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#### Question 1:

Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.      (i) No, it doesnt lie on the same base and between the same parallels.
(ii) No, it doesnt lie on the same base and between the same parallels.
(iii) Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
(iv) No, it doesnt lie on the same base and between the same parallels.
(v) Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
(vi) Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.

#### Question 2:

In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.  Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB   (extend DC and AB). Join AC, the other diagonal of ABCD.

Proof: ar(quad. ABCD) = ar(∆ABD) + ar(​∆DCB)
= 2 ar(​∆ABD)                    [∵ ar​(∆ABD) = ar(​∆DCB)]
∴ ar(​∆ABD) = $\frac{1}{2}$ar(quad. ABCD)                 ...(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(​∆CDA)
= 2 ar(​∆ ABC)                    [∵ ar​(∆ABC) = ar(​∆CDA)]
∴ ar(​∆ABC) = $\frac{1}{2}$ar(quad. ABCD)                ...(ii)
From (i) and (ii), we have:
ar(​∆ABD) = ar(​∆ABC) = $\frac{1}{2}$ AB ⨯ BD = $\frac{1}{2}$ AB ⨯ CL
⇒ CL = BD
⇒ DC |​​| AB
Hence, ABCD is a paralleogram.
∴ ar(​|​| gm ABCD) = base ​⨯ height = 5 ​⨯ 7 = 35 cm2

#### Question 3:

In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm, respectively. Find AD. ar(parallelogram ABCD) = base ​⨯ height
AB ​⨯DL = AD ​⨯ BM
⇒ 10 ​​⨯ 6 = AD ​⨯ BM
⇒ AD ​⨯ 8 = 60 cm2

#### Question 4:

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm. Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]

Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]

#### Question 5:

Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.

ar(trapezium) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
= $\frac{1}{2}$ ⨯ (9 + 6) ⨯ 8
= 60 cm2
Hence, the area of the trapezium is 60 cm2.

#### Question 6:

(i) Calculate the area of quad. ABCD, given in Fig. (i).
(ii) Calculate the area of trap. PQRS, given in Fig. (ii).  (i) In $△$BCD,

Ar($△$BCD) =
In $△$BAD,

Ar($△$DAB) =

Area of quad. ABCD = Ar($△$DAB) + Ar($△$BCD) = 54 + 60 = 114 cm.

(ii) Area of trap(PQRS) =

#### Question 7:

In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD. ADL is a right angle triangle.
So, DL =
Similarly, in ∆BMC, we have:
MC
DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = $\frac{1}{2}$⨯ (sum of parallel sides) ⨯ (distance between them)
=$\frac{1}{2}$ ⨯ (7 + 13) ⨯ 4
= 40 cm2
​Hence, DC = 13 cm and area of trapezium = 40 cm2

#### Question 8:

BD is one of the diagonals of a quad. ABCD. If AL BD and CM ⊥ BD, show that . ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)
ar(∆ABD) = $\frac{1}{2}$⨯ base ⨯ height = $\frac{1}{2}$BD ⨯ AL             ...(i)
ar(∆DBC) = $\frac{1}{2}$BD ⨯ CL                  ...(ii)
From (i) and (ii), we get:
​ar(quad ABCD) = $\frac{1}{2}$ ⨯BD ⨯​ AL + $\frac{1}{2}$ ⨯ BD ⨯​ CL
$⇒$​ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ ​(AL + CL)
Hence, proved.

#### Question 9:

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm2, find ar(∆ABC). Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$
M is the midpoint of AB. So, CM is the median.
CM divides $△$ABC in two triangles with equal area.
$\mathrm{ar}\left(△\mathrm{AMC}\right)=\mathrm{ar}\left(△\mathrm{BMC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
ar(AMCD) = ar($△$ACD) + ar($△$AMC) = ar($△$ABC) + ar($△$AMC) = ​ar($△$ABC) + $\frac{1}{2}$​ar($△$ABC)

#### Question 10:

In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If AL BD and CMBD such that AL = 8 cm and CM = 6 cm, find the area of quad. ABCD. ​ar(quad ABCD) = ar($△$ABD) + ar($△$BDC)
= $\frac{1}{2}$ ⨯BD ⨯​ AL  +$\frac{1}{2}$ ⨯BD ⨯​ CM
$\frac{1}{2}$ ⨯BD ⨯​ ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = ​$\frac{1}{2}$ ⨯ 14 ⨯ ( 8 + 6)
= 7 ​⨯14
= 98 cm2

#### Question 11:

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(∆APB) = ar(∆BQC). We know
ar(∆APB) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(1)                     [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 half the area of the parallelogram]
Similarly,
ar(∆BQC) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved

#### Question 12:

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
(i) ar(MNPQ) = are(ABPQ)
(ii) ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ). (i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = are(ABPQ)                   (Same base PQ and MB || PQ)                      .....(1)

(ii) If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram.
So, ar(∆ATQ) = $\frac{1}{2}$ar(ABPQ)                 (Same base AQ and AQ || BP)                       .....(2)
From (1) and (2)
ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ)

#### Question 13:

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar(AOD) = ar(BOC). CDA and ​∆CBD lies on the same base and between the same parallel lines.
So, ar(​∆CDA) = ar(CDB)            ...(i)
Subtracting ar(​∆OCD) from both sides of equation (i), we get:
ar(​∆CDA) $-$ ar(​∆OCD) = ar(​​∆CDB) $-$ ar (​​∆OCD)
⇒ ar(​​∆AOD) = ar(​​∆BOC)

#### Question 14:

In the adjoining figure, DE || BC. Prove that
(i) ar(ACD) = ar(ABE),
(ii) ar(OCE) = ar(OBD), DEC and ​∆DEB lies on the same base and between the same parallel lines.
So, ar(​∆DEC) = ar(∆DEB)                      ...(1)

(i) On adding​ ar(∆ADE)​ in both sides of equation (1), we get:
⇒ ar(​​∆ACD) = ar(​​∆ABE

(ii) On subtracting​ ar(ODE)​ from both sides of equation (1), we get:​
ar(​∆DEC) $-$ ar(∆ODE)​ = ar(∆DEB) $-$ ar(∆ODE)​ ​      ​
⇒ ar(​​∆OCE) = ar(​∆OBD)

#### Question 15:

Prove that a median divides a triangle into two triangles of equal area. Let AD is a median of
ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ADC.
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with $\frac{1}{2}$AL on both sides, we get:
$\frac{1}{2}$ × BD × AL$\frac{1}{2}$ × DC × AL

#### Question 16:

Show that a diagonal divides a parallelogram into two triangles of equal area. Let ABCD be a parallelogram and BD be its diagonal.

To prove: ar(∆ABD) = ar(∆CDB)

Proof:
In ∆ABD and ∆CDB, we have:
AB = CD                    [Opposite sides of a parallelogram]
AD = CB                   [Opposite sides of a parallelogram]​

BD  = DB                  [Common]
i.e., ∆ABD  CDB           [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)

#### Question 17:

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD) Line segment CD is bisected by AB at O                   (Given)
CO = OD                                .....(1)
In ΔCAO,
AO is the median.                (From (1))
So, arΔCAO = arΔDAO          .....(2)
Similarly,
In ΔCBD,
BO is the median                 (From (1))
So, arΔCBO = arΔDBO          .....(3)
From (2) and (3) we have
arΔCAO + arΔCBO = arΔDBO + arΔDAO
$⇒$ar(ΔABC) = ar(ΔABD)

#### Question 18:

D and E are points on sides AB and AC respectively of ∆ABC such that ar(∆BCD) = ar(∆BCE). Prove that DE || BC. ar(∆BCD) = ar(∆BCE)                     (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC.

#### Question 19:

P is any point on the diagonal AC of a parallelogram ABCD. Prove that ar(∆ADP) = ar(∆ABP). Join BD.
Let BD and AC intersect at point O.
O is thus the midpoint of DB and AC.
PO is the median of $△$DPB,
So,

Case II: $\mathrm{ar}\left(△\mathrm{ADO}\right)+\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)+\mathrm{ar}\left(△\mathrm{BPO}\right)\phantom{\rule{0ex}{0ex}}$

#### Question 20:

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
If BO = OD, prove that Given:  BO = OD
Proof
Since BO = OD, O is the mid point of BD.
We know that a median of a triangle divides it into two triangles of equal areas.
CO is a median of ∆BCD.
i.e., ar(∆COD) = ar (∆COB)            ...(i)

AO is a median of ∆ABD.
i.e., ar(∆AOD) = ar(∆AOB)              ...(ii)

From (i) and (ii), we have:
ar(∆COD) + ar(∆AOD) ar(∆COB) + ar(∆AOB)

#### Question 21:

The vertex A of ABC is joined to a point D on the side BC. The midpoint of AD is E.
Prove that $\mathrm{ar}\left(∆BEC\right)=\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$. Given:  D is the midpoint of BC and E is the midpoint of AD.
To prove: $\mathrm{ar}\left(∆BEC\right)=\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = $\frac{1}{2}$ar(∆ABD)                 ...(i)
Also, ar(∆CDE ) =$\frac{1}{2}$ ar(∆ADC)             ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ $\frac{1}{2}$ ⨯​ ar(∆ABD)​ + $\frac{1}{2}$ ⨯​ ar(∆ADC)
⇒ ar(∆BEC )​ = $\frac{1}{2}$⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ​ar(∆BEC )​ =​ $\frac{1}{2}$ ⨯​ ar(∆ABC)

#### Question 22:

D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE, prove that ar(∆BOE) = $\frac{1}{8}$ar(∆ABC). D is the midpoint of side BC of ∆ABC.
$⇒$AD is the median of ∆ABC.
$⇒$$\mathrm{ar}\left(△\mathrm{ABD}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
E is the midpoint of BD of ∆ABD,
$⇒$AE is the median of ∆ABD
$⇒$$\mathrm{ar}\left(△\mathrm{ABE}\right)=\mathrm{ar}\left(△\mathrm{AED}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Also, O is the midpoint of AE,
$⇒$BO is the median of ∆ABE,
$⇒$$\mathrm{ar}\left(△\mathrm{ABO}\right)=\mathrm{ar}\left(△\mathrm{BOE}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABE}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{8}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Thus, ar(∆BOE) = $\frac{1}{8}$ar(∆ABC)

#### Question 23:

In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD). In $△$MQC and $△$MPB,
MC = MB                            (M is the midpoint of BC)
$\angle$CMQ = $\angle$BMP                (Vertically opposite angles)
$\angle$MCQ = $\angle$MBP                (Alternate interior angles on the parallel lines AB and DQ)
Thus, $△$MQC $\cong$ $△$MPB   (ASA congruency)
$⇒$ar($△$MQC) = ar($△$MPB)
$⇒$ar($△$MQC) + ar(APMCD) = ar($△$MPB) + ar(APMCD)
$⇒$ar(APQD) = ar(ABCD)

#### Question 24:

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(ABP) = ar(quad. ABCD). We have:
​ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar(∆ACP)​​ + ar(∆ABC)

ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ ACP)​
By adding ar(∆ABC) on both sides, we get:
ar(∆ACD) ar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)
⇒​ ar (quad. ABCD) = ar(∆ABP)
Hence, proved.

#### Question 25:

In the adjoining figure, ABC and ∆DBC are on the same base BC with A and D on opposite sides of BC such that ar(∆ABC) = ar(∆DBC). Show that BC bisects AD.  GivenABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMO, we have:
AL = DM
ALO = ∠DMO =  90o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
​​
∴​ OA = OD

#### Question 26:

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersects CD at M. If ar(DMB) = 7 cm2, find the area of parallelogram ABCD. In $△$MDA and $△$MCP,
$\angle$DMA = $\angle$CMP                  (Vertically opposite angles)
$\angle$MDA = $\angle$MCP                  (Alternate interior angles)
So, $△$MDA $\cong$ $△$MCP         (ASA congruency)
$⇒$DM = MC                         (CPCT)
$⇒$M is the midpoint of DC
$⇒$BM is the median of $△$BDC
$⇒$ar($△$BMC) = ar($△$DMB) = 7 cm2
ar($△$BMC) + ar($△$DMB) = ar($△$DBC) = 7 + 7 = 14 ${\mathrm{cm}}^{2}$
Area of parallelogram ABCD = $×$ ar($△$DBC) = 2 $×$ 14 = 28 ${\mathrm{cm}}^{2}$

#### Question 27:

In a parallelogram ABCD, any point E is taken on the side BC. AE and DC when produced meet at a point M. Prove that Join BM and AC.
ar(∆ADC) = $\frac{1}{2}bh$ = $\frac{1}{2}×\mathrm{DC}×h$
ar(∆ABM) = $\frac{1}{2}×\mathrm{AB}×h$
AB = DC                               (Since ABCD is a parallelogram)
$⇒$ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
$⇒$ar(∆ADM) = ar(ABMC)
Hence Proved

#### Question 28:

P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of || gm ABCD. Show that PQRS is a parallelogram and also show that
. Given:  ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof:
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC              [ By midpoint theorem]
Again, in
DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = $\frac{1}{2}$ × AC                                   [ By midpoint theorem]
Now, PQ
|| AC and SR || AC
​PQ || SR
Also, PQ = SR =
$\frac{1}{2}$ × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram​ PQRS) = ar(∆PSQ) + ar(∆SRQ)                       ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD)            ...(ii)

PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) =$\frac{1}{2}$ × ar(parallelogram ABQS)                         ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ × ar(parallelogram SQCD)                        ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABQS)​ + $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABCD)

#### Question 29:

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(∆BCG) = ar(AFGE).

Figure
CF is median of $△$ABC.
$⇒$ar($△$BCF) = $\frac{1}{2}$($△$ABC)                     .....(1)
Similarly, BE is the median of $△$ABC,
$⇒$ar($△$ABE) = $\frac{1}{2}$($△$ABC)                     .....(2)
From (1) and (2) we have
ar($△$BCF) = ar($△$ABE)
$⇒$ar($△$BCF) $-$ ar($△$BFG) = ar($△$ABE) $-$ ar($△$BFG)
$⇒$ar(∆BCG) = ar(AFGE)

#### Question 30:

The base BC of ABC is divided  at D such that $BD=\frac{1}{2}DC$. Prove that $\mathrm{ar}\left(∆ABD\right)=\frac{1}{3}×\mathrm{ar}\left(∆ABC\right)$. Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$DC
To prove:  ar(∆ABD) = $\frac{1}{3}$ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof:
In ∆ABC, we have:
BC = BD + DC

⇒​ BD​ + 2 BD = 3 × BD
Now, we have:
ar(∆ABD)​ = $\frac{1}{2}$​ ×​ BD ×​ AL
ar(∆ABC)​ = $\frac{1}{2}$​ ×​ BC ×​ AL
⇒  ar(∆ABC) = $\frac{1}{2}$ ×​ 3BD ×​ AL = 3 ×​ $\left(\frac{1}{2}×BD×AL\right)$
⇒ ar(∆ABC)​ = 3 × ar(∆ABD)
∴ ​ar(∆ABD) = ​$\frac{1}{3}$​ar(∆ABC)

#### Question 31:

In the adjoining figure, BD || CA, E is the midpoint of CA and BD = $\frac{1}{2}$ CA. Prove that ar(∆ABC) = 2ar(∆DBC). E is the midpoint of CA.
So, AE = EC                            .....(1)
Also, BD = $\frac{1}{2}$ CA                    (Given)
So, BD = AE                            .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of $△$ABC
so, ar($△$BCE) = ar($△$ABE) = $\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$                  .....(1)
ar($△$DBC) = ar($△$BCE)                  .....(2)                  (Triangles on the same base and between the same parallels are equal in area)
From (1) and (2)
ar(∆ABC) = 2ar(∆DBC)

#### Question 32:

The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar(pentagon ABCDE) = ar(DGF). Given:  ABCDE is a pentagon.  EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) =  ar( DGF)
Proof:
ar(pentagon ABCDE )​ = ar(∆DBC) + ar(∆ADE ) + ar(∆ABD)               ...(i)
Also, ar(DGF) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD )                ...(ii)

Now, ∆DBC and ∆DBF lie on the same base and between the same parallel lines.
∴ ar(∆DBC) = ar(∆DBF)                         ...(iii)
Similarly, ∆ADE and ∆ADG lie on same base and between the same parallel lines.

From (iii) and (iv), we have:
Adding ar(∆ABD) on both sides, we get:
ar(∆DBC) + ar(∆ADE) + ar(∆ABD) = ar (∆DBF) + ar(∆ADG) + ar(∆ABD
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) =  ar(DGF)

#### Question 33:

In the adjoining figure, CE || AD and CF || BA. Prove that ar(∆CBG) = ar(∆AFG). $\mathrm{ar}\left(△\mathrm{CFA}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)$                            (Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
$⇒\mathrm{ar}\left(△\mathrm{CFA}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(△\mathrm{AFG}\right)=\mathrm{ar}\left(△\mathrm{CBG}\right)$
Hence Proved

#### Question 34:

In the adjoining figure, the point D divides the side BC of ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n. Given: D is a point on BC of ∆ ABC, such that BD : DC =  m : n
To prove:  ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(∆ABD)​ = $\frac{1}{2}$ ×​ BD ×​ AL                     ...(i)
ar(∆ADC)​ = $\frac{1}{2}$​ ×​ DC ×​ AL                   ...(ii)
Dividing (i) by (ii), we get:
$\frac{\mathrm{ar}\left(△ABD\right)}{\mathrm{ar}\left(∆ADC}=\frac{\frac{1}{2}×BD×AL}{\frac{1}{2}×DC×AL}\phantom{\rule{0ex}{0ex}}=\frac{BD}{DC}\phantom{\rule{0ex}{0ex}}=\frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = mn

#### Question 35:

In a trapezium ABCD, AB || DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).  Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.
Let length DQ = h
Given, M and N are the midpoints of AD and BC respectively.
So, MN || AB || DC and MN = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
M is the mid point of AD and MP || AB so by converse of mid point theorem,
MP || AQ and P will be the mid point of DQ.
$\mathrm{ar}\left(\mathrm{DCNM}\right)=\frac{1}{2}×\mathrm{DP}\left(\mathrm{DC}+\mathrm{MN}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right)\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\mathrm{MNBA}\right)=\frac{1}{2}×\mathrm{PQ}\left(\mathrm{AB}+\mathrm{MN}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right)\phantom{\rule{0ex}{0ex}}$
ar(DCNM) : ar(MNBA) = (a + 3b) : (3a + b)

#### Question 36:

ABCD is a trapezium in which AB || CD, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of AD and BC, prove that ar(ABFE) = $\frac{9}{11}$ ar(EFCD). Construction: Draw a perpendicular from point D to the opposite side CD, meeting CD at Q and EF at P.
Let length AQ = h
Given, E and F are the midpoints of AD and BC respectively.
So, EF || AB || DC and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
E is the mid point of AD and EP || AB so by converse of mid point theorem,
EP || DQ and P will be the mid point of AQ.
$\mathrm{ar}\left(\mathrm{ABFE}\right)=\frac{1}{2}×\mathrm{AP}\left(\mathrm{AB}+\mathrm{EF}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right)\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\mathrm{EFCD}\right)=\frac{1}{2}×\mathrm{PQ}\left(\mathrm{CD}+\mathrm{EF}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right)\phantom{\rule{0ex}{0ex}}$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{\mathrm{ar}\left(\mathrm{ABEF}\right)}{\mathrm{ar}\left(\mathrm{EFCD}\right)}=\frac{24+3×16}{16+3×24}=\frac{9}{11}$

#### Question 37:

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of ∆ABC. If PQ || BC and CDP and BEQ are straight lines then prove that ar(∆ABQ) = ar(∆ACP). In $△$PAC,
PA || DE and E is the midpoint of AC
So, D is the midpoint of PC by converse of midpoint theorem.
Also, $\mathrm{DE}=\frac{1}{2}\mathrm{PA}$                         .....(1)
Similarly, $\mathrm{DE}=\frac{1}{2}\mathrm{AQ}$                  .....(2)
From (1) and (2) we have
PA = AQ
∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC
ar(∆ABQ) = ar(∆ACP)

#### Question 38:

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB). In ∆RSC and ∆PQB
$\angle$CRS = $\angle$BPQ                       (CD || AB) so, corresponding angles are equal)
$\angle$CSR = $\angle$BQP                        ( SC || QB so, corresponding angles are equal)
SC = QB                                    (BQSC is a parallelogram)
So, ∆RSC $\cong$ ∆PQB                   (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)

#### Question 1:

Out of the following given figures which are on the same base but not between the same parallels?  In this figure, both the triangles are on the same base (QR) but not on the same parallels.

#### Question 2:

In which of the following figures, you find polynomials on the same base and between the same parallels?  In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ​​​Parallelogram ABPQ

#### Question 3:

The median of a triangle divides it into two
(a) triangles of equal areas
(b) congruent triangles
(c) isosceles triangles
(d) right triangles

(a)  triangles of equal areas

#### Question 4:

The area of quadrilateral ABCD in the given figure is
(a) 57 cm2
(b) 108 cm2
(c) 114 cm2
(d) 195 cm2 (c)114 cm2

ar (quad. ABCD) =  ar (∆ ABC)  +  ar (∆ ACD)
In right angle triangle ACD, we have:
AC  =
In right angle triangle ABC, we have:
BC
Now, we have the following:
ar(∆ABC) = $\frac{1}{2}$ × 12 × 9 = 54 cm2
ar(∆ADC) = $\frac{1}{2}$ × 15 × 8 = 60 cm2
ar(quad. ABCD) =​ 54 + 60 = 114 cm2

#### Question 5:

The area of trapezium ABCD in the given figure is
(a) 62 cm2
(b) 93 cm2
(c) 124 cm2
(d) 155 cm2 (c)124 cm2

In the right angle triangle BEC, we have:
EC  =
ar(trapez. ABCD) = cm2

#### Question 6:

In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
(a) 17 cm2
(b) 25 cm2
(c) 34 cm2
(d) 68 cm2 (c) 34 cm2

ar(parallelogram ABCD) = base × height = 5 ​× 6.8 =  34 cm2

#### Question 7:

In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm2, then the ar(OAB) = ?
(a) 26 cm2
(b) 18.5 cm2
(c) 39 cm2
(d) 13 cm2 (d) 13 cm2
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of ∆OAB$\frac{1}{4}$ ⨯ ar(||gm ABCD)
⇒ ar(∆OAB) = ​​$\frac{1}{4}$ ⨯​ 52 = 13 cm2

#### Question 8:

In the given figure, ABCD is a || gm in which DL AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
(a) 40 cm2
(b) 80 cm2

(c) 20 cm2
(d) 196 cm2 (a) 40 cm2
ar(||gm ABCD) = base × height =  10 ​× 4 =  40 cm2

#### Question 9:

The area of ||gm ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL Area of a parallelogram is base into height.
Height = DL = NB
Base = AB = CD
So, area of parallelogram ABCD = DC $×$ DL
Hence, the correct answer is option (c).

#### Question 10:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1

Parallelograms on the same base and between the same parallels are equal in area.
So, the ratio of their areas will be 1 : 1.
Hence, the correct answer is option (b).

#### Question 11:

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Then, ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABCD) is
(a) true
(b) false We know parallelogram on the same base and between the same parallels are equal in area.
Here, AB is the common base and AB || PD
Hence, ar(ABCD) = ar(ABPQ)                             .....(1)
Also, when a triangle and a parallelogram are on the same base and between the same parallels then the
area of triangle is half the area of the parallelogram.
Here, for the ∆BMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABPQ)                      .....(2)
From (1) and (2) we have
ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABCD)
Thus, the given statement is true.
Hence, the correct answer is option (a).

#### Question 12:

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
(a) $\frac{1}{2}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(b) $\frac{1}{3}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(c) $\frac{1}{4}\left(\mathrm{ar}∆\mathrm{ABC}\right)$  D, E and F are the midpoints of sides BC, AC and AB respectively.
On joining FE, we divide $△$ABC into 4 triangles of equal area.
Also, median of a triangle divides it into two triangles with equal area
$\mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{ar}\left(△\mathrm{AFE}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Hence, the correct answer is option (a).

#### Question 13:

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is
(a) 192 cm2
(b) 96 cm2
(c) 64 cm2
(d) 80 cm2

(b) 96 cm2
Area of the rhombus = $\frac{1}{2}$ × product of diagonals = $\frac{1}{2}$ ×​ 12 ​× 16 = 96 cm2

#### Question 14:

Two parallel sides of a trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm. The area of the trapezium is
(a) 74 cm2
(b) 32.5 cm2
(c) 65 cm2
(d) 130 cm2
Figure

(c) 65 cm2
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
$\frac{1}{2}$ ×​ ( 12 + 8) ​× 6.5
= 65 cm2

#### Question 15:

In the given figure ABCD is a trapezium such that AL DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD) = ?
(a) 24 cm2
(b) 40 cm2
(c) 55 cm2
(d) 27.5 cm2 (b) 40 cm2

In right angled triangle MBC, we have:
MC
In right angled triangle ADL, we have:
DL

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
$\frac{1}{2}$ ×​ ( 13 + 7) ​× 4
= 40 cm2

#### Question 16:

In a quadrilateral ABCD, it is given that BD = 16 cm. If AL BD and CMBD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?
(a) 256 cm2
(b) 128 cm2
(c) 64 cm2
(d) 96 cm2 (b)128 cm2

ar(quad ABCD) = ar (∆ ABD) + ​ar (∆ DBC)

We have the following:
ar(∆ABD) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$× 16 ​× 9 = 72 cm2
ar(∆DBC) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$ × 16 ​× 7 = 56 cm2

∴ ar(quad ABCD) =​ 72 + 56 = 128 cm2

#### Question 17:

ABCD is a rhombus in which C = 60°. Then, AC : BD = ?
(a) $\sqrt{3}:1$
(b) $\sqrt{3}:\sqrt{2}$
(c) 3 : 1
(d) 3 : 2 (a)$\sqrt{3}:1$

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC
⇒ ∠BDC = ∠DBC = xo   (Angles opposite to equal sides are equal)
Also, ∠BCD = 60o
xo + xo + 60o = 180o
⇒​ 2xo = 120o
⇒​ xo = 60o
i.e., ∠BCD = BDC = ∠DBC 60o
So, ​∆BCD is an equilateral triangle.
BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:

#### Question 18:

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17 cm2 and ar(||gm ABCD) = 25 cm2. Then, ar(BCF) = ?
(a) 4 cm2
(b) 4.8 cm2
(c) 6 cm2
(d) 8 cm2 (d) 8 cm2

Since ||gm  ABCD and ||gm ABFE are on the same base and between the same parallel lines, we have:
ar(||gm ABFE​) = ar(||gm ABCD​) = 25 cm2
⇒ ar(∆BCF )ar(||gm ABFE​)​ $-$ ar(quad EABC​)​ = ( 25 $-$ 17) = 8 cm2

#### Question 19:

ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC. Then, ar(∆BDE) : ar(∆ABC) = ?
(a) 1 : 2
(b) 1 : 4
(c) $\sqrt{3}:2$
(d) 3 : 4 (b) 1:4

ABC and ∆BDE are two equilateral triangles​ and D is the midpoint of BC.
Let AB = BC = AC =  a
Then BD = BE = ED =  $\frac{a}{2}$
∴
So, required ratio = 1 : 4

#### Question 20:

In a || gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(|| gm ABCD) = 16 cm2, then ar(|| gm APQD) = ?
(a) 8 cm2
(b) 12 cm2
(c) 6 cm2
(d) 9 cm2 (a) 8 cm2

Let the distance between AB and CD be h cm.
Then ar(||gm APQD​) = AP ×​ h
= $\frac{1}{2}$ ×​ AB ×​h               (AP = $\frac{1}{2}$AB )
$\frac{1}{2}$ ×​ ar(||gm ABCD)                [ ar(|| gm ABCD) = AB ×​h )
∴ ar (||gm APQD​) = $\frac{1}{2}$ ×​ 16 = 8 cm2

#### Question 21:

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a
(a) rectangle of area 24 cm2
(b) square of area 24 cm2
(c) trapezium of area 24 cm2
(d) rhombus of area 24 cm2 (d) rhombus of 24 cm2

We know that the figure formed by joining the midpoints of the adjacent sides of a rectangle is a rhombus.
So, PQRS is a rhombus and SQ and PR are its diagonals.
i.e., SQ = 8 cm and PR = 6 cm
∴ ar(rhombus PQRS​) = $\frac{1}{2}$ ×​ product of diagonals = $\frac{1}{2}$ ×​ 8  ×​ 6 = 24 cm2

#### Question 22:

In ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{2}{3}\mathrm{ar}\left(∆ABC\right)$ (c) $\frac{1}{4}$ar (∆ ABC )

Since D is the mid point of BC, AD is a median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(ABD ) =
$\frac{1}{2}$ ar(ABC)                      ...(i)

⇒ ar(BED) =
$\frac{1}{2}$ ar(ABD)                      ...(ii)

From (i) and (ii), we have:
ar(BED)
$\frac{1}{2}$⨯ $\frac{1}{2}$​ ⨯​ ar(∆ABC)
∴​ ar(∆BED)​ =
$\frac{1}{4}$⨯ ar(∆ABC)
ar(ABC)

#### Question 23:

The vertex A of ∆ABC is joined to a point D on BC. If E is the midpoint of AD, then ar(∆BEC) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$ (a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$

Since E is the midpoint of AD, BE is a median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED) $\frac{1}{2}$ar(∆ABD)                ...(i)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆CED ) $\frac{1}{2}$ ar(ADC)               ...(ii)

Adding (i) and (ii), we have:
ar(∆BED ) + ar(∆CED ) = $\frac{1}{2}$ ar(∆ABD) + $\frac{1}{2}$ ar(∆ADC)
⇒ ar (∆ BEC ) = $\frac{1}{2}\left(∆ABD+ADC\right)=\frac{1}{2}∆ABC$

#### Question 24:

In ∆ABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(∆BOE) = ?
(a) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{8}\mathrm{ar}\left(∆ABC\right)$ (d) $\frac{1}{8}$ ar (∆ ABC)

Given: D is the midpoint of BC, E is the midpoint of BD and O is the mid point of AE.
Since D is the midpoint of BC, AD is the median of ∆ABC.
E is the midpoint of BC, so AE is the median of ∆ABD. O is the midpoint of AE, so BO is median of ABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD ) = $\frac{1}{2}$ ⨯ ​ar(∆ABC)                ...(i)
ar(∆ABE ) =$\frac{1}{2}$​ ⨯​ ar(∆ABD)                        ...(ii)
ar(∆BOE) = $\frac{1}{2}$⨯​ ar(∆ABE)                       ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE ) = $\frac{1}{2}$ar(∆ABE)
ar(∆BOE = $\frac{1}{2}$ ⨯​ $\frac{1}{2}$​ ⨯​ $\frac{1}{2}$​ ⨯​ ar(∆ABC)​
∴​​  ar(∆BOE )​ = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)

#### Question 25:

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 3 : 4

Figure

(a) 1:2

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangles is half the area of the parallelogram.
i.e., area of triangle = $\frac{1}{2}$× area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram =$\frac{1}{2}$ : 1 = 1 : 2

#### Question 26:

In the given figure ABCD is a trapezium in which AB || DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b) (c) (3a +b) : (a +3b)

Clearly, EF$\frac{1}{2}$ (a + b)                   [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.

#### Question 27:

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is
(a) a rectangle
(b) a || gm
(c) a rhombus
(d) all of these

(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas.

#### Question 28:

In the given figure, a || gm ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of (c) perimeter of ABCD > perimeter of ABEF

Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
∴ Perimeter of ABCD > perimeter of ABEF

#### Question 29:

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If $AD=2\sqrt{5}$ cm, then area of the rectangle is
(a) 32 cm2
(b) 40 cm2
(c) 44 cm2
(d) 48 cm2 (b) 40 cm2

Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm

#### Question 30:

Which of the following is a false statement?
(a) A median of a triangle divides it into two triangles of equal areas.
(b) The diagonals of a || gm divide it into four triangles of equal areas.
(c) In a ABC, if E is the midpoint of median AD, then $\mathrm{ar}\left(∆BED\right)=\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$. (d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then, ar(∆AOB) = ar(∆COD). (d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(∆AOB) = ar(∆COD).

Consider ∆ADB and ∆ADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(∆ADB)$\ne$ ar(∆ADC)
Subtracting ar(∆AOD) from both sides, we get:
ar(∆ADB) $-$ ar(∆AOD) $\ne$ ar(∆ADC) $-$ ar(∆AOD)
Or ar(∆ AOB)$\ne$ ar(∆ COD)

#### Question 31:

Which of the following is a false statement?
(a) If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm2.
(b) Area of
(c) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(d) If the area of a || gm with one side 24 cm and corresponding height h cm is 192 cm2, then h = 8 cm.

(b)

Area of a parallelogram  = ​base ​× corresponding height

#### Question 32:

Look at the statements given below:
I. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
II. In a || gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.
III. Area of a || gm $=\frac{1}{2}×\mathrm{base}×\mathrm{altitude}.$
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III (c) I and II

Statement I is true, because if a parallelogram and a rectangle lie on the same base and between the same parallel lines, then they have the same altitude and therefore equal areas.
Statement II is also true as area of a parallelogram = base × height
$⇒$AB × DE = AD × BF
$⇒$10 × 6 = 8 × AD
$⇒$ AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.

#### Question 33:

Assertion: In a trapezium ABCD we have AB || DC and the diagonals AC and BD intersect at O. Then, ar(AOD) = ar(BOC) Reason: Triangles on the same base and between the same parallels are equal in areas.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

In trapezium ABCD, ∆ABC and ABD  are on the same base and between the same parallel lines.
∴ ar(∆ABC) = ar(∆ABD)
⇒ ar(∆ABC) $-$ ar(∆AOB) = ar(∆ABD) $-$ ar(∆AOB)
⇒ ​ar(BOC) = ar(AOD
∴ Assertion (A) is true and, clearly, reason (R) gives (A).

#### Question 34:

Assertion: If ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is $\sqrt{3}:1$.
Reason: Median of triangle divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Reason (R) is clearly true.
The explanation of assertion (A)​ is as follows:
ABCD is a rhombus. So, all of its sides are equal.
Now, BC = DC
⇒∠BDC = ∠DBC = x$°$
Also, ∠BCD = 60$°$
x$°$ + x$°$ + 60$°$ = 180$°$
⇒​2x$°$ = 120$°$
⇒​ x$°$ = 60$°$
∴ ∠BCD = ∠BDC = ∠DBC =  60$°$
So, ​∆BCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB = $\frac{a}{2}$
Now, in ∆ OAB, we have:

Thus,
assertion (A)​ is also true, but reason (R) does not give (A).
​Hence, the correct answer is (b).

#### Question 35:

Assertion: The diagonals of a || gm divide it into four triangles of equal area.
Reason: A diagonal of a || gm divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 36:

Assertion: The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm2.
Reason: The area of an equilateral triangle of side 8 cm is .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not a correct explanation of Assertion (A).

Explanation:
Reason (R):
∴ ar(∆ABC ) =
Thus, reason (R) is true.

Assertion (A):
Area of trapezium =
Thus, assertion (A) is true, but reason (R) does not give assertion (A).

#### Question 37:

Assertion: In the given figure, ABCD is a || gm in which DE AB and BFAD. If AB = 16 cm, DE = 8 cm and BF = 10 cm, then AD is 12 cm. Reason: Area of a || gm = base × height.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(d) Assertion is false and Reason is true.

Clearly, reason (R) is true.
Assertion: Area of a parallelogram = base × height
$⇒$AB ×​ DE = AD × BF
$⇒$AD = (16 × 8) ÷ 10 = 12.8 cm

So, the assertion is ​false.

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