myhelper: RS AGGARWAL CLASS 9 CHAPTER 11 AREAS OF PARALLELOGRAMS AND TRIANGLES EXERCISE 11

EXERCISE 11

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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.

Answer 1:

(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.

Answer 2:

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) = $\frac{1}{2}$ $\frac{1}{2}$ ar(quad. ABCD) ...(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(∆CDA)
= 2 ar(∆ ABC) [∵ ar(∆ABC) = ar(∆CDA)]
∴ ar(∆ABC) = $\frac{1}{2}$ $\frac{1}{2}$ ar(quad. ABCD) ...(ii)
From (i) and (ii), we have:
ar(∆ABD) = ar(∆ABC) = $\frac{1}{2}$ $\frac{1}{2}$ AB ⨯ BD = $\frac{1}{2}$ $\frac{1}{2}$ AB ⨯ CL
⇒ CL = BD
⇒ DC || AB
Similarly, AD || BC.
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.

Answer 3:

ar(parallelogram ABCD) = base ⨯ height
⇒ AB ⨯DL = AD ⨯ BM
⇒ 10 ⨯ 6 = AD ⨯ BM
⇒ AD ⨯ 8 = 60 cm²
⇒ AD = 7.5 cm
∴ AD = 7.5 cm

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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.

Answer 4:

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]
$PQ = \frac{1}{2} \times 16 = 8 cm$
Again, in ∆DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$ AC [By midpoint theorem]
$SR = \frac{1}{2} \times 12 = 6 cm$
$Area of PQRS = length \times breadth = 6 \times 8 = 48 {cm}^{2}$

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.

Answer 5:

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).

Answer 6:

(i) In $△$ BCD,
${DB}^{2} + {BC}^{2} = {DC}^{2} \Rightarrow {DB}^{2} = 17^{2} - 8^{2} = 225 \Rightarrow DB = 15 cm$
Ar( $△$ BCD) = $\frac{1}{2} \times b \times h = \frac{1}{2} \times 8 \times 15 = 60 {cm}^{2}$
In $△$ BAD,
${DA}^{2} + {AB}^{2} = {DB}^{2} \Rightarrow {AB}^{2} = 15^{2} - 9^{2} = 144 \Rightarrow AB = 12 cm$
Ar( $△$ DAB) = $\frac{1}{2} \times b \times h = \frac{1}{2} \times 9 \times 12 = 54 {cm}^{2}$

Area of quad. ABCD = Ar( $△$ DAB) + Ar( $△$ BCD) = 54 + 60 = 114 cm².

(ii) Area of trap(PQRS) = $\frac{1}{2} (8 + 16) \times 8 = 96 {cm}^{2}$

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.

Answer 7:

∆ADL is a right angle triangle.
So, DL = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
Similarly, in ∆BMC, we have:
MC = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
∴ 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 $ar (quad . A B C D) = \frac{1}{2} \times B D \times (A L + C M)$ .

Answer 8:

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
$\Rightarrow$ 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).

Answer 9:

Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
$ar (△ ABC) = ar (△ ACD) = \frac{1}{2} ar (ABCD)$
M is the midpoint of AB. So, CM is the median.
CM divides $△$ ABC in two triangles with equal area.
$ar (△ AMC) = ar (△ BMC) = \frac{1}{2} ar (△ ABC)$
ar(AMCD) = ar( $△$ ACD) + ar( $△$ AMC) = ar( $△$ ABC) + ar( $△$ AMC) = ar( $△$ ABC) + $\frac{1}{2}$ ar( $△$ ABC)
$\Rightarrow 24 = \frac{3}{2} ar (△ ABC) \Rightarrow ar (△ ABC) = 16 {cm}^{2}$

Question 10:

In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 8 cm and CM = 6 cm, find the area of quad. ABCD.

Answer 10:

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).

Answer 11:

We know
ar(∆APB) = $\frac{1}{2} ar (ABCD)$ .....(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) = $\frac{1}{2} ar (ABCD)$ .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved

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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) = $\frac{1}{2}$ $\frac{1}{2}$ ar(MNPQ).

Answer 12:

(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) = $\frac{1}{2}$ $\frac{1}{2}$ ar(ABPQ) (Same base AQ and AQ || BP) .....(2)
From (1) and (2)
ar(∆ATQ) = $\frac{1}{2}$ $\frac{1}{2}$ 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).

Answer 13:

∆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),

Answer 14:

∆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(∆DEC) + ar(∆ADE) = ar(∆DEB) + ar(∆ADE)
⇒ 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.

Answer 15:

Let AD is a median of ∆ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ∆ADC.
To prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with $\frac{1}{2}$ $\frac{1}{2}$ AL on both sides, we get:
$\frac{1}{2}$ $\frac{1}{2}$ × BD × AL = $\frac{1}{2}$ $\frac{1}{2}$ × DC × AL
⇒ ar(∆ABD) = ar(∆ADC)

Question 16:

Show that a diagonal divides a parallelogram into two triangles of equal area.

Answer 16:

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)

Answer 17:

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
$\Rightarrow$ $\Rightarrow$ 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.

Answer 18:

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).

Answer 19:

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,
$ar (△ DPO) = ar (△ BPO) . . . . . (1) ar (△ ADO) = ar (△ ABO) . . . . . (2) Case 1 : (2) - (1) \Rightarrow ar (△ ADO) - ar (△ DPO) = ar (△ ABO) - ar (△ BPO)$ $ar (△ DPO) = ar (△ BPO) . . . . . (1) ar (△ ADO) = ar (△ ABO) . . . . . (2) Case 1 : (2) - (1) \Rightarrow ar (△ ADO) - ar (△ DPO) = ar (△ ABO) - ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

Case II:

$ar (△ ADO) + ar (△ DPO) = ar (△ ABO) + ar (△ BPO)$ $ar (△ ADO) + ar (△ DPO) = ar (△ ABO) + ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

Question 20:

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
If BO = OD, prove that
ar(∆ABC) = ar(∆ADC),

Answer 20:

Given: BO = OD
To prove: ar(∆ABC) = ar(∆ADC)
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)
∴ ar(∆ADC ) = ar(∆ABC)

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 $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$ $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$ .

Answer 21:

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$ $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$
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 ) = $\frac{1}{2}$ $\frac{1}{2}$ ar(∆ABD)                 ...(i)
Also, ar(∆CDE ) = $\frac{1}{2}$ $\frac{1}{2}$ ar(∆ADC)             ...(ii)

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

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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) = $\frac{1}{8}$ $\frac{1}{8}$ ar(∆ABC).

Answer 22:

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

Answer 23:

In $△$ $△$ MQC and $△$ $△$ MPB,
MC = MB (M is the midpoint of BC)
$∠$ $∠$ CMQ = $∠$ $∠$ BMP (Vertically opposite angles)
$∠$ $∠$ MCQ = $∠$ $∠$ MBP (Alternate interior angles on the parallel lines AB and DQ)
Thus, $△$ $△$ MQC $≅$ $≅$ $△$ $△$ MPB (ASA congruency)
$\Rightarrow$ $\Rightarrow$ ar( $△$ $△$ MQC) = ar( $△$ $△$ MPB)
$\Rightarrow$ $\Rightarrow$ ar( $△$ $△$ MQC) + ar(APMCD) = ar( $△$ $△$ MPB) + ar(APMCD)
$\Rightarrow$ $\Rightarrow$ 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).

Answer 24:

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.

Answer 25:

Given: ∆ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)
To prove: BC bisects AD
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 = 90^o
∠AOL = ∠DOM (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
∴ OA = OD
Hence, BC bisects AD.

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.

Answer 26:

In $△$ $△$ MDA and $△$ $△$ MCP,
$∠$ $∠$ DMA = $∠$ $∠$ CMP (Vertically opposite angles)
$∠$ $∠$ MDA = $∠$ $∠$ MCP (Alternate interior angles)
AD = CP (Since AD = CB and CB = CP)
So, $△$ $△$ MDA $≅$ $≅$ $△$ $△$ MCP (ASA congruency)
$\Rightarrow$ $\Rightarrow$ DM = MC (CPCT)
$\Rightarrow$ $\Rightarrow$ M is the midpoint of DC
$\Rightarrow$ $\Rightarrow$ BM is the median of $△$ $△$ BDC
$\Rightarrow$ $\Rightarrow$ ar( $△$ $△$ BMC) = ar( $△$ $△$ DMB) = 7 cm²
ar( $△$ $△$ BMC) + ar( $△$ $△$ DMB) = ar( $△$ $△$ DBC) = 7 + 7 = 14 ${cm}^{2}$ ${cm}^{2}$
Area of parallelogram ABCD = 2 $\times$ $\times$ ar( $△$ $△$ DBC) = 2 $\times$ $\times$ 14 = 28 ${cm}^{2}$ ${cm}^{2}$

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
ar(∆ADM) = ar(ABMC)

Answer 27:

Join BM and AC.
ar(∆ADC) = $\frac{1}{2} b h$ $\frac{1}{2} b h$ = $\frac{1}{2} \times DC \times h$ $\frac{1}{2} \times DC \times h$
ar(∆ABM) = $\frac{1}{2} \times AB \times h$ $\frac{1}{2} \times AB \times h$
AB = DC (Since ABCD is a parallelogram)
So, ar(∆ADC) = ar(∆ABM)
$\Rightarrow$ $\Rightarrow$ ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
$\Rightarrow$ $\Rightarrow$ 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
$ar (| | gm P Q R S) = \frac{1}{2} \times ar (∣ ∣ ∣ ∣ gm A B C D)$ $ar (| | gm P Q R S) = \frac{1}{2} \times ar (| | gm A B C D)$ .

Answer 28:

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 ) = $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof:
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ $\frac{1}{2}$ × 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 = $\frac{1}{2}$ $\frac{1}{2}$ × AC [ By midpoint theorem]
Now, PQ || AC and SR || AC
⇒ PQ || SR
Also, PQ = SR = $\frac{1}{2}$ $\frac{1}{2}$ × 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 ) = $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram ABQS)                         ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram SQCD)                        ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
  ar(parallelogram PQRS) = $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram ABQS) + $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram PQRS) = $\frac{1}{2}$ $\frac{1}{2}$ × ar(parallelogram ABCD)

Question 29:

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

Answer 29:

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

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Question 30:

The base BC of ∆ABC is divided at D such that $B D = \frac{1}{2} D C$ $B D = \frac{1}{2} D C$ . Prove that $ar (∆ A B D) = \frac{1}{3} \times ar (∆ A B C)$ $ar (∆ A B D) = \frac{1}{3} \times ar (∆ A B C)$ .

Answer 30:

Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$ $\frac{1}{2}$ DC
To prove: ar(∆ABD) = $\frac{1}{3}$ $\frac{1}{3}$ ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof:
In ∆ABC, we have:
BC = BD + DC
⇒ BD + 2 BD = 3 × BD
Now, we have:
ar(∆ABD) = $\frac{1}{2}$ $\frac{1}{2}$ × BD × AL
ar(∆ABC) = $\frac{1}{2}$ $\frac{1}{2}$ × BC × AL
⇒ ar(∆ABC) = $\frac{1}{2}$ $\frac{1}{2}$ × 3BD × AL = 3 × $(\frac{1}{2} \times B D \times A L)$ $(\frac{1}{2} \times B D \times A L)$
⇒ ar(∆ABC) = 3 × ar(∆ABD)
∴ ar(∆ABD) = $\frac{1}{3}$ $\frac{1}{3}$ ar(∆ABC)

Question 31:

In the adjoining figure, BD || CA, E is the midpoint of CA and BD = $\frac{1}{2}$ $\frac{1}{2}$ CA. Prove that ar(∆ABC) = 2ar(∆DBC).

Answer 31:

E is the midpoint of CA.
So, AE = EC .....(1)
Also, BD = $\frac{1}{2}$ $\frac{1}{2}$ 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) = $\frac{1}{2} ar (△ ABC)$ $\frac{1}{2} ar (△ ABC)$ .....(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).

Answer 32:

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.
∴ ar(∆ADE) = ar(∆ADG)                       ...(iv)

From (iii) and (iv), we have:
ar(∆DBC) + ar(∆ADE) = ar(∆DBF) + ar(∆ADG)
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).

Answer 33:

$ar (△ CFA) = ar (△ CFB)$ $ar (△ CFA) = ar (△ CFB)$ (Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
$\Rightarrow ar (△ CFA) - ar (△ CFG) = ar (△ CFB) - ar (△ CFG) \Rightarrow ar (△ AFG) = ar (△ CBG)$ $\Rightarrow ar (△ CFA) - ar (△ CFG) = ar (△ CFB) - ar (△ CFG) \Rightarrow ar (△ AFG) = ar (△ CBG)$
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.

Answer 34:

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) = $\frac{1}{2}$ $\frac{1}{2}$ × BD × AL ...(i)
ar(∆ADC) = $\frac{1}{2}$ $\frac{1}{2}$ × DC × AL ...(ii)
Dividing (i) by (ii), we get:
$\frac{ar (△ A B D)}{ar (∆ A D C} = \frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L} = \frac{B D}{D C} = \frac{m}{n}$ $\frac{ar (△ A B D)}{ar (∆ A D C} = \frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L} = \frac{B D}{D C} = \frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = m : n

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).

Answer 35:

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 = $\frac{1}{2} (AB + DC) = (\frac{a + b}{2})$ $\frac{1}{2} (AB + DC) = (\frac{a + b}{2})$
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.
$ar (DCNM) = \frac{1}{2} \times DP (DC + MN) = \frac{1}{2} h (b + \frac{a + b}{2}) = \frac{h}{4} (a + 3 b) ar (MNBA) = \frac{1}{2} \times PQ (AB + MN) = \frac{1}{2} h (a + \frac{a + b}{2}) = \frac{h}{4} (b + 3 a)$ $ar (DCNM) = \frac{1}{2} \times DP (DC + MN) = \frac{1}{2} h (b + \frac{a + b}{2}) = \frac{h}{4} (a + 3 b) ar (MNBA) = \frac{1}{2} \times PQ (AB + MN) = \frac{1}{2} h (a + \frac{a + b}{2}) = \frac{h}{4} (b + 3 a)$
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) = $\frac{9}{11}$ $\frac{9}{11}$ ar(EFCD).

Answer 36:

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 = $\frac{1}{2} (AB + DC) = (\frac{a + b}{2})$ $\frac{1}{2} (AB + DC) = (\frac{a + b}{2})$
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.
$ar (ABFE) = \frac{1}{2} \times AP (AB + EF) = \frac{1}{2} h (b + \frac{a + b}{2}) = \frac{h}{4} (a + 3 b) ar (EFCD) = \frac{1}{2} \times PQ (CD + EF) = \frac{1}{2} h (a + \frac{a + b}{2}) = \frac{h}{4} (b + 3 a)$ $ar (ABFE) = \frac{1}{2} \times AP (AB + EF) = \frac{1}{2} h (b + \frac{a + b}{2}) = \frac{h}{4} (a + 3 b) ar (EFCD) = \frac{1}{2} \times PQ (CD + EF) = \frac{1}{2} h (a + \frac{a + b}{2}) = \frac{h}{4} (b + 3 a)$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{ar (ABEF)}{ar (EFCD)} = \frac{24 + 3 \times 16}{16 + 3 \times 24} = \frac{9}{11}$ $\frac{ar (ABEF)}{ar (EFCD)} = \frac{24 + 3 \times 16}{16 + 3 \times 24} = \frac{9}{11}$

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).

Answer 37:

In $△$ $△$ PAC,
PA || DE and E is the midpoint of AC
So, D is the midpoint of PC by converse of midpoint theorem.
Also, $DE = \frac{1}{2} PA$ $DE = \frac{1}{2} PA$ .....(1)
Similarly, $DE = \frac{1}{2} AQ$ $DE = \frac{1}{2} AQ$ .....(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)

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Question 38:

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB).

Answer 38:

In ∆RSC and ∆PQB
$∠$ $∠$ CRS = $∠$ $∠$ BPQ (CD || AB) so, corresponding angles are equal)
$∠$ $∠$ CSR = $∠$ $∠$ BQP ( SC || QB so, corresponding angles are equal)
SC = QB (BQSC is a parallelogram)
So, ∆RSC $≅$ $≅$ ∆PQB (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)

RS AGGARWAL CLASS 9 CHAPTER 11 AREAS OF PARALLELOGRAMS AND TRIANGLES EXERCISE 11