Table of contents | |
Introduction | |
What are the Properties of a Parallelogram? | |
Theorems on Properties of a Parallelogram | |
Solved Examples |
It is a four-sided closed figure with opposite sides and equal angles. A parallelogram is a convex polygon with 4 edges and 4 vertices. The opposite sides are equal and parallel; the opposite angles are also equal.
Let's learn more about the properties of parallelograms in detail in this lesson.
The four important properties of a parallelogram are as follows.
Sides and Angles of a Parallelogram
Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.
First, we will recall the meaning of a diagonal. Diagonals are line segments that join the opposite vertices. In parallelogram PQRS, PR and QS are the diagonals. The properties of diagonals of a parallelogram are as follows:
Diagonals of a Parallelogram
Theorem 2: In a Parallelogram the Opposite Sides Are Equal. This means, in a parallelogram, the opposite sides are equal.
Given: ABCD is a parallelogram.
To Prove: The opposite sides are equal, AB=CD, and BC=AD.
Proof: In parallelogram ABCB, compare triangles ABC and CDA. In these triangles AC = CA (common sides). Also ∠BAC =∠DCA (alternate interior angles), and ∠BCA = ∠DAC (alternate interior angles). Hence by the ASA criterion, both the triangles are congruent and the corresponding sides are equal. Therefore we have AB = CD, and BC = AD.
Theorem 3: If the opposite sides in a quadrilateral are equal, then it is a parallelogram. If AB = CD and BC = AD in the given quadrilateral ABCD, then it is a parallelogram.
Given: The opposite sides in a quadrilateral ABCD are equal, AB = CD, and BC = AD.
To Prove: ABCD is a parallelogram.
Proof: In the quadrilateral ABCD we are given that AB = CD, and AD = BC. Now compare the two triangles ABC, and CDA. Here we have AC = AC (Common sides), AB = CD (since alternate interior angles are equal), and AD = BC (given). Thus by the SSS criterion both the triangles are congruent, and the corresponding angles are equal. Hence we can conclude that ∠BAC = ∠DCA, and ∠BCA = ∠DAC. Therefore AB // CD, BC // AD, and ABCD is a parallelogram.
Theorem 4: In a Parallelogram, the Opposite Angles Are Equal.
Given: ABCD is a parallelogram, and ∠A, ∠B, ∠C, ∠D are the four angles.
To Prove: ∠A =∠C and ∠B=∠D
Proof: Let us assume that ABCD is a parallelogram.
Now compare triangles ABC, and CDA.
Here we have AC=AC (common sides), ∠1=∠4 (alternate interior angles), and ∠2=∠3 (alternate interior angles).
Thus, the two triangles are congruent, which means that ∠B=∠D. Similarly, we can show that ∠A=∠C.
This proves that opposite angles in any parallelogram are equal.
Theorem 5: If the opposite angles in a quadrilateral are equal, then it is a parallelogram.
Given: ∠A=∠C and ∠B=∠D in the quadrilateral ABCD. To Prove: ABCD is a parallelogram.
Proof: Assume that ∠A = ∠C and ∠B = ∠D in the parallelogram ABCD given above.
We have to prove that ABCD is a parallelogram.
We have: ∠A + ∠B + ∠C + ∠D = 360º;2(∠A + ∠B) =360º; ∠A + ∠B = 180º.
This must mean that AD // BC. Similarly, we can show that AB//CD. Hence, AD//BC, and AB//CD.
Therefore ABCD is a parallelogram.
Theorem 6: Diagonals of a Parallelogram Bisect Each Other. That means, in a parallelogram, the diagonals bisect each other.
Given: PQTR is a parallelogram. PT and QR are the diagonals of the parallelogram.
To Prove: The diagonals PT, and RQ bisect each other. PE=ET and ER=EQ
Proof: First, let us assume that PQTR is a parallelogram. Compare triangles RET, and triangle PEQ.
We have PQ = RT (opposite sides of the parallelogram), ∠QRT = ∠PQR (alternate interior angles), and ∠PTR = ∠QPT (alternate interior angles).
By the ASA criterion, the two triangles are congruent, which means that PE = ET, and RE = EQ.
Thus, the two diagonals PT, and RQ bisect each other, and PE=ET and ER=EQ
Theorem 7: If the diagonals in a quadrilateral bisect each other, then it is a parallelogram. In the quadrilateral PQTR, if PE=ET and ER=EQ, then it is a parallelogram.
Given: The diagonals PT and QR bisect each other.
To Prove: PQRT is a parallelogram.
Proof: Suppose that the diagonals PT and QR bisect each other. Compare triangle RET, and triangle PEQ once again.
We have: RE = EQ, ET = PE (Diagonals bisect each other), ∠RET =∠PEQ (vertically opposite angles).
Hence by the SAS criterion, the two triangles are congruent.
This means that ∠QRT = ∠PQR, and ∠PRT = ∠QPT .
Hence, PQ//RT, and RT//QT.
Thus PQRT is a parallelogram.
Theorem 8: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Proof: You can prove this theorem using the following clue:
Observe above figure in which E and F are mid-points of AB and AC respectively and CD || BA.
∆ AEF ≅ ∆ CDF (ASA Rule)
So, EF = DF and BE = AE = DC (Why?)
Therefore, BCDE is a parallelogram. (Why?)
This gives EF || BC.
In this case, also note that EF = 1/2ED = 1/2BC.
Can you state the converse of Theorem 8.8? Is the converse true?
You will see that converse of the above theorem is also true which is stated as below:
Theorem 9: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
In above figure, observe that E is the mid-point of AB, line l is passing through E and is parallel to BC and CM || BA.
Sol: A rectangle is a parallelogram in which each angle is 90° and parallelogram on the same base and between same parallels are equal in area.
∴ ar. (||gmABCD) = ar. (rectangle ABPQ)
But ar. (||gmABPQ) = AB×AQ
= (8 × 6) cm2 = 48 cm2
∴ Area of parallelogram ABCD = 48 cm2.
Example 2: In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 m and 5 cm. Find AD.
Sol: We know that the area of a parallelogram = Base × Corresponding altitude.
∴ Area of parallelogram ABCD = AD × BN = AB × DM
⇒ AD × 5 = 8 × 4
⇒ AD =
= 6.4 cm.
Example 3: ABCD is a quadrilateral and BD is one its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.
Sol: In ΔABD and ΔCDB
∠ABD = ∠CBD = 90o (alternate interior angles are equal)
∠ADB = ∠CBD
∠BDA = ∠BDC
Hence, ABCD is a parallelogram.
Now, area of parallelogram ABCD = (base ×corresponding altitude)
⇒area of parallelogram ABCD = AB × BD = 5 ×12 sq. units = 60 sq. cms.
Example 4: Prove that parallelograms on the same base and between the same parallel lines are equal in area.
Sol: Given: Two parallelogram ABCD and PBCQ having same base BC and between the same parallel lines BC and AQ.
To Prove: ar(parallelogram ABCD)= ar (parallelogram PBCQ)
Proof: In triangles ABP and DCQ,
∠BAP = ∠CDQ
(Corresponding angles when AQ intersects parallel lines AB and DC).
∠BPA = ∠CDQ
(Corresponding angles when AQ intersects parallel lines BP and CQ).
AB = DC (Opposite sides of a parallelogram)
∴ΔABP ≅ ΔDCQ (AAS)
Hence area ΔABP = area ΔDCQ
or, area ΔABP + area BPDC = area ΔDCQ + area Δ BPDC
∴ar (parallelogram ABCD) = ar (parallelogram PBCQ).
44 videos|412 docs|54 tests
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1. What are the main properties of a parallelogram? |
2. How can you prove that a quadrilateral is a parallelogram? |
3. What is the significance of the diagonals in a parallelogram? |
4. Are the opposite angles in a parallelogram always equal? |
5. How do the properties of a parallelogram apply in real-life situations? |
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