CHAPTER 9 AREAS OF PARALLELOGRAMS AND TRIANGLES
NCERT Class 9 Mathematics Textbook 2018 made Screen Readable for Blind Students by Professor T K Bansal.
“God is so great that he has manifested himself in to an infinite number of living and non-living objects. For instance, the God is present in front of me in the form of my students. So, when I make my chapters, I make sure that I do not leave any mistakes in my work, and do it so perfectly, that my students enjoy studying these chapters, and feel happy. My work is my worship . In case you face any difficulties, or find any mistakes in my work, please do write to me at Blind2Visionary@Gmail.com. By T K Bansal.
Table of symbols used in this chapter.
| Symbol | Known As | Used for | Description |
|---|---|---|---|
| ar(A) | Ar(A) | Area of figure A | Ar stands for area and value in parenthesis tells us about the figure. |
| ≅ | Approximately equal to | Congruent to | A tilda sign over a = sign. |
| ∥ | Parallel to | Parallel to | 2 vertical bars. |
| ∠ | Angle | Angle | A horizontal line and a tilted line meeting on the left. |
9.1 Introduction
NCERT Class 9 Mathematics Textbook made Screen Readable for Blind Students by Professor T K Bansal.
In Chapter 5, we learnt that the study of Geometry originated with the measurement of earth (lands) in the process of recasting boundaries of the fields and dividing them into appropriate parts. For example, an old farm lady had a triangular field and she wanted to divide it equally among her two daughters and one son. Without actually calculating the area of the field, she just divided one side of the triangular field into three equal parts and joined the two points of division to the opposite vertex. In this way, the field was divided into three parts, and she gave one part to each of her children. Do you think that all the three parts so obtained by her were equal in area? To get answers to this type of questions and other related problems, there is a need to have a relook at areas of plane figures, which we have already studied in our earlier classes.
We may recall that the part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area. This magnitude or measure is always expressed with the help of a number (in some unit) such as 5 square centimeter, 8 square meter,\) 3 hectares etc. So, we can say that area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure.
From our earlier classes and from Chapter 7, we are familiar with the concept of congruent figures. Two figures are called congruent if they have the same shape and the same size. In other words, if two figures A and B are congruent (see Figure 9.1) , then using a tracing paper, we can superpose one figure over the other such that it will cover the other completely.
Figure 9.1 Congruent Figures.
So, if two figures A and B are congruent, they must have equal areas. However, the converse of this statement is not true. In other words, two figures having equal areas need not be congruent. For example, in Figure Figure 9.2, rectangles ABCD and EFGH have equal areas (9 × 4 cm^2 and 6 × 6 square cm) but clearly, they are not congruent. (Why?)
Figure 9.2 Non-Congruent Figures with same Area.
Now, let us look at Figure 9.3 given below:
Figure 9.3. A figure joined by 2 figures.
We observe that planar region formed by figure T is made up of two planar regions formed by figures P and Q. We can easily see that
Area of figure T = Area of figure P + Area of figure Q.
We may denote the area of figure A as \(ar(A),\) area of figure B as \(ar(B),\) and area of figure T as \(ar(T),\) and so on. Now we can say that the area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure with the following two properties:
(1) If A and B are two congruent figures (as in figure 9.1), then \(ar(A)\ =\ ar(B);\) and
(2) if a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q (as in figure 9.3), then
\[ar(T)\ =\ ar(P)\ +\ ar(Q).\]
In our earlier classes, We have learnt some formulae for finding the areas of different figures such as rectangle, square, parallelogram, triangle etc. In this chapter, attempt will be made to consolidate the knowledge about these formulae by studying some relationship between the areas of these geometric figures under the condition when they are constructed on the same base and between the same parallel lines. This study will also be useful in the understanding of some results on ‘similarity of triangles.
9.2 Figures on the Same Base and Between the Same Parallels
NCERT Class 9 Mathematics Textbook made Screen Readable for Blind Students by Professor T K Bansal.
Look at the following figures:
Figure 9.4(i) Trapezium and parallelogram on the same base.
In Figure 9.4(i), trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC.
Similarly, in Figure 9.4 (ii), parallelograms PQRS and MNRS are on the same base SR.
Figure 9.4(ii) Two parallelograms on the same base.
in Figure 9.4(iii), triangles ABC and DBC are on the same base BC
Figure 9.4(iii) Two triangles on the same base.
and in Figure 9.4(iv), parallelogram ABCD and triangle PDC are on the same base DC.
Figure 9.4(iv) A parallelogram and a triangle on the same base.
Now, look at the following figures:
In Figure 9.5(i), trapezium ABCD and parallelogram EFCD are on the same base DC. In addition to the above, the vertices A and B (of trapezium ABCD) opposite to base DC and the vertices E and F (of parallelogram EFCD) opposite to base DC lie on a line AF parallel to DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC and between the same parallels AF and DC.
Figure 9.5(i) A Trapezium and parallelogram on the same base and between the same parallels.
Similarly, parallelograms PQRS and MNRS are on the same base SR and between the same parallels PN and SR [see Figure 9.5 (ii)] as vertices P and Q of PQRS and vertices M and N of MNRS lie on a line PN parallel to base SR.
Figure 9.5(ii) 2 parallelograms on the same base and between the same parallels.
In the same way, triangles ABC and DBC lie on the same base BC and between the same parallels AD and BC [see Figure 9.5 (iii)]
Fig 9.5(iii) 2 triangles on the same base and between the same parallels.
parallelogram ABCD and triangle PCD lie on the same base DC and between the same parallels AP and DC [see Figure 9.5(iv)].
Figure 9.5(iv) A parallelogram and a triangle on the same base and between the same parallels.
So, two figures are said to be on the same base and between the same parallels if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
Keeping in view of the above statement, you cannot say that triangle PQR and triangle DQR of Figure 9.6(i) lie between the same parallels l and QR.
Figure 9.6(i)
Similarly, we cannot say that parallelograms EFGH and MNGH of Figure 9.6(ii) lie between the same parallels EF and HG and that parallelograms ABCD and EFCD of Figure 9.6(iii) lie between the same parallels AB and DC (even though they have a common base DC and lie between the parallels AD and BC).
Figure 9.6(ii)
Figure 9.6(iii)
So, it should clearly be noted that out of the two parallels, one must be the line containing the common base.
Note that triangle ABC and triangle DBE of Figure 9.7(i) are not on the common base.
Figure 9.7 (i)
Similarly, triangle ABC and parallelogram PQRS of Figure 9.7(ii) are also not on the same base.
Figure 9.7 (ii)
EXERCISE 9.1
Q1. Which of the following figures lie on the same base and between the same parallels? In such a case, write the common base and the two parallels.
Figure 9.8 (i)
Figure 9.8 (ii)
Figure 9.8 (iii)
Figure 9.8 (iv)
Figure 9.8 (v
Figure 9.8 (vi)
A1.
(i) Base DC, parallels DC and AB.
(iii) Base QR, parallels QR and PS.
(v) Base AD, parallels AD and BQ
9.3 Parallelograms on the same Base and Between the same Parallels
NCERT Class 9 Mathematics Textbook made Screen Readable for Blind Students by Professor T K Bansal.
Now, let us try to find a relation, if any, between the areas of two parallelograms on the same base and between the same parallels. For this, let us perform the following activities:
Activity 1
Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Figure 9.9.
Figure 9.9
The above two parallelograms are on the same base DC and between the same parallels PB and DC. You may recall the method of finding the areas of these two parallelograms by counting the squares.
In this method, the area is found by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure. The squares whose less than half parts are enclosed by the figure are ignored. You will find that areas of both the parallelograms are (approximately) 15cm^2. Repeat this activity* [*This activity can also be performed by using a Geoboard.] by drawing some more pairs of parallelograms on the graph sheet. What do you observe? Are the areas of the two parallelograms different or equal? If fact, they are equal. So, this may lead us to conclude that parallelograms on the same base and between the same parallels are equal in area. However, remember that this is just a verification.
Activity 2
Draw a parallelogram ABCD on a thick sheet of paper or on a cardboard sheet. Now, draw a line segment DE as shown in Figure 9.10.
Figure 9.10
Next, cut a triangle A′D′E′ congruent to triangle ADE on a separate sheet with the help of a tracing paper and place triangle A′D′E′ in such a way that A′D′coincides with BC as shown in Figure 9.11.
Note that there are two parallelograms ABCD and EE′CD on the same base DC and between the same parallels, AE′ and DC. What can you say about their areas?
Figure 9.11
As ΔADE congruent to ΔA′D′E′
\[∴\ ar(ADE)\ =\ ar(A′D′E′)\]
Also,
\[ar(ABCD)\ =\ ar(ADE)\ +\ ar(EBCD)\]
\]=\ ar(A′D′E′)\ +\ ar(EBCD)\]
\[=\ ar(EE′CD)\]
So, the two parallelograms are equal in area.
Let us now try to prove this relation between the two such parallelograms.
Theorem 9.1
Parallelograms on the same base and between the same parallels are equal in area.
Proof :
Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given (see Figure 9.12).
Figure 9.12
We need to prove that ar (ABCD) = ar (EFCD).
In ΔADE and ΔBCF,
∠DAE = ∠CBF (Corresponding angles from AD ∥ BC and transversal AF) (1)
∠AED = ∠BFC (Corresponding angles from ED ∥ FC and transversal A F) (2)
Therefore, ∠ADE = ∠BCF (Angle sum property of a triangle ) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, ΔADE ≅ ΔBCF [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB) [From(5)]
= ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.
Let us now take some examples to illustrate the use of the above theorem.
Example 1
In Figure 9.13, ABCD is a parallelogram and EFCD is a rectangle.
Figure 9.13
Also, AL perpendicular to DC. Prove that
(i) ar (ABCD) = ar (EFCD)
(ii) ar (ABCD) = DC × AL
Solution :
(i) As a rectangle is also a parallelogram,
therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1)
(ii) From above result,
ar (ABCD) = DC × FC (Area of the rectangle = length × breadth) (1)
As AL perpendicular to DC, therefore, AFCL is also a rectangle
So, AL = FC (2)
Therefore, ar (ABCD) = DC × AL [From (1) and (2)]
Can you see from the Result (ii) above that area of a parallelogram is the product of its any side and the corresponding altitude? Do you remember that you studied this formula for area of a parallelogram in Class 7. On the basis of this formula, Theorem 9.1 can be rewritten as parallelograms on the same base or equal bases and between the same parallels are equal in area.
Can you write the converse of the above statement? It is as follows:
Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. Is the converse true? Prove the converse using the formula for area of the parallelogram.
Example 2
If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.
Solution :
Let ΔABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (see Figure 9.14).
Figure 9.14
We wish to prove that ar (P AB) = ½ ar (ABCD)
Draw BQ ∥ AP to obtain another parallelogram ABQP.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)
But ΔPAB ≅ ΔBQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.)
So, ar (PAB) = ar (BQP) (2)
Therefore, ar (P AB) = ½ ar (ABQP) [From (2)] (3)
This gives ar (P AB) = ½ ar (ABCD) [From (1) and (3)]
EXERCISE 9.2."
Q1. In Figure 9.15, ABCD is a parallelogram, AE perpendicular to DC and CF perpendicular to AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, find AD.
Figure 9.15
A1. 12.8 cm.
Q2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ½ ar (ABCD).
A2. Join EG; Use result of Example 2.
Q3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Q4. In Figure 9.16, P is a point in the interior of a parallelogram ABCD.
Figure 9.16
Show that
(i) ar (APB) + ar (PCD) = ½ ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint : Through P, draw a line parallel to AB.]
Q5. In Figure 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
Figure 9.17
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AX S) = ½ ar (PQRS)
Q6. A farmer had a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
A6. Wheat in triangle APQ and pulses in other two triangles or pulses in triangle APQ and wheat in other two triangles.
9.4 triangles on the same Base and between the same Parallels
NCERT Class 9 Mathematics Textbook made Screen Readable for Blind Students by Professor T K Bansal.
Let us look at Figure 9.18.
Figure 9.18
In it, we have two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. What can you say about the areas of such triangles? To answer this question, we may perform the activity of drawing several pairs of triangles on the same base and between the same parallels on the graph sheet and find their areas by the method of counting the squares. Each time, we will find that the areas of the two triangles are (approximately) equal. This activity can be performed using a geoboard also. We will again find that the two areas are (approximately) equal.
To obtain a logical answer to the above question, we may proceed as follows:
In Figure 9.18, draw CD ∥ BA and CR ∥ BP such that D and R lie on line AP(see Figure 9.19).
Figure 9 .19
From this, we obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.
Therefore, ar (ABCD) = ar (PBCR) (Why?)
Now ΔABC ≅ ΔCDA and ΔPBC ≅ ΔCRP (Why?)
So, ar (ABC) = ½ ar (ABCD) and ar (PBC) = ½ ar (PBCR) (Why?)
Therefore, ar (ABC) = ar (PBC)
In this way, we have arrived at the following theorem:
Theorem 9.2
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Now, suppose ABCD is a parallelogram one of whose diagonals is AC.
(see Figure 9.20). Let A N perpendicular to DC. Note that
ΔADC ≅ ΔCBA (Why?)
Figure 9.20
So, ar (ADC) = ar (CBA) (Why?)
Therefore, ar (ADC) = ½ ar (ABCD) = ½ (DC × AN) (Why?)
So, area of ΔADC = ½ × base DC × corresponding altitude AN
In other words, the area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).
Do you remember that you have learnt this formula for area of a triangle in Class 7 ? From this formula, you can see that two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
For having equal corresponding altitudes, the triangles must lie between the same parallels. From this, you arrive at the following converse of Theorem 9.2 .
Theorem 9.3
Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
Let us now take some examples to illustrate the use of the above results.
Example 3
Show that a median of a triangle divides it into two triangles of equal areas.
Solution :
Let ABC be a triangle and let AD be one of its medians (see Figure 9.21).
Figure 9.21
We wish to show that
\[ar(ABD)\ =\ ar(ACD).\]
Since the formula for area involves altitude, let us draw AN perpendicular to BC.
Now ar(ABD) = ½ × base × altitude (of triangle ABD)
= ½ BD × AN
= ½ CD × AN (As BD = CD)
= ½ × base × altitude (of triangle ACD)
= ar(ACD)
Example 4
In Figure 9.22, ABCD is a quadrilateral and BE ∥ AC and also BE meets DC produced at E. Show that area of triangle ADE is equal to the area of the quadrilateral ABCD.
Fig 9.22
Solution :
Observe the figure carefully .
ΔBAC and ΔEAC lie on the same base AC and between the same parallels AC and BE.
Therefore, ar(BAC) = ar(EAC) (By Theorem 9.2)
So, ar(BAC) + ar(ADC) = ar(EAC) + ar(ADC) (Adding same areas on both sides)
or ar(ABCD) = ar (ADE)
EXERCISE 9.3
Q1. In Figure 9.23, E is any point on median AD of a triangle ABC. Show that ar (ABE) = ar (ACE).
Figure 9.23
Q2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = ¼ ar(ABC)
Q3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Q4. In Figure 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).
Figure 9.24
A4. Draw CM perpendicular to AB and DN perpendicular to AB. Show CM = DN.
Q5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a triangle ABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = ¼ ar (ABC)
(iii) ar (BDEF) = ½ ar (ABC)
Q6. In Figure 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
Figure 9.25
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA ∥ CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Q7. D and E are points on sides AB and AC respectively of triangle ABC such that ar (DBC) = ar (EBC). Prove that DE ∥ BC.
Q8. XY is a line parallel to side BC of a triangle ABC. If BE ∥and CF ∥ AB meet XY at E and F respectively, show that
ar (AB E) = ar (ACF)
Q9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and ∥ CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure 9.26). Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Figure 9.26
Q10. Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at O. Prove that ar (A O D) = ar (B O C).
Q11. In Figure 9.27, ABCD E is a pentagon. A line through B ∥ AC meets DC produced at F. Show that
Figure 9.27
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCD E)
Q12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
A12. See Example 4.
Q13. ABCD is a trapezium with AB ∥ DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint : Join CX.]
Q14. In Figure 9.28, AP ∥ BQ ∥ CR. Prove that ar (A QC) = ar (PBR).
Figure 9.28
Q15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (B O C). Prove that ABCD is a trapezium.
Q16. In Figure 9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (A R C). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Figure 9.29
EXERCISE 9.4 (Optional)*
Q1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Q2. In Figure 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
Figure 9.30
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide triangle ABC into n triangles of equal areas.]
Q3. In Figure 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
Figure 9.31
Q4. In Figure 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint : Join AC.]
Figure 9.32
Q5. In Figure 9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
Figure 9.33
(i) ar (BDE) = ¼ ar (ABC)
(ii) ar (BDE) = ½ ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE ∥ AC and DE ∥ AB, etc.]
Q6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
Q7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar (PRQ) = ½ ar (A R C)
(ii) ar (RQC) = 3/8 ar (ABC)
(iii) ar (PBQ) = ar (A R C)
A7. Use result of Example 3 repeatedly.
Q8. In Figure 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment A X perpendicular to DE meets BC at Y. Show that:
Figure 9.34
(i) triangle MBC congruent to triangle ABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) triangle FCB congruent to triangle ACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. We will learn a simpler proof of this theorem in Class 10.
9.5 Summary
NCERT Class 9 Mathematics Textbook made Screen Readable for Blind Students by Professor T K Bansal.
In this chapter, we have studied the following points :
1.
Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure.
2.
Two congruent figures have equal areas, but the converse need not be true.
3.
If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X.
4.
Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
5.
Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
6.
Area of a parallelogram is the product of its base and the corresponding altitude.
7.
Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
8.
If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
9.
triangles on the same base (or equal bases) and between the same parallels are equal in area.
10.
Area of a triangle is half the product of its base and the corresponding altitude.
11.
triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
12.
A median of a triangle divides it into two triangles of equal areas.
Note:
Great! I, Dr T K Bansal, is thankful to you for studying this chapter with so much of efforts and patience! I am sure that this lesson was fully accessible and is of immense help to you. If you still have difficulties, study the chapter repeatedly. Immense efforts have been made to make this chapter mistake-free, still if you find any mistakes, or have any difficulties, please do write to us at blind2visionary@gmail.com. Your efforts will be highly appreciated.
End of the Chapter.