# 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:Â  A parallelogram ABCD in which BD is one of the diagonals.
To prove: Â Â $ar(\Delta&space;ABD)=ar(\Delta&space;CDB)$
Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that $ar(\Delta&space;ABD)=ar(\Delta&space;CDB)$ 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, $\angle DAF = \angle CBE$Â Â Â Â Â Â Â Â  [ â¸ª $AD\parallel BC$ and $AF\parallel BE$]
So, by SAS criterion of congruence, we have
$\Delta ADF \cong \Delta BCE$
$ar\left( {\Delta ADF} \right) = ar\left( {\Delta BCE} \right)$Â Â  â€¦..(i)
Now, $ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {sq.ABED} \right) + ar\left( {\Delta BCE} \right)$
$ar\left( {parallelogram{\text{ }}ABCD} \right) = ar\left( {sq.ABED} \right) + ar\left( {\Delta ADF} \right)$ Â  Â [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.
$ar\left( {parallelogram{\text{ }}ABCD} \right)$
$= ar\left( {rect.ALMB} \right)$
$= AB \times AL$Â Â Â  [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 $PM \bot DR$Proof: Since $AB \bot DR$, $AL \bot DR$ and $PM \bot DR$
$AL = PM$
Now, $ar\left( {parallelogram{\text{ }}ABCD} \right) = AB \times AL$
$ar\left( {parallelogram{\text{ }}ABCD} \right) = PQ \times PM$ Â [$AB = PQ$ 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.
$ar\left( {parallelogram{\text{ BCQP}}} \right) = ar\left( {parallelogram{\text{ BCAD}}} \right)$Â Â Â  â€¦.(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
$ar\left( {\Delta PBC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCQP}}} \right)$ Â  â€¦..(ii)
And, $ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCAQ}}} \right)$ Â ....(iii)
Now, $ar\left( {parallelogram{\text{ BCQP}}} \right) = ar\left( {parallelogram{\text{ BCAD}}} \right)$ [ From (i)]
$\frac{1}{2}ar\left( {parallelogram{\text{ BCQP}}} \right) = \frac{1}{2}ar\left( {parallelogram{\text{ BCAD}}} \right)$
$ar\left( {\Delta ABC} \right) = ar\left( {\Delta PBC} \right)$Â Â Â  [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 $BC\parallel AD$
$AL = DM$Â Â Â  â€¦..(i)
Now,
$ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times AL} \right)$
$ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {BC \times DM} \right)$Â Â  [$AL = DM$ (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,
$ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = ar\left( {\Delta ABC} \right) + ar\left( {\Delta ACD} \right)$
$ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}\left( {AB \times CM} \right) + \frac{1}{2}\left( {DC \times AL} \right)$
$ar\left( {{\text{trap}}{\text{. }}ABCD} \right) = \frac{1}{2}ah \times \frac{1}{2}bh$ Â [$AB =a$ 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: $CN = RT$Proof: In ${\Delta ABC}$, CN is the altitude corresponding to side AB.
$ar\left( {\Delta ABC} \right) = \frac{1}{2}\left( {AB \times CN} \right)$Â Â Â  â€¦.(i)
Similarly, we have,
$ar\left( {\Delta PQR} \right) = \frac{1}{2}\left( {PQ \times RT} \right)$Â Â  â€¦..(ii)
Now, $ar\left( {\Delta ABC} \right) = ar\left( {\Delta PQR} \right)$
$\frac{1}{2}\left( {AB \times CN} \right) = \frac{1}{2}\left( {PQ \times RT} \right)$
$\left( {AB \times CN} \right) = \left( {PQ \times RT} \right)$
$\left( {PQ \times CN} \right) = \left( {PQ \times RT} \right)$ Â  Â [ $AB = PQ$ (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,
$ar\left( {\Delta ABC} \right) = ar\left( {\Delta ACD} \right)$Â Â Â  â€¦..(i)
But, $ar\left( {\Delta ABC} \right) + ar\left( {\Delta ACD} \right) = ar\left( {quad.ABCD} \right)$
$2ar\left( {\Delta ABC} \right) = ar\left( {quad.ABCD} \right)$Â Â Â  [Using (i)]
$ar\left( {\Delta ABC} \right) = \frac{1}{2}ar\left( {quad.ABCD} \right)$ â€¦.(ii)
Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.
$ar\left( {\Delta ABD} \right) = ar\left( {\Delta BCD} \right)$Â Â  â€¦.(iii)
But, $ar\left( {\Delta ABD} \right) + ar\left( {\Delta BCD} \right) = ar(quad.ABCD)$
$2ar\left( {\Delta ABD} \right) = ar(quad.ABCD)$Â Â Â  [Using(iii)]

$ar\left( {\Delta ABD} \right) = \frac{1}{2}ar\left( {quad.ABCD} \right)$Â Â  â€¦..(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.
$ar\left( {\Delta OAB} \right) = ar\left( {\Delta OBC} \right)$Â  â€¦.(i)
In ${\Delta BCD}$, CO is the median
$ar\left( {\Delta OBC} \right) = ar\left( {\Delta OCD} \right)$Â Â Â  â€¦..(ii)
In ${\Delta ACD}$, DO is the median
$ar\left( {\Delta OCD} \right) = ar\left( {\Delta AOD} \right)$Â Â  â€¦.(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.
$BD = DC$
$BD \times AL = DC \times AL$Â  [ 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.
$ar\left( {\Delta ABD} \right) = ar\left( {\Delta ADC} \right)$

• Anonymous

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• aur Nani hai