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Areas of Parallelograms And Triangles - Class 9 : Notes


 

(1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.
Given:  A parallelogram ABCD in which BD is one of the diagonals.
To prove:   
Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that it is sufficient to show that
\Delta ABD \cong \Delta CDB
In \Delta s ABD and CDB, we have
AB = CD
AD = CB
And, BD = DB
So, by SSS criterion of congruence, we have
\Delta ABD \cong \Delta CDB
Hence, ar\left( {\Delta ABD} \right) = ar\left( {\Delta CDB} \right)

(2) Prove that parallelograms on the same base and between the same parallels are equal in area.
Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.
To prove: ar(parallelogram ABCD) = ar(parallelogram ABCD)Proof: In \Delta s ADF and BCE, we have
AD = BC
AF = BE
And, \angle DAF = \angle CBE         [ ⸪ AD\parallel BC and AF\parallel BE]
So, by SAS criterion of congruence, we have
\Delta ADF \cong \Delta BCE
ar\left( {\Delta ADF} \right) = ar\left( {\Delta BCE} \right)   …..(i)
Now, ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {sq.ABED} \right) + ar\left( {\Delta BCE} \right)
ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {sq.ABED} \right) + ar\left( {\Delta ADF} \right)    [Using(i)]
ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {parallelogram{\text{ }}ABEF} \right)
Hence, ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {parallelogram{\text{ }}ABEF} \right)

(3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.
Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.
To prove: ar\left( {parallelogram{\text{ }}ABCD} \right) = AB \times AL
Construction: Complete the rectangle ALMB by drawing BM \bot CD.Proof: Since ar\left( {parallelogram{\text{ }}ABCD} \right) and rectangle ALMB are on the same base and between the same parallels.
ar\left( {parallelogram{\text{ }}ABCD} \right)
 = ar\left( {rect.ALMB} \right)
 = AB \times AL    [By rect. Area axiom area of a rectangle = Base X Height]
Hence, ar\left( {parallelogram{\text{ }}ABCD} \right) = AB \times AL

(4) Prove that parallelograms on equal bases and between the same parallels are equal in area.
Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.
To prove: ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {parallelogram{\text{ PQRS}}} \right)
Construction: Draw AL \bot DR and PM \bot DRProof: Since AB \bot DR, AL \bot DR and PM \bot DR
AL = PM
Now, ar\left( {parallelogram{\text{ }}ABCD} \right) = AB \times AL
ar\left( {parallelogram{\text{ }}ABCD} \right) = PQ \times PM  [AB = PQ and AL = PM]
ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {parallelogram{\text{ PQRS}}} \right)

(5) Prove that triangles on the same bases and between the same parallels are equal in area.
Proof: We have,
BD\parallel CA
And, BC\parallel DA
sq.BCAD is a parallelogram.Similarly, sq.BCQP is a parallelogram.
Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.
ar\left( {parallelogram{\text{ BCQP}}} \right) = ar\left( {parallelogram{\text{ BCAD}}} \right)    ….(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
ar\left( {\Delta PBC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCQP}}} \right)   …..(ii)
And, ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCAQ}}} \right)  ....(iii)
Now, ar\left( {parallelogram{\text{ BCQP}}} \right) = ar\left( {parallelogram{\text{ BCAD}}} \right) [ From (i)]
\frac{1}{2}ar\left( {parallelogram{\text{ BCQP}}} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCAD}}} \right)
ar\left( {\Delta ABC} \right) = ar\left( {\Delta PBC} \right)    [From (ii) and (iii)]
Hence, ar\left( {\Delta ABC} \right) = ar\left( {\Delta PBC} \right)

(6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.
Given: A {\Delta ABC} in which AL is the altitude to the side BC.
To prove: ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times AL} \right)
Construction: Through C and A draw CD\parallel BA and AD\parallel BC respectively, intersecting each other at D.Proof: We have,
BA\parallel CD
And, AD\parallel BC
BCDA is a parallelogram.
Since AC is a diagonal of parallelogram BCDA.
ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCAD}}} \right)
ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times AL} \right) [BC is the base and AL is the corresponding altitude of parallelogram BCDA]

(7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.
Given: A {\Delta ABC} and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.
To prove: ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCDE}}} \right)
Construction: Draw AL \bot BC and DM \bot BC, meeting BC produced in M.Proof: Since A, E and D are collinear and BC\parallel AD
AL = DM    …..(i)
Now,
ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times AL} \right)
ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times DM} \right)   [AL = DM (from (i)]
ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCDE}}} \right)

(8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.
Given: A trapezium ABCD in which AB\parallel DC; AB = a, DC = b and AL = CM = h, where AL \bot DC and CM \bot AB.
To prove: ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}h \times \left( {a + b} \right)
Construction: Join ACProof: We have,
ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = ar\left( {\Delta ABC} \right) + ar\left( {\Delta ACD} \right)
ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}\left( {AB \times CM} \right) + \frac{1}{2}\left( {DC \times AL} \right)
ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}ah \times \frac{1}{2}bh  [AB =a and DC = b]
ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}h \times \left( {a + b} \right)

(9) Prove that triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal.
Given: Two triangles ABC and PQR such that:
ar\left( {\Delta ABC} \right) = ar\left( {\Delta PQR} \right)
AB = PQ
CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.
To prove: CN = RTProof: In {\Delta ABC}, CN is the altitude corresponding to side AB.
ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {AB \times CN} \right)    ….(i)
Similarly, we have,
ar\left( {\Delta PQR} \right) = \frac{1}{2}\left( {PQ \times RT} \right)   …..(ii)
Now, ar\left( {\Delta ABC} \right) = ar\left( {\Delta PQR} \right)
\frac{1}{2}\left( {AB \times CN} \right) = \frac{1}{2}\left( {PQ \times RT} \right)
\left( {AB \times CN} \right) = \left( {PQ \times RT} \right)
\left( {PQ \times CN} \right) = \left( {PQ \times RT} \right)    [ AB = PQ (Given)]
CN = RT

(10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.
Given: A quadrilateral ABCD such that its diagonals AC and BD are such that
ar\left( {\Delta ABD} \right) = ar\left( {\Delta CDB} \right) and ar\left( {\Delta ABC} \right) = ar\left( {\Delta ACD} \right)
To prove: Quadrilateral ABCD is a parallelogram.Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,
ar\left( {\Delta ABC} \right) = ar\left( {\Delta ACD} \right)    …..(i)
But, ar\left( {\Delta ABC} \right) + ar\left( {\Delta ACD} \right) = ar\left( {quad.ABCD} \right)
2ar\left( {\Delta ABC} \right) = ar\left( {quad.ABCD} \right)    [Using (i)]
ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {quad.ABCD} \right) ….(ii)
Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.
ar\left( {\Delta ABD} \right) = ar\left( {\Delta BCD} \right)   ….(iii)
But, ar\left( {\Delta ABD} \right) + ar\left( {\Delta BCD} \right) = ar(quad.ABCD)
2ar\left( {\Delta ABD} \right) = ar(quad.ABCD)    [Using(iii)]

ar\left( {\Delta ABD} \right) = \frac{1}{2}ar\left( {quad.ABCD} \right)   …..(iv)
From (ii) and (iv), we get
ar\left( {\Delta ABC} \right) = ar\left( {\Delta ABD} \right)
Since \Delta s ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.
i.e. Altitude from C of {\Delta ABC} = Altitude from D of {\Delta ABD}
DC\parallel AB
Similarly, AD\parallel BC
Hence, quadrilateral ABCD is a parallelogram.

(11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.
Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
To prove: ar(rhombus ABCD)  = \frac{1}{2}\left( {AC \times BD} \right)Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,
OB \bot AC and OD \bot AC
ar(rhombus)  = ar\left( {\Delta ABC} \right) + ar\left( {\Delta ADC} \right)
ar(rhombus)  = \frac{1}{2}\left( {AC \times BO} \right) + \frac{1}{2}\left( {AC \times DO} \right)
ar(rhombus) = \frac{1}{2}\left( {AC \times \left( {BO + DO} \right)} \right)
ar(rhombus)  = \frac{1}{2}\left( {AC \times BD} \right)

(12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.
Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.
To prove: ar\left( {\Delta OAB} \right) = ar\left( {\Delta OBC} \right) = ar\left( {\Delta OCD} \right) = ar\left( {\Delta AOD} \right)Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.
OA = OC and OB = OD
Also, the median of a triangle divides it into two equal parts.
Now, in {\Delta ABC}, BO is the median.
ar\left( {\Delta OAB} \right) = ar\left( {\Delta OBC} \right)  ….(i)
In {\Delta BCD}, CO is the median
ar\left( {\Delta OBC} \right) = ar\left( {\Delta OCD} \right)    …..(ii)
In {\Delta ACD}, DO is the median
ar\left( {\Delta OCD} \right) = ar\left( {\Delta AOD} \right)   ….(iii)
From (i), (ii) and (iii), we get
ar\left( {\Delta OAB} \right) = ar\left( {\Delta OBC} \right) = ar\left( {\Delta OCD} \right) = ar\left( {\Delta AOD} \right)

(13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.
Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that
ar\left( {\Delta OAB} \right) = ar\left( {\Delta OBC} \right) = ar\left( {\Delta OCD} \right) = ar\left( {\Delta AOD} \right)
To prove: Quadrilateral ABCD is a parallelogram.Proof: We have,
ar\left( {\Delta AOD} \right) = ar\left( {\Delta BOC} \right)
ar\left( {\Delta AOD} \right) + ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) + ar\left( {\Delta AOB} \right)
ar\left( {\Delta ABD} \right) = ar\left( {\Delta ABC} \right)
Thus, \Delta s ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.
Altitude from {\Delta ABD} Altitude from C of {\Delta ABC}
DC\parallel AB
Similarly, we have, AD\parallel BC.
Hence, quadrilateral ABCD is a parallelogram.

(14) Prove that a median of a triangle divides it into two triangles of equal area.
Given: A {\Delta ABC} in which AD is the median.
To prove: ar\left( {\Delta ABD} \right) = ar\left( {\Delta ADC} \right)
Construction: Draw AL \bot BC.Proof: Since AD is the median of {\Delta ABC}. Therefore, D is the mid point of BC.
BD = DC
BD \times AL = DC \times AL  [ Multiplying both sides by AL]
\frac{1}{2}\left( {BD \times AL} \right) = \frac{1}{2}\left( {DC \times AL} \right)
ar\left( {\Delta ABD} \right) = ar\left( {\Delta ADC} \right)
ALITER Since \Delta s ABD and ADC have equal bases and the same altitude AL.
ar\left( {\Delta ABD} \right) = ar\left( {\Delta ADC} \right)



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