PSEB Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
PSEB 9th Class Maths Solutions Chapter 9 Areas of Parallelograms and Triangles
Ex 9.1
Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
(i)
Answer:
In figure (i), trapezium ABCD and ∆ PDC lie on the same base and between the same parallels.
Here, DC is the common base and DC and AB are two parallels.
(ii)
Answer:
In figure (ii), no two figures lie on the same base and between the same parallels.
(iii)
Answer:
In figure (iii), parallelogram PQRS and ∆ TQR lie on the same base and between the same parallels.
Here, QR is the common base and QR and PS are two parallels.
(iv)
Answer:
In figure (iv), no two figures lie on the same base and between the same parallels.
(v)
Answer:
In figure (v), parallelograms ABCD and APQD as well as trapeziums ABQD and APCD, all the four figures, lie on the same base and between the same parallels.
Here, AD is the common base and AD and BQ are two parallels.
(vi)
Answer:
In figure (vi), no two figure lie on the same base and between the same parallels.
Ex 9.2
Question 1.
In the given figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Answer:
The area of a parallelogram is the product of its base and the altitude corresponding to that base.
Here, in parallelogram ABCD, the altitude corresponding to base DC is AE and the altitude corresponding to base AD is CF.
∴ ar(ABCD) = DC × AE = AD × CF
∴ DC × AE = AD × CF
∴ AB × AE = AD × CF (∵ AB = DC In parallelogram ABCD)
∴ 16 × 8 = AD × 1O
∴ AD = 16×8/10
∴ AD = 128/10
∴ AD = 12.8 cm
Question 2.
If E, F, G and H are respectively the midpoints of the sides of a parallelogram ABCD, show that ar(EFGH) = 1/2 ar (ABCD).
Answer:
In parallelogram ABCD. E, F, G and H are the midpoints of AB, BC. CD and DA respectively.
Draw GE.
Question 3.
p and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Question 4.
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that,
(i) ar (APB) + ar (PCD) = 1/2 ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through P, draw a line parallel to AB.]
Answer:
Through P, draw a line parallel to AB which intersects BC at Q and AD at R.
Now, in quadrilateral ABQR,
AB || QR (By construction)
BQ || AR (In parallelogram ABCD, BC || AD)
∴ Quadrilateral ABQR is a parallelogram.
Similarly, DCQR is a parallelogram.
∆ APB and parallelogram ABQR are on the same base AB and between the same parallels AB and QR.
Question 5.
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)
(ii) ar(AXS) = 1/2 ar (PQRS)
Answer:
Parallelograms PQRS and ABRS are on the same base RS and between the same parallels PB and SR.
∴ ar (PQRS) = ar (ABRS) …………….. (1)
∆ AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
∴ ar (AXS) = 1/2 ar (ABRS) ………………… (2)
From (1) and (2),
ar (AXS) = 1/2 ar (PQRS)
Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field Is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Answer:
By taking any point A on RS and joining it to points P and Q, the field is divided into three triangular parts as ∆ PSA, ∆ APQ and ∆ QRA.
Here, ∆ APQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and SR.
∴ ar (APQ) = 1/2 ar (PQRS)
∴ ar (PSA) + ar(QRA) = 1/2 ar (PQRS)
Thus, ar (APQ) = ar (PSA) + ar (QRA)
Now, as the farmer wants to sow wheat and pulses in equal portions of the field separately.
she has two options as below to do so:
(1) She should sow wheat in ∆ APQ and pulses in ∆ PSA as well as ∆ QRA.
(2) She should sow pulses in ∆ APQ and wheat in ∆ PSA as well as ∆ QRA.
Ex 9.3
Question 1.
In the given figure, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
Answer:
As AD is a median of ∆ ABC, it divides ∆ ABC into two triangles with equal areas.
∴ ar (ADB) = ar (ADC) …………… (1)
In ∆ ABC, AD is a median.
∴ BD = CD
Draw a line l through E and parallel to BC. Then, ∆ EDB and ∆ EDC are on the equal bases and between the same parallels l and BC.
∴ ar (EDB) = ar (EDC) ……………… (2)
Subtracting (2) from (1),
ar (ADB) – ar (EDB) = ar (ADC) – ar (EDC)
∴ ar (ABE) = ar (ACE)
Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer:
The diagonals of a parallelogram bisect each other.
Hence, in parallelogram ABCD, diagonals AC and BD bisect each other at M. Thus, M is the midpoint of AC as well as BD.
In ∆ ABC, BM is a median.
∴ ar (ABM) = ar (CBM) ……………. (1)
In ∆ BCD, CM is a median.
∴ ar (CBM) = ar (CDM) ……………. (2)
In ∆ ABD, AM is a median.
∴ ar (ABM) = ar (DAM) ……………. (3)
From (1), (2) and (3),
ar (ABM) = ar (CBM) = ar (CDM) = ar (DAM)
Thus, the diagonals of a parallelogram divide it into four triangles of equal areas.
Question 4.
In the given 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 (ABC) = ar (ABD).
Answer:
Drawing a line through A and parallel to CD, ∆ AOC and A AOD are on the equal bases and between the same parallels.
∴ ar (AOC) = ar (AOD) ……………… (1)
Similarly, drawing a line through B and parallel to CD, A BOC and A BOD are on the equal bases and between the same parallels.
∴ ar (BOC) = ar (BOD) ……………. (2)
Adding (1) and (2),
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
∴ ar (ABC) = ar (ABD)
Question 5.
D, E and F are respectively the midpoints of the sides BC, CA and AB of a A ABC. Show that:
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
Answer:
In ∆ ABC, F and E are the midpoints of AB and AC respectively.
∴ FE || BC
∴ FE || BD
In ∆ ABC, E and D are the midpoints of AC and BC respectively.
∴ ED || AB
∴ ED || FB
In quadrilateral BDEF, FE || BD and ED || FB.
∴ Quadrilateral BDEF is a parallelogram. …… Result (i)
Similarly, quadrilaterals AFDE and FDCE are parallelograms.
In parallelogram BDEF, FD is a diagonal.
∴ ar (BDF) = ar (DEF) ……………….. (1)
In parallelogram AFDE, EF is a diagonal.
∴ ar (AFE) = ar (DEF) ……………….. (2)
In parallelogram FDCE, ED is a diagonal.
∴ ar (DCE) = ar (DEF) ………………. (3)
∆ ABC is made up of four non-overlapping triangles, ∆ BDF, ∆ AFE, ∆ DCE and ∆ DEF.
∴ ar (ABC)
= ar (BDF) + ar (AFE) + ar (DCE) + ar (DEF)
∴ ar (ABC)
= ar (DEF) + ar (DEF) + ar (DEF) + ar (DEF) [From (1), (2) and (3)]
∴ ar (ABC) = 4ar (DEF)
∴ ar (DEF) = 1/4ar (ABC) …………. Result (ii)
Now, ar (BDEF) = ar (BDF) + ar (DEF)
∴ ar (BDEF) = ar (DEF) + ar (DEF)
∴ ar (BDEF) = 2ar (DEF)
∴ ar (BDEF) = 2 × 1/4ar (ABC)
∴ ar (BDEF) = 1/2ar (ABC) ……. Result (iii)
Question 6.
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB and ABCD is a parallelogram. [Hint: From D and B, draw perpendiculars to AC.]
Answer:
Draw DM ⊥ AC and BN ⊥ AC, where M and N are points on AC.
In ∆ DMO and ∆ BNO,
DO = BO (Given)
∠ DOM = ∠ BON (Vertically opposite angles)
∠ DMO = ∠ BNO (Right angles)
∴ ∆ DMO ≅ ∆ BNO (AAS rule)
∴ DM = BN (CPCT)
Now, in ∆ DMC and ∆ BNA,
DM = BN
hypotenuse DC = hypotenuse BA (Given)
∠ DMC = ∠ BNA (Right angles)
∴ ∆ DMC = ∆ BNA (RHS rule)
∴ ar (DMC) = ar (BNA) ……………… (1)
Moreover, ar (DMO) = ar (BNO) (∵ ∆ DMO ≅ ∆ BNO) ………………… (2)
Adding (1) and (2),
ar (DMC) + ar (DMO) = ar (BNA) + ar (BNO)
∴ ar (DOC) = ar (AOB) (Non-overlapping triangles) … Result (i)
Now, adding ar (COB) on both the sides,
ar (DOC) + ar (COB) = ar (AOB) + ar (COB)
∴ ar (DCB) = ar (ACB) ……… Result (ii)
∆ DCB and ∆ ACB are on the same base BC and their areas are equal.
∴ ∆ DCB and ∆ ACB are between the same parallels DA and CB.
∴ DA || CB ……………… (3)
∆ DMO = ∆ BNO gives OM = ON
∆ DMC = ∆ BNA gives CM = AN
∴ OM + CM = ON + AN
∴ OC = OA
Also, OD = OB (Given)
Thus, in quadrilateral ABCD, diagonals AC and BD bisect each other at O.
∴ ABCD is a parallelogram. ……………….. (4)
Taking (3) and (4) together,
DA || CB and ABCD is a parallelogram. ………. Result (iii)
Question 7.
D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Answer:
ar (DBC) = ar (EBC)
∴ ∆ DBC and ∆ EBC have equal areas.
Here, ∆ DBC and ∆ EBC are on the same base BC and their areas are equal.
∴ ∆ DBC and ∆ EBC are between the same parallels DE and BC.
∴ DE || BC
Question 8.
XY is a line parallel to side BC of a ‘triangle ABC. If BE II AC and CF II AB meet ( XY at E and F respectively, show that ar (ABE) = ar (ACF).
Answer:
Suppose line XY drawn parallel to BC intersects AB and AC at M and N respectively.
In quadrilateral EBCN, EN || BC and BE || CN.
∴ EBCN is a parallelogram.
In quadrilateral MBCF, MF || BC and BM || CF.
∴ MBCF is a parallelogram.
Parallelograms EBCN and MBCF are on the same base BC and between the same parallels BC and EF
∴ ar (EBCN) = ar (MBCF)
∴ ar (BME) + ar (MBCN) = ar (MBCN) + ar (CFN)
∴ ar (BME) = ar (CFN) ……………….. (1)
Now, ∆ BME and ∆ CFN are between the same parallels EF and BC and their areas are equal.
∴ Their bases are equal.
∴ EM = NF
Through A, draw line PQ parallel to EF.
Then, ∆ AEM and ∆ ANF are on equal bases s and between the same parallels PQ and EF. !>
∴ ar (AEM) = ar (ANF) ……………… (2)
Adding (1) and (2),
ar (BME) + ar (AEM) = ar (CFN) + ar (ANF)
∴ ar (ABE) = ar (ACF)
Question 9.
The side AB of a parallelogram ABCD is ‘produced to any point P. A line through A and parallel to CP meets CB produced at and then parallelogram PBQR is completed (see the given figure). Show that: ar (ABCD) = ar (PBQR)
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Answer:
Join AC and PQ.
∆ CAQ and ∆ PAQ are on the same base AQ and between the same parallels CP and AQ.
∴ ar (CAQ) = ar(PAQ)
∴ ar (ACB) + ar (ABQ) = ar (PBQ) + ar (ABQ) (Non-overlapping triangles)
∴ ar (ACB) = ar (PBQ) ……………. (1)
In parallelogram ABCD, AC is a diagonal.
Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
Answer:
∆ DAB and ∆ CBA are on the same base AB and between the same parallels AB and CD.
∴ ar (DAB) = ar (CBA)
∆ DAB and ∆ CBA are both formed by two non-overlapping triangles.
∴ ar (DAB) = ar (AOD) + ar (OAB) and
ar (CBA) = ar (BOC) + ar (OAB)
∴ ar (AOD) + ar (OAB) = ar (BOC) + ar (OAB)
∴ ar (AOD) = ar (BOC)
Question 11.
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
Answer:
∆ ACB and ∆ ACF are on the same base AC and between the same parallels AC and BF.
∴ ar (ACB) = ar (ACF)
Adding ar (AEDC) on both the sides,
ar (ACF) + ar (AEDC) = ar (ACB) + ar (AEDC)
∴ ACF and quadrilateral AEDC are two non-overlapping figures and they form quadrilateral AEDF.
Similarly, ∆ ACB and quadrilateral AEDC are two non-overlapping figures and they form pentagon ABCDE.
∴ ar (AEDF) = ar (ABCDE)
Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the comers to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Answer:
Suppose, quadrilateral ABCD is the plot of Itwaari in the shape of a quadrilateral.
Draw diagonal AC. Through D draw a line parallel to AC to intersect BC produced at P
Join PA to intersect CD at Q.
Here, ∆ DAC and ∆ PAC are on the same base AC and between the same parallels DP and AC.
∴ ar (DAC) = ar (PAC)
∴ ar (DAQ) + ar (QAC) = ar (PQC) + ar (QAC)
∴ ar (DAQ) = ar (PQC)
Thus, we get two triangles – ∆ DAQ and ∆ PQC of equal areas.
So, the Gram Panchayat should take triangular region DAQ from the plot of Itwaari and in exchange give him triangular region PQC. Then, the plot of Itwaari will be of the triangular shape PAB and its area will be the same as the area of original plot ABCD.
Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.)
Answer:
Join CX.
∆ ADX and ∆ ACX are on the same base AX and between the same parallels AB and CD.
∴ ar (ADX) = ar (ACX) …………… (1)
A ACX and A ACY are on the same base AC and between the same parallels AC and XY.
∴ ar (ACX) = ar (ACY) ……………. (2)
From (1) and (2),
ar (ADX) = ar (ACY)
Question 14.
In the given figure, AP || BQ || OR. Prove that ar (AQC) = ar (PBR).
Answer:
∆ CBQ and ∆ RBQ are on the same base BQ and between the same parallels BQ and CR.
∴ ar (CBQ) = ar (RBQ) ……. (1)
∆ ABQ and ∆ PBQ are on the same base BQ and between the same parallels BQ and AE
∴ ar (ABQ) = ar (PBQ) ……… (2)
Adding (1) and (2),
ar (CBQ) + ar (ABQ) = ar (RBQ) + ar (PBQ)
∴ ar (AQC) = ar(PBR) (Non-overlapping triangles)
Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a , trapezium.
Answer:
ar (AOD) = ar (BOC)
Adding ar (OAB) on both the sides,
ar (AOD) + ar (OAB) = ar (BOC) + ar (OAB)
∴ ar (DAB) = ar (CAB)
Thus, ∆ DAB and ∆ CAB are on the same base AB and their areas are equal.
Hence, by theorem 9.3, ∆ DAB and ∆ CAB are between the same parallel.
∴ DC || AB
In quadrilateral ABCD, DC || AB.
Hence, ABCD is a trapezium.
Question 16.
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Answer:
ar (DRC) = ar (DPC)
Thus, ∆ DRC and ∆ DPC are on the same base DC and their areas are equal.
Hence, by theorem 9.3, they are between the same parallels.
∴ DC || RP
In quadrilateral DCPR, DC || RE .
∴ Quadrilateral DCPR is a trapezium.
Now, ar (BDP) = ar (ARC)
∴ ar (DPC) + ar (BDC) = ar (DRC) + ar (ADC) (Non-overlapping triangles)
But, ar (DPC) = ar (DRC)
∴ ar (BDC) = ar (ADC)
Thus, ∆ BDC and ∆ ADC are on the same base DC and their areas are equal.
Hence, by theorem 9.3, they are between the same parallels.
∴ AB || DC
In quadrilateral ABCD, AB || DC.
∴ Quadrilateral ABCD is a trapezium.
Ex 9.4
Question 1.
Parallelogram ABCD and rectangle ABEF are on the, same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Rectangle ABEF is a parallelogram too.
Now, parallelograms ABCD and ABEF are on the same base AB and they have equal areas. Hence, they are between the same parallels FC and AB.
In ∆ AFD, ∠F, being an angle of rectangle ABEF, is a right angle and so, AD is the hypotenuse.
∴ AD > AF
∴ AD + AB > AF + AB
∴ 2 (AD + AB) > 2 (AF + AB)
∴ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF
Question 2.
In the given figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark : Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide AABC into n triangles of equal areas.]
Answer:
Here, in ∆ ABE, D is a point on BE such that BD = DE.
So, in ∆ ABE, D is the midpoint of BE and AD is a median.
∴ ar (ABD) = ar (ADE) ……………… (1)
Similarly, in A ADC, E is the midpoint of DC and AE is a median.
∴ ar (ADE) = ar (AEC) ……………. (2)
From (1) and (2),
ar (ABD) = ar (ADE) = ar (AEC)
Thus, in ∆ ABC, by joining the points of trisection of BC, i.e., D and E to vertex A, the triangle is divided into ∆ ABD, ∆ ADE and ∆ AEC which have the same area.
Now, the answer to the question which was left unanswered in the ‘Introduction’ is ‘Yes’. The manner in which Budhia divided her field, the area of all the three parts are equal.
Question 3.
In the given figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
Answer:
Opposite sides of a parallelogram are equal.
∴ In parallelogram ABCD, AD = BC, in parallelogram DCFE, DE = CF and in parallelogram ABFE, AE = BF.
Now, in ∆ ADE and ∆ BCE
AD = BC, DE = CF and AE = BE
∴ By SSS rule, ∆ ADE = ∆ BCF
∴ ar (ADE) = ar (BCF)
Question 4.
In the given figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at E show that ar (BPC) = ar (DPQ).
[Hint: Join AC.]
Answer:
Join AC.
In parallelogram ABCD, BC || AD and BC = AD.
BC is produced to point Q such that AD = CQ.
Thus, AD = CQ and AD || CQ.
∴ Quadrilateral ACQD is a parallelogram.
Diagonals of a parallelogram divide it into four triangles of equal areas.
∴ ar (DPQ) = ar (DPA) = ar (APC) = ar (CPQ)
∴ ar (DPQ) = ar (APC) ……………. (1)
Now, ∆ APC and ∆ BPC are on the same base PC and between the same parallels PC and AB.
∴ ar (APC) = ar (BPC) ………….. (2)
From (1) and (2),
ar (BPC) = ar (DPQ)
Question 5.
In the given figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that
Answer:
Join EC and AD.
In equilateral ∆ ABC, ∠ ACB = 60°
In equilateral ∆ BDE, ∠ DBE = 60°
∴ ∠ CBE = 60°
Thus, ∠ ACB = ∠ CBE
But, ∠ ACB and ∠ CBE are alternate angles formed by transversal BC of AC and BE and they are equal.
∴BE || AC .
Similarly, ∠ ABD = ∠ BDE = 60°
∴ DE || AB
Now, in ∆ ABC, D is the midpoint of BC.
Hence, AD is a median of ∆ ABC.
Now, in ∆ BFE and ∆ FED, the altitudes corresponding to base BF and FD respectively are the same.
∴ ar (BFE) = 2ar (FED) … Result (v)
Suppose, in ∆ ABD, the altitude on base BD = h.
∴ In ∆ AFC, the altitude on base FC = h.
Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at E Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[(Hint: From A and C, draw perpendiculars to BD.]
Answer:
Question 7.
P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the midpoint of AR show that
Question 8.
In the given figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN „ are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:
(i) ∆ MBC S ∆ ABD
(ii) ar (BYXD) = 2ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ∆ FCB ≅ ∆ ACE
( v ) ar (CYXE) = 2ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler j! proof of this theorem in Class X.
Answer:
(i) ∠ ABM = ∠ CBD = 90°
∴ ∠ABM + ∠ABC = ∠CBD + ∠ABC
∴ ∠ MBC = ∠ ABD
In ∆ MBC and ∆ ABD,
MB = AB, ∠ MBC = ∠ ABD and BC = BD
∴ By SAS rule, ∆ MBC ≅ ∆ ABD
(ii) ar (BYXD) = 2ar (ABD)
∴ ar (BYXD) = 2ar (MBC) [∆ MBC ≅ ∆ ABD]
(iii) ar (BYXD) = 2ar (ABD)
ar (ABMN) = 2ar (MBC)
But, ar (MBC) = ar (ABD)
∴ ar (BYXD) = ar (ABMN)
(iv) ∠ FCA = ∠ ECB = 90°
∴ ∠FCA + ∠ACB = ∠ECB + ∠ACB
∴ ∠FCB = ∠ACE
In ∆ FCB and ∆ ACE,
FC = AC, ∠ FCB = ∠ACE and CB = CE
∴By SAS rule, ∆ FCB ≅ ∆ ACE
(v) ar (CYXE) = 2ar (ACE)
∴ar (CYXE) = 2ar (FCB) [∵ ∆ FCB ≅ ∆ ACE]
(vi) ar (CYXE) = 2ar (FCB)
and ar (ACFG) = 2ar (FCB)
∴ ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (CYXE) + ar (BYXD)
∴ ar (BCED) = ar (ACFG) + ar (ABMN) [By result (iii) and (vi)]
∴ ar (BCED) = ar (ABMN) + ar (ACFG)
MCQ
Multiple Choice Questions and Answer
Answer each question by selecting the proper alternative from those given below each question to make the statement true:
Question 1.
Area of a parallelogram = ………………….
A. 1/2 × base × corresponding altitude
B. 1/2 × the product of diagonals
C. base × corresponding altitude
D. 1/2 × the product of adjacent sides.
Answer:
C. base × corresponding altitude
Question 2.
Area of a triangle = ……………………
A. base × corresponding altitude
B. base + corresponding altitude
C. 1/2 × base × corresponding altitude
D. 2 × base × corresponding altitude
Answer:
C. 1/2 × base × corresponding altitude
Question 3.
ABCD is a rectangle. If AB = 10 cm and ar (ABCD) = 150 cm2, then BC = ………………….. cm.
A. 7.5
B. 15
C. 30
D. 12
Answer:
B. 15
Question 4.
ABCD is a square. If ar (ABCD) = 36 cm2, then AB = ………………… cm.
A. 18
B. 9
C. 6
D. 12
Answer:
C. 6
Question 5.
In ∆ ABC, BC = 10 cm and the length of altitude AD is 5 cm. Then, ar (ABC) = …………………. cm2.
A. 50
B. 100
C. 25
D. 15
Answer:
C. 25
Question 6.
In ∆ ABC, AD is an altitude. If BC = 8 cm and ar (ABC) = 40 cm2, then AD = …………………. cm.
A. 5
B. 10
C. 15
D. 20
Answer:
B. 10
Question 7.
In ∆ PQR, QM is an altitude and PR is the hypotenuse. If PR = 12 cm and QM = 6 cm, then ar (PQR) = ……………………. cm2.
A. 18
B. 72
C. 36
D. 24
Answer:
C. 36
Question 8.
In ∆ XYZ, XZ is the hypotenuse. If XY = 8cm and YZ = 12 cm, then ar (XYZ) = ……………….. cm2.
A. 20
B. 40
C. 96
D. 48
Answer:
D. 48
Question 9.
In parallelogram ABCD, AM is an altitude corresponding to base BC. If BC = 8 cm and AM = 6 cm, then ar (ABCD) = …………………. cm2.
A. 48
B. 24
C. 12
D. 96
Answer:
A. 48
Question 10.
In parallelogram PQRS, QR = 10 cm and ar (PQRS) = 120 cm2. Then, the length of altitude PM corresponding to base QR is ……………………… cm.
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12
Question 11.
For parallelogram ABCD, ar (ABCD) = 48 cm2.
Then, ar (ABC) = …………………….. cm2.
A. 96
B. 48
C. 24
D. 12
Answer:
C. 24
Question 12.
ABCD is a rhombus. If AC = 6 cm and BD = 9 cm, then ar (ABCD) = ………………….. cm2.
A. 15
B. 7.5
C. 54
D. 27
Answer:
D. 27
Question 13.
PQRS is a rhombus. If ar (PQRS) = 40 cm2 and PR = 8 cm, then QS = ………………….. cm.
A. 20
B. 10
C. 25
D. 40
Answer:
B. 10
Question 14.
In ∆ PQR, ∠Q = 90°, PQ = 5 cm and PR = 13 cm.
Then, ar (PQR) = …………………….. cm2.
A. 15
B. 30
C. 45
D. 60
Answer:
B. 30
Question 15.
In ∆ ABC, P Q and R are the midpoints of AB, BC and CA respectively. If ar (ABC) = 32 cm2,
then ar (PQR) = ………………………. cm2.
A. 128
B. 16
C. 8
D. 64
Answer:
C. 8
Question 16.
In ∆ ABC, P, Q and R are the midpoints of AB, BC and CA respectively. If ar (ABC) = 32 cm2, then ar (PBCR) = ………………….. cm2.
A. 10
B. 20
C. 30
D. 40
Answer:
C. 30
Question 17.
In ∆ ABC, P, Q and R are the midpoints of AB, BC and CA respectively. If ar (PBQR) = 36 cm2, then ar (ABC) = ……………………….. cm2.
A. 18
B. 36
C. 54
D. 72
Answer:
D. 72