divides the triangle into two smaller triangles
• The point of intersection of this line with the opposite side divides the opposite side into two smaller line segments
• If the lengths of these smaller line segments are in the ratio x:y
• Then the areas of the two smaller triangles will also be in the ratio x:y
• While taking ratios, x and y should be assigned to the corresponding triangles
We did the above discussion in terms of x and y. Let us check with actual numbers:
Fig.14.26(b) shows a ΔPQR. A line RS is drawn from the vertex R to the base PQ. Point R divides the line PQ into PS and RS in such a way that PS = 6 cm and RS = 4 cm. The height of PQR is 5 cm. Let us calculate the areas:
• ar (PSR) = 1⁄2 × b × h = 1⁄2 × 6 × 5 = 15 cm2
• ar (QSR) = 1⁄2 × 4 × 5 = 10 cm2
• Ratio of areas = ar (PSR) : ar (QSR) = 15 :10 = 3:2
• Ratio of lengths = length of PS : Length of QS = 6 : 4 = 3 :2
■ Both ratios are the same
Solved example 14.16
A median AD is drawn in the triangle ABC. The midpoint of this median is marked as E. Show that (i) ar (AEC) = ar (BEC) (ii) ar (AED) = 1/4 ar (ABC)
Solution:
The rough sketch is shown in fig.14.27(a)
 |
| Fig.14.27 |
• Since the areas are the same, let us denote each area as X
• The sum of the two areas = Total area of ABC = ar (ABC) = X + X = 2X
■ From this we get X = ar (ABC)/2. = ar (ADC) = ar (BDC). That is., each area is half of the total area of ABC - - - (1)
• ar (AEC) = ar (AED) [∵ E is the midpoint of CD]
• Let us denote each area as Y
• The sum of the two areas = Total area of ADC = ar (ADC) = Y + Y = 2Y
■ From this we get Y = ar (ADC)/2. That is., each area is half of the total area of ΔADC. But from (1), ar (ADC) = X
So we get: Y = ar (AEC) = ar (AED) = ar (ADC)/2 = X/2 - - - (2)
• ar (BDE) = ar (BCE) [∵ E is the midpoint of CD]
• Let us denote each area as Z
• The sum of the two areas = Total area of BDC = ar (BDC) = Z + Z = 2Z
• From this we get Z = ar (BDC)/2. That is., each area is half of the total area of ΔBDC
• But from (1), ar (BDC) = X
■ So we get: Z = ar (BDE) = ar (BCE) = ar(BDC)/2 = X/2 - - - (3)
• From (2) we get ar (AEC) = X/2
• From (3) we get ar (BCE) = X/2
■ So we can write ar (AEC) = ar (BCE) - - - Answer 1
• From (2) we get AED = X/2. But from (1) X = ar (ABC)/2
■ So ar (AED) = [ar (ABC)/2]/2 = ar (ABC)/4 - - - Answer 2
Solved example 14.17
In the fig.14.27(b), CD is bisected by AB at O. Show that ar (ABC) = ar (ABD)
Solution:
• AB bisects CD at O. So we get OC = OD
• So in the ΔADC, AO is a median, and thus, ar (AOD) = ar (AOC)
• Since the areas are equal, we will denote both the areas as X
• That is., ar (AOD) = ar (AOC) = X
• In a similar way, we can write: ar (BOD) = ar (BOC) = Y
■ ar (ABC) = ar (AOC) + ar (BOC) = X + Y
■ ar (ABD) = ar (AOD) + ar (BOD) = X + Y
■ So they are equal
Solved example 14.18
In the ΔABC in fig(c), D is any point on AB, and E is any point on AC. If ar (CBD) = ar (BCE), prove that DC is parallel to BC
Solution:
• Given that ar (CBD) = ar (BCE)
• These two triangles have the same base BC
• So if they have the same area, it means that, they have the same height
• So the third vertex D of CBD and, the third vertex E of BCE are at the same height from the base BC
• That means the line joining D and E will be parallel to BC as shown in fig(d)
In the next section we continue our discussion on Division of Triangles.
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