Question 1: In the adjoining figure, compute the area of quadrilateral ABCD .


ar(ABCD) = ar(\triangle ABD) + ar(\triangle ABD)

DB = \sqrt{17^2 - 8^2} = 15 \ cm

Therefore AB = \sqrt{15^2 - 9^2} = 12 \ cm

Therefore ar(ABCD) = \frac{1}{2} \times 12 \times 9 + \frac{1}{2} \times 8 \times 15 = 54 + 60 = 114 \ cm^2


Question 2: In the adjoining figure, PQRS is a square and T and U are, respectively, the mid points of PS and QR . Find the area of \triangle OTS if PQ = 8 \ cm .


PQRS is a square

\therefore PT = TS = 4 \ cm

QU = UR = 4 \ cm

RS = 4 \ cm

Since T \ \& \ U are mid points of PA \ \& \ QR respectively, TO \parallel P

\therefore TQ = \frac{1}{2} PQ = 4 \ cm

Similarly, TS = \frac{1}{2} PS = 4 \ cm

\therefore ar( \triangle OTS) = \frac{1}{2} \times 4 \times 4 = 8 \ cm^2


Question 3: Compute the area of trapezium PQRS in the adjoining figure.


ar (PQRS) = ar (PTRS) + ar( \triangle RTQ)

RT = \sqrt{17^2 - 8^2} = 15

\therefore ar (PQRS)

= 8 \times 15 + \frac{1}{2} \times 8 \times 15 = 120 + 60 = 180 \ cm^2


Question 4: In the adjoining figure, \angle AOB = 90^o, AC = BC, OA = 12 \ cm and OC = 6.5 \ cm . Find the area of  \triangle AOB .2019-02-04_22-10-18


Since mid point of hypotenuse is equidistant from all three vertices

AC = CB = OC = 6.5 \ cm

\therefore AB = 2 \times 6.5 = 13 \ cm

\therefore OB = \sqrt{13^2 - 12^2} = 5 \ cm

\therefore ar (\triangle AOB) = \frac{1}{2} \times 5 \times 12 = 30 \ cm^2


Question 5: In the adjoining figure, ABCD is a trapezium in which AB = 7 \ cm, AD = BC = 5 \ cm, DC = x \ cm and the distance between AB and DC is 4 \ cm . Find the value of x and area of trapezium ABCD .


DF = \sqrt{5^2 - 4^2} = 3 \ cm

EC = \sqrt{5^2 - 4^2} = 3 \ cm

\therefore x = 3 + 7 + 3 = 13 \ cm

ar (ABCD) = ar (\triangle ADF) + ar (ABEF) + ar (\triangle BCE)

= \frac{1}{2} \times 3 \times 4 + 7 \times 4 + \frac{1}{2} \times 3 \times 4

= 6 + 28 + 6 = 40 \ cm^2


Question 6: In the adjoining figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 \ cm . If OE = 2 \sqrt{5} , find the area of the rectangle.2019-02-04_22-35-29


OCDE is a rectangle.

ED = \sqrt{10^2 - (2\sqrt{5})^2} = \sqrt{100=20} = \sqrt{80}

\therefore ar(OCED) = 2 \sqrt{5} \times \sqrt{80} = 2 \sqrt{400} = 40 \ cm^2


Question 7: In the adjoining figure, ABCD is a trapezium in which AB \parallel DC . Prove that ar( \triangle ADE) = ar( \triangle BOC)


Given AB \parallel DC

Since \triangle ABD and \triangle ABC are between the same parallels and have the same base, therefore

ar (\triangle ABD)= ar (\triangle ABC)

\Rightarrow ar (\triangle ABD) - ar (\triangle ABO) = ar (\triangle ABC) - ar (\triangle ABO)

\Rightarrow ar (\triangle AOD) = ar (\triangle BOC)


Question 8: In the adjoining figure, ABCD and CDEF are parallelograms. Prove that ar( \triangle ADE) = ar( \triangle BCF) 2019-02-04_22-51-34


ADCB is a parallelogram

\therefore AD = BC

Similarly, DEFC is a parallelogram

\therefore DE = CF

Since AB \parallel DC \parallel EF \Rightarrow AE = BF

\therefore \triangle ADE \cong \triangle BCF (By SSS criterion)

Hence ar(\triangle ADE) = ar (\triangle BCF)


Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P . Show that ar( \triangle APB) \times ar( \triangle CPD) = ar( \triangle APD) \times ar( \triangle BPC)


ar( \triangle APD) \times ar( \triangle BPC)

= \{ \frac{1}{2} \times PD \times AL\} \times \{ \frac{1}{2} \times BP \times CM \}

= \{ \frac{1}{2} \times BP \times AL\} \times \{ \frac{1}{2} \times PD \times CM \}

= ar( \triangle APB) \times ar( \triangle CPD)


Question 10: In the adjoining figure, ABC and ABD are two triangles on the base AB . If the line segment CD bisected by AB at O , show that ar( \triangle ABC) = ar( \triangle ABD) 2019-02-05_21-04-56


ar (\triangle ABC) = \frac{1}{2} \times AB \times CL

ar (\triangle ABD) = \frac{1}{2} \times AB \times DM

Now consider \triangle CLO and \triangle DMO

CO = OD (given)

\angle COL = \angle DOM (Vertically opposite angles)

\angle CLO = \angle DMO = 90^o (altitudes)

\therefore \triangle CLO \cong \triangle DMO

\Rightarrow CL = DM

\therefore ar( \triangle ABC) = ar( \triangle ABD)


Question 11: If P is any point in the interior of a parallelogram ABCD , then prove that the area of the \triangle APB is less than half the area of the parallelogram.2019-02-05_21-10-05


ar ({\parallel}^{gm} ABCD) = AB \times DL

ar(\triangle APB) = \frac{1}{2} \times AB \times PM

We know PM < DL

\Rightarrow AB \times PM < AB \times DL

\Rightarrow \frac{1}{2} \times AB \times PM < \frac{1}{2} \times AB \times D

\Rightarrow ar (\triangle APB) < \frac{1}{2} ar ({\parallel}^{gm} ABCD)


Question 12: If AD is the median of \triangle ABC , then prove that \triangle ADB and \triangle ADC are equal in area. If G is the mid point of median AD , prove that ar( \triangle BGC) = 2 \ ar( \triangle AGC)


ar(\triangle ABD) = \frac{1}{2} \times BD \times AL

ar(\triangle ADC) = \frac{1}{2} \times DC \times AL

Since BD = DC

\therefore ar(\triangle ABD) = ar(\triangle ADC)


ar(\triangle BGD) = \frac{1}{2} \times GD \times BO

ar(\triangle AGB) = \frac{1}{2} \times AG \times BO

Since GD = AG

\Rightarrow ar(\triangle BGD) = ar(\triangle AGB)

Now ar(\triangle BGC)= ar(\triangle BGD) + ar(\triangle AGB)

\Rightarrow ar(\triangle BGC)= ar(\triangle AGB) + ar(\triangle AGC)

Since ar(\triangle AGB) = ar(\triangle AGC)

\Rightarrow ar( \triangle BGC) = 2 \ ar( \triangle AGC)


Question 13: A point D is taken on the side of BC and of a \triangle ABC such that BD = 2 DC . Prove that ar( \triangle ABD) = 2 \ ar( \triangle ADC) 2019-02-05_21-31-28.jpg


ar(\triangle ABD) = \frac{1}{2} \times 2x \times AL

ar(\triangle ADC) = \frac{1}{2} \times x \times AL

\therefore ar(\triangle ABD) = 2 ar (\triangle ADC)


Question 14: ABCD is a parallelogram whose diagonals intersect at O . If P is any point on BP , prove that

i)    ar( \triangle ADO) = ar( \triangle CDO)       ii)    ar( \triangle APB) = ar( \triangle CBP) 2019-02-05_21-44-37


i) Since diagonals of a parallelogram bisect each other. Therefore O is the mid point of AC as well as BD

In \triangle ACD, \ DO is the median,

\therefore ar( \triangle ADO) = ar( \triangle DOC)

ii) In \triangle ABC , since OB is the median

ar( \triangle AOB) = ar( \triangle BOC)

In \triangle PAC , since PO is the median

ar( \triangle APO) = ar( \triangle POC)

\Rightarrow ar( \triangle AOB) - ar( \triangle APO) = ar( \triangle BOC) - ar( \triangle POC)

\Rightarrow ar( \triangle APB) = ar( \triangle BPC)


Question 15: ABCD is a parallelogram in which BC is produced to E such that CE = BC . AE intersects CD at F .

i) Prove that ar( \triangle ADF) = ar( \triangle ECF)

ii) If the area of \triangle DFB = 3 \ cm^2 , find the area of {\parallel}^{gm} ABCD 2019-02-05_21-49-31.jpg


Given BC = CE

Consider \triangle ADF and \triangle CEF

AD = CE ( since BC = CE)

\angle AFD = \angle CFE

\angle ADF = \angle FCE

\therefore \triangle ADF \cong \triangle CEF ( By AAS criterion)

\therefore ar( \triangle ADF) = ar( \triangle ECF)

\Rightarrow DF = FC

Since BF is median in \triangle BCD

\Rightarrow  ar( \triangle BCF) = ar( \triangle BFD)

\Rightarrow ar( \triangle BCD) = 6 \ cm^2

\therefore ar(ABCD) = 12 \ cm^2


Question 16: ABCD is a parallelogram whose diagonals AC and BD intersect at O . A line through O intersect AB at P and DC at Q . Prove that ar( \triangle POA) = ar( \triangle QOC)


Consider \triangle POA and \triangle QCO

AO = OC (diagonals bisect each other)

\angle QOC = \angle AOP (vertically opposite angles)

\angle QCO = \angle PAO (since DC\parallel AB and AC is a transversal)

\therefore \triangle POA \cong \triangle QCO

\Rightarrow ar( \triangle POA) = ar( \triangle QOC)


Question 17: ABCD is a parallelogram.  E is a point on BA such that BE= 2 EA and F is the point on DC such that DF = 2 FC . Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD .


ABCD is a {\parallel}^{gm}

\therefore AB = DC

Also AE = FC (since DC \parallel AB \Rightarrow FC \parallel AE )

\therefore AF \parallel FC and equal to each other.

\therefore AFCE is a {\parallel}^{gm}

ar ({\parallel}^{gm} ABCD) = 3x \times h

ar ({\parallel}^{gm} AECF) = x \times h

\therefore ar ({\parallel}^{gm} ABCD) = 3\ ar ({\parallel}^{gm} AECF)


Question 18:  In a \triangle ABC, P and Q are respectively the mid points of AB and BC and R is the mid point of AP . Prove that:

i) ar( \triangle PQB) = ar( \triangle ARC)      ii) ar( \triangle PRQ) = \frac{1}{2} ar( \triangle ARC)

iii) ar( \triangle RQC) = \frac{3}{8} ar( \triangle ABC)


i) In \triangle APC, CR is the median

\Rightarrow ar( \triangle CPR) = ar( \triangle ARC)    … … … … … i)

In \triangle ABC, CP is the median

\Rightarrow ar( \triangle APC) = ar( \triangle BPC)    … … … … … ii)

In \triangle BPC, PQ is the median

\Rightarrow ar( \triangle BPQ) = ar( \triangle PQC)    … … … … … iii)

From i), ii) and iii) we get

ar( \triangle BPC) = 2 \ ar( \triangle BPQ)    … … … … … iv)

Similarly, ar( \triangle APC) = 2 \ ar( \triangle ARC)    … … … … … v)

From iv) and v) we get ar( \triangle BPQ) = ar( \triangle ARC)

ii) In \triangle PQA, QR is the median

\Rightarrow ar( \triangle PQR) = ar( \triangle ARQ)

In \triangle APC, CR is the median

\Rightarrow ar( \triangle CPR) = ar( \triangle ARC)

From (i) we have ar( \triangle BPQ) = ar( \triangle ARC)

and ar( \triangle BPQ) = ar( \triangle APQ)

\therefore ar( \triangle APQ) = ar( \triangle ARC)

\Rightarrow 2\ ar( \triangle PQR) = ar( \triangle ARC)

\Rightarrow ar( \triangle PQR) = \frac{1}{2} ar( \triangle ARC)

iii) Since AQ is the median of \triangle ABC

\Rightarrow ar( \triangle ARC) = \frac{1}{2} ar( \triangle CAP)

\Rightarrow C = \frac{1}{4} ar( \triangle ABC)

Since RQ is the median of \triangle RBC

\Rightarrow ar( \triangle RQC) = \frac{1}{2} ar( \triangle RBC)

= \frac{1}{2} [ ar( \triangle ABC) - ar( \triangle ARC) ]

= \frac{1}{2} [ ar( \triangle ABC) - \frac{1}{4} ar( \triangle ABC) ]

= \frac{3}{8} ar( \triangle ABC)


Question 19: ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2 DE and F is a point of BC such that BF = 2 FC . Prove that 

i) ar( ADGE) = ar( GBCE)      ii) ar( \triangle EGB) = \frac{1}{6} ar( ABCD)

iii) ar( \triangle EFC) = \frac{1}{2} ar( \triangle EBF)    


i) Given ABCD is a {\parallel}^{gm}

ar (AGED) = \frac{1}{2} \times 3x \times h

ar (GBCE) = \frac{1}{2} \times 3x \times h

\therefore ar (AGED) = ar (GBCE)

ii) ar( \triangle EGB) = \frac{1}{2} \times x \times h

ar (ABCD) = 3x \times h

\therefore ar( \triangle EGB) = \frac{1}{6} ar (ABCD)

iii) In \triangle ECB , construct altitude h from E on EB

\therefore ar( \triangle ECF) = \frac{1}{2} \times y \times h

and  ar( \triangle EBF) = \frac{1}{2} \times 2y \times h

\therefore ar( \triangle ECF) = \frac{1}{2} ar( \triangle EBF)


Question 20: In the adjoining figure, CD \parallel AE and CY \parallel BA

i) Name a triangle equal in area of \triangle CBX

ii) Prove that ar( \triangle ZDE) = ar( \triangle CZA)

iii) Prove that ar( BCZY) = ar( \triangle EDZ)


Given CD \parallel AE and CY \parallel BA

i) Consider ABCY

\triangle CYB and \triangle CYA have the same base are between the same parallels.

\therefore ar( \triangle CYB) = ar( \triangle CYA)

\Rightarrow ar( \triangle CYB) - ar( \triangle CXY) = ar( \triangle CYA) - ar( \triangle CXY)

\Rightarrow ar( \triangle CXB) = ar( \triangle XAY)

ii) \triangle CDA and \triangle CDE are between the same parallels

\therefore ar( \triangle CDA) = ar( \triangle CDE)

\Rightarrow ar( \triangle CDA) - ar( \triangle CZD) = ar( \triangle CDE) - ar( \triangle CZD)

\Rightarrow ar( \triangle DZE) = ar( \triangle CZA)

iii) From i)

ar( \triangle BCX) = ar( \triangle XAY)

\therefore ar( \triangle CYA) = ar( \triangle CXY) + ar( \triangle AXY)

\Rightarrow ar( \triangle CYA) = ar( \triangle CXY) + ar( \triangle BXC)

\Rightarrow ar( \triangle CYA) = ar( BYZC)

From ii) ar( \triangle EDZ) = ar( BYZC )


Question 21: In the adjoining area, PSDA is a parallelogram in which PQ = QR = RS and AP \parallel BQ \parallel CR . Prove that ar( \triangle PQE) = ar( \triangle CFD) 2019-02-08_7-50-06.jpg


PSDA is a {\parallel}^{gm}

Also PQ = QR = RS

Since AP \parallel QB \parallel RC \parallel SD

AD = PS and AD \parallel PS


\angle EPQ = \angle FDC

\angle QEP = \angle CFD

\therefore \triangle PQE \cong \triangle CDF

\therefore ar( \triangle PQE) = ar( \triangle CFD)


Question 22: In the adjoining figure, ABCD is a trapezium in which AB \parallel DC and DC = 40 \ cm and AB = 60 \ cm . If X and Y are, respectively the mind points of AD and BC , prove that i) XY = 50 \ cm    ii) DCYX is a trapezium   iii) ar( DCYX) = \frac{9}{11} ar( XYBA)


i) Construction: Join DY and extend it meeting AB at P

Consider \triangle DCY and \triangle BPY

CY = YB (given)

\angle CYD = \angle BYP (vertically opposite angles)

\angle DCY = \angle YBP (since DC \parallel BA )  (Alternate angles)

\therefore \triangle DCY \cong \triangle BPY

\Rightarrow DY = YP and DC = BP

Since X and Y are mid points of DA and DP respectively, 

XY = \frac{1}{2} (AP) = \frac{1}{2} (60+40) = 50 \ cm

ii) From i) we have XY \parallel AB and AB \parallel DC (Given)

\therefore XY \parallel DC

\therefore XYCD is a trapezium

iii) Since X and Y are mid points of DA and CB  respectively, and DC \parallel AB , the distance between DC and XY   and XY and AB are the same.

Let the distance between XY and DC be h

ar (DCXY) = \frac{1}{2} (DC + XY)h = \frac{1}{2} (50 + 40)h = 45h

ar (ABXY) = \frac{1}{2} (AB + XY)h = \frac{1}{2} (50 + 60)h = 55h

Therefore \frac{ar (DCXY)}{ar (ABXY)} = \frac{45h}{55h} = \frac{9}{11}


Question 23: In the adjoining figure, ABC and BDE are two equilateral triangles such that D is the mid point of BC . AE intersects BC at F . Prove that

i) ar( \triangle BDE) = \frac{1}{4} ar( \triangle ABC)      ii) ar( \triangle BDE) = \frac{1}{2} ar( \triangle BAE)

iii) ar( \triangle BFE) =  ar( \triangle AFD) 2019-02-09_7-31-32


i) \triangle ABC and \triangle BDE are equilateral triangles (given)

BD = DC (given)

Altitude of  \triangle BDE = \sqrt{x^2 - (\frac{x}{2})^2} = \frac{\sqrt{3}}{2} x

ar (\triangle BDE) = \frac{1}{2} \times x \times \frac{\sqrt{3}}{2} x = \frac{\sqrt{3}}{4} x^2

Altitude of  \triangle ABC = \sqrt{(2x)^2 - x^2} = \sqrt{3} x

\therefore ar (\triangle ABC) = \frac{1}{2} \times 2x \times \sqrt{3}x = \sqrt{3} x^2

\therefore ar (\triangle BDE) = \frac{1}{4} ar (\triangle ABC)

ii) Since \angle CBE = \angle BCA = 60^o (alternate angles)

\therefore BE \parallel AC

Since DE is median in \triangle BEC

ar (\triangle BED) =  ar (\triangle DEC)

Since \triangle BAE and \triangle BCE are between the same parallels and have the same base

ar (\triangle BAE) =  ar (\triangle BEC)

\Rightarrow ar (\triangle BAE) =  2 \ ar (\triangle BDE)

\Rightarrow ar (\triangle BDE) = \frac{1}{2} ar (\triangle BEA)

iii) Since \angle DBA = \angle BDE = 60^o (alternate angles)

AB \parallel DE

\therefore ar (\triangle BED) = ar (\triangle DEA)

\Rightarrow ar (\triangle BED) - ar (\triangle EFD) = ar (\triangle DEA) - ar (\triangle EFD)

\Rightarrow ar (\triangle BFE) = ar (\triangle AFD)


Question 24: D is the mid point of side BC of  \triangle ABC and E is the mid point of BD . If O is the mid point of AE , prove that ar( \triangle BOE) = \frac{1}{8} ar( \triangle ABC)


In \triangle ABE , since BO is the median

ar (\triangle ABO) = ar (\triangle BOE)    … … … … … i)

In \triangle ABD , since AE is the median

ar (\triangle ABE) = ar (\triangle AED)    … … … … … ii)

In \triangle ABC , since AD is the median

ar (\triangle ABD) = ar (\triangle ADC)    … … … … … iii)

\therefore ar (\triangle ABC) = 2 \ ar (\triangle ABD)

\Rightarrow ar (\triangle ABC) = 2 [ 2 \ ar (\triangle ABE) ]

\Rightarrow ar (\triangle ABC) = 2 [ 2 [ 2 \ ar (\triangle BOE) ] ]

\Rightarrow ar (\triangle ABC) = 8 \ ar (\triangle BOE)

\Rightarrow ar (\triangle BOE) = \frac{1}{8} \ ar (\triangle ABC)


Question 25: In the adjoining figure, X and Y are mid points of AC and AB respectively, QP \parallel BC and CYQ and BXP are straight lines. Prove that ar( \triangle ABP) =  ar( \triangle ACQ)


In \triangle ABC,  \ X is the mid point of AC and Y is the mid point of $latex AB

\therefore XY \parallel BC

Since BC \parallel QP \Rightarrow XY \parallel QP also

Also ar (\triangle BYC) = ar (\triangle BXC)   (Between the same parallels)

\Rightarrow ar (\triangle BOY) + ar (\triangle BOC) = ar (\triangle COX) + ar (\triangle BOC)

\Rightarrow ar (\triangle BOY) = ar (\triangle COX)

Similarly, ar (\triangle QAC) = ar (\triangle APB)

\Rightarrow ar (\triangle AQY) + ar (\triangle AXY) + ar (\triangle XYO) + ar (\triangle COX) = \\ \ \ \ \ \ \ \ \ \ \ ar (\triangle AXP) + ar (\triangle AYX) + ar (\triangle XYO) + ar (\triangle BOY)

\Rightarrow ar (\triangle AQY) = ar (\triangle AXP)

\therefore AQ = AP

Therefore since the bases are equal and triangles are between the same parallels, 

ar (\triangle ABP) = ar (\triangle ACQ)


Question 26: In the adjoining figure, ABCD and AEFD are two parallelogram. Prove that

i) PE = FQ       ii) ar( \triangle APE) : ar( \triangle PFA) =  ar( \triangle QFD) :  ar( \triangle PFD)

iii) ar( \triangle PEA) =  ar( \triangle QFD)


i) Consider \triangle APE and \triangle DFQ

\angle AEP = \angle DFQ (corresponding angles)

EA = FD (opposite sides)

\angle EPA = \angle FQD (corresponding angles)

\therefore \triangle APE \cong \triangle QFD

\Rightarrow EP = FQ

ii) From i) ar( \triangle APE) =  ar( \triangle DFQ)    … … … … … i)

ar( \triangle APF) =  ar( \triangle PFD)    … … … … … ii)

(Same base and between same two parallels)

Dividing i) by ii)

\frac{ar( \triangle APE)}{ar( \triangle APF)} = \frac{ar( \triangle DFQ)}{ar( \triangle PFD)} 

iii) From i) since \triangle PEA \cong \triangle QFD

\Rightarrow ar(\triangle PEA) = ar (\triangle QFD)


Question 27: In the adjoining figure, ABCD is a {\parallel}^{gm} . O is any point on AC . PQ \parallel AB and LM \parallel AD . Prove that ar( {\parallel}^{gm} DLOP) =  ar( {\parallel}^{gm} BMOQ)


Since DC = AB and DC \parallel AB

ar(\triangle ADC) = ar (\triangle ABC)

ar(\triangle APO) +ar(DLOP) +ar(\triangle LOC) = ar(\triangle AOM) +ar(BMOQ) +ar(\triangle QOC)

Since LM \parallel DA and PQ \parallel AB

ar(\triangle APO) = ar (\triangle AOM)

Similarly, ar(\triangle LOC) = ar (\triangle QOC)

Since DC \parallel PQ and LM \parallel BC

\therefore from i) and ii) and iii) we get

ar({\parallel}^{gm} DLOP) = ar({\parallel}^{gm} BMOQ)


Question 28: In a \triangle ABC , if L and M are point on AB and AC respectively such that LM \parallel BC . Prove that:

i) ar( \triangle LCM) =  ar( \triangle LBM)      ii) ar( \triangle LBC) =  ar( \triangle MBC)

iii) ar( \triangle ABM) =  ar( \triangle ACL)      iv)  ar( \triangle LOB) =  ar( \triangle MOC) 2019-02-09_8-45-04


i) Since \triangle LMC and \triangle LMB are on the same base and between two parallels,

ar( \triangle LMC) =  ar( \triangle LMB)

ii) Similarly, ar( \triangle BCM) =  ar( \triangle BCL)

Base is the same and between same parallels

iii) ar( \triangle ALM) + ar( \triangle LMB) =  ar( \triangle ALM) + ar( \triangle LMC)

ar( \triangle ABM) =  ar( \triangle ALC)

iv) ar( \triangle LMB) =  ar( \triangle LMC)

ar( \triangle LMO) + ar( \triangle LOB) =  ar( \triangle LMO)  + ar( \triangle MOC)

ar( \triangle LOB) = ar( \triangle MOC)


Question 29: In the adjoining figure, D and E are two points on BC such that BD = DE = EC . Show that ar( \triangle ABD) =  ar( \triangle ADE) =  ar( \triangle AEC) 2019-02-09_8-48-01


ar (\triangle ABD) = \frac{1}{2} xh

ar (\triangle ADE) = \frac{1}{2} xh

ar (\triangle AEC) = \frac{1}{2} xh

\therefore ar (\triangle ABD) = ar (\triangle ADE) =ar (\triangle AEC)

All the three triangles have equal bases and are between the same parallels.


Question 30: In the adjoining figure, ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides of BC, CA and AB respectively. Line segment AX \perp DE meets BC at Y . Show that:

i) \triangle MBC \cong \triangle ABD      ii) ar( BYXD) =  2 ar( \triangle MBC)

iii) ar( BYXD) =  ar( ABMN)      iv) \triangle MBC \cong \triangle ABD

v) ar( CYXE) =  2 ar( \triangle FCB)      vi) ar( CYXE) =  ar( ACFG)

vii) ar( BCED ) =  ar( ABMN) + ar(ACFG) 2019-02-09_8-57-27


i) Consider \triangle MBC and \triangle ABD



\angle MBC = \angle ABC

(since 90^o + \angle ABC = \angle ABC + 90^o  )

\therefore \triangle MBC \cong \triangle ABD    … … … … … i)   (By SAS criterion)

ii) Consider \triangle ABD and \triangle BDY

Since \triangle ABD and \triangle BDY are on the same base an BD and between the same parallels

ar( \triangle ABD) =  ar( \triangle BYD)

2 ar( \triangle BYD) =  ar( BYXD)

\therefore ar( BYXD) = 2 ar( \triangle ABD)

\Rightarrow ar( BYXD) = 2 ar( \triangle MBC) from i)

iii) Consider MB \parallel NC

\Rightarrow ar( \triangle MBC) =  ar( \triangle MBA)

ar( \triangle MBC) =  \frac{1}{2} ar( MNBA)

\Rightarrow ar( MNBA) = 2 ar( \triangle MBC)

iv) Consider \triangle FCB and \triangle ACE



\angle FCB = \angle ACE

\therefore \triangle FCB \cong \triangle ACE (By SAS criterion)

v) ar( \triangle ACE) =  ar( \triangle YCE)

ar( \triangle ACE) =  \frac{1}{2} ar( {\parallel}^{gm} CYXE)

\Rightarrow ar( {\parallel}^{gm} CYXE) = 2 ar( \triangle ACE)    … … … … … iv)

vi) \triangle FCB and rectangle FGAC are having the same base FC and are between the same parallels

\therefore 2 \ ar( \triangle FCB) =  ar( ACFG)    … … … … … v)

From iv) and v) we get ar( CYXE) = ar( ACFG)

v) BC^2 = AB^2 + AC^2

BC \times BD = AB \times NB + AC \times FC

\Rightarrow ar( BCED) = ar( ABMN)  + ar( ACFG)


Question 31: In the adjoining figure, PQRS and PXYZ are two parallelograms of equal area. Prove that SX is parallel to YR .2019-02-09_11-27-07


Given ar( {\parallel}^{gm} PQRS) =ar( {\parallel}^{gm} PXYZ)

\Rightarrow ar( {\parallel}^{gm} PQRS) - ar( {\parallel}^{gm} PSOX)  =ar( {\parallel}^{gm} PXYZ) - ar( {\parallel}^{gm} PSOX)

\Rightarrow ar( {\parallel}^{gm} XORQ) =ar( {\parallel}^{gm} SOYZ)

\Rightarrow \frac{1}{2} ar( {\parallel}^{gm} XORQ) = \frac{1}{2} ar( {\parallel}^{gm} SOYZ)

\Rightarrow ar( \triangle XOR) =  ar( \triangle SOY)

\Rightarrow ar( \triangle XOR) + ar( \triangle ROY) =  ar( \triangle SOY)+ ar( \triangle ROY)

\Rightarrow ar( \triangle SRY) =  ar( \triangle XRY)

Since the two triangles have the same base, SX \parallel RY


Question 32: Prove that the area of the quadrilateral formed by joining the mid points  of the adjacent sides of a quadrilateral is half the area of the given quadrilateral.


Construction: Join AC and AR .

AR is the median in \triangle ADC

\Rightarrow \frac{1}{2} ar( \triangle ADC) =  ar( \triangle ARD)    … … … … … i)

RS is median in \triangle ARD

\Rightarrow \frac{1}{2} ar( \triangle ARD) =  ar( \triangle SRD)    … … … … … ii)

From i) and ii)  we get

ar( \triangle SRD) = \frac{1}{4} ar( \triangle ACD)    … … … … … iii)

Similarly, ar( \triangle PQB) = \frac{1}{4} ar( \triangle ABC)    … … … … … iv)

Adding iii) and iv) we get

ar( \triangle SRD) + ar( \triangle PQB) = \frac{1}{4} [ ar( \triangle ACD) + ar( \triangle ABC) ]

\Rightarrow ar( \triangle SRD) + ar( \triangle PQB) = \frac{1}{4} ar( ABCD)    … … … … … v)

Similarly we can prove that

ar( \triangle APS) + ar( \triangle QCR) = \frac{1}{4} ar( ABCD)    … … … … … vi)

Adding v) and vi) we get

ar( \triangle SRD) + ar( \triangle PQB) + ar( \triangle APS) + ar( \triangle QCR) = \frac{1}{2} ar( ABCD)

\Rightarrow ar( ABCD) - ar( PQRS)= \frac{1}{2} ar( ABCD)

\Rightarrow ar( PQRS) = \frac{1}{2} ar( ABCD)


Question 33: In the adjoining figure, ABCD is a parallelogram. P is the mid point of AB and CP meets diagonal BD at Q . If area of \triangle PBQ = 10 \ cm^2 .

i) PQ : QC      ii) area of  \triangle PBC      iii) area of {\parallel}^{gm} ABCD 2019-02-09_11-35-43


i) Given ABCD is a {\parallel}^{gm}

Consider \triangle CDQ and \triangle PBQ

\angle CDQ = \angle BQP (vertically opposite angles)

\angle DCQ = \angle QPB (alternate angles)

\angle CDQ = \angle QBP (alternate angles)

\therefore \triangle CDQ \sim \triangle PBQ

\Rightarrow \frac{PQ}{QC} = \frac{PB}{DC} 

\Rightarrow \frac{PQ}{QC} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2}

\therefore PQ : QC = 1:2

ii) Since bases CP and CQ of \triangle PBC and \triangle PBQ lie on the same line, and have the same height,

\frac{ar( \triangle PBC)}{ar( \triangle PBQ)} = \frac{PC}{PQ} = \frac{PQ+QC}{PQ} = \frac{2PQ}{PQ} = 3

\Rightarrow ar( \triangle PBC) = 3 ar( \triangle PBQ) = 3 \times 10 = 30 \ cm^2

iii) ar( \triangle PBC) = \frac{1}{2} ar( \triangle ACB)

\Rightarrow ar( \triangle ACB) = 60 \ cm^2

Similarly, ar( \triangle ACB) = \frac{1}{2} ar( {\parallel}^{gm} ABCD)

\Rightarrow ar( {\parallel}^{gm} ABCD) = 2 \times 60 = 120 \ cm^2


Question 34: If E and F are mid points of the sides AB and AC respectively of  \triangle ABC , prove that EBCF is a trapezium. Also find it’s area if area of  \triangle ABC is 100 \ cm^2 .


Since E and F are mid points of AB and AC respectively, EF \parallel BC

\therefore EBCF is a trapezium.

ar (\triangle ABC) = 100 \ cm^2

\Rightarrow \frac{1}{2} BC \times AM = 100 \ cm^2

Draw altitude from A .

\therefore AM \perp BC , and since EF \parallel BC, \Rightarrow AN \perp EF

We can prove \triangle AEN \sim \triangle ABM (by AAA criterion)

\therefore \frac{AN}{AM} = \frac{AE}{AB} = \frac{1}{2}

\therefore ar (\triangle AEF) = \frac{1}{2} \times EF \times AN = \frac{1}{2} \Big( \frac{1}{2} BC \Big) \times \Big( \frac{1}{2} AM \Big)

= \frac{1}{4} \Big( \frac{1}{2} BC \times AM \Big) = 25 \ cm^2

\therefore ar (EBFC) = 100 - 25 = 75 \ cm^2


Question 35: In the adjoining figure, P is any point on median AD of \triangle ABC . Prove that, i) ar( \triangle PBD) =  ar( \triangle PDC)      ii) ar( \triangle APB) =  ar( \triangle ACP)


i) Since D is the mid point of BC and \triangle BPD and \triangle DPC have the same height,

ar( \triangle PBD) =  ar( \triangle PDC)

ii) Since AD is the median

ar( \triangle ABD) =  ar( \triangle ACD)

\therefore ar( \triangle ABD) - ar( \triangle PBD) =  ar( \triangle ACD) - ar( \triangle PDC)

\Rightarrow ar( \triangle APB) =  ar( \triangle ACP)


Question 36: In the adjoining figure, if DE \parallel BC , prove that

i) ar( \triangle ACD) =  ar( \triangle ABE)      ii) ar( \triangle OBD) =  ar( \triangle OCE) 2019-02-09_23-07-57


i) Given DE \parallel BC

ar( \triangle BCE) =  ar( \triangle BCD)

\Rightarrow ar( \triangle BCE)  - ar( \triangle BOC)=  ar( \triangle BCD)- ar( \triangle BOC)

\Rightarrow ar( \triangle COE) =  ar( \triangle BOD)    … … … … … i)

\Rightarrow ar( \triangle COE) + ar (ADOE)  =  ar( \triangle BOD) + ar (ADOE)

\Rightarrow ar( \triangle ABE) =  ar( \triangle ACD)

ii) from i) ar( \triangle OCE) =  ar( \triangle OBD)


Question 37: If in a quadrilateral ABCD , diagonal AC bisects the diagonal BD , then prove that ar( \triangle ABC) =  ar( \triangle ACD)  


In \triangle ADB, AO is a median

\therefore ar( \triangle ADO) =  ar( \triangle ABO)    … … … … … i)

Similarly, ar( \triangle CDO) =  ar( \triangle CBO)    … … … … … ii)

Adding i) and ii) ar( \triangle ABC) =  ar( \triangle ACD)


Question 38: In the adjoining figure, P is a point on the side BC of  \triangle ABC such that PC = 2 BP and Q is a pint on AP such that QA = 5 PQ . Find ar( \triangle AQC) :  ar( \triangle ABC) 2019-02-09_23-01-48


PC = 2 BP \Rightarrow ar( \triangle APC) =  \frac{2}{3} ar( \triangle ABC)

QA = 5 PQ \Rightarrow ar( \triangle AQC) =  \frac{5}{6} ar( \triangle APC)

\Rightarrow ar( \triangle AQC) =  \frac{5}{6} ar( \triangle APC) \times \frac{2}{3} ar( \triangle ABC)

\Rightarrow ar( \triangle AQC) = \frac{5}{9} ar( \triangle ABC)


Question 39: In the adjoining figure, AD is the median of the \triangle ABC and P is the point on AC such that ar( \triangle ADP) :  ar( \triangle ABD) = 2:3 . Find

i) AP : PC      ii)  ar( \triangle PDC) :  ar( \triangle ABC) 2019-02-09_22-58-37


Given AD is median

\frac{ar( \triangle ADP)}{ar( \triangle ABD)} = \frac{2}{3}

i) Also ar( \triangle ABD) =  ar( \triangle ADC)

Given, ar( \triangle ABD) =  \frac{3}{2} ar( \triangle ADP)

\Rightarrow \frac{ar( \triangle ADC)}{ar( \triangle ADP)} = \frac{3}{2}

\Rightarrow \frac{AC}{AP} = \frac{3}{2}

\Rightarrow \frac{AP + PC}{AP} = \frac{3}{2}

\Rightarrow \frac{PC}{AP} = \frac{1}{2}

or \frac{AP}{PC} = \frac{2}{1}

ii) from i) \frac{ar( \triangle ADC)}{ar( \triangle ADP)} = \frac{ar( \triangle ADP) + ar(\triangle PDC) }{ar( \triangle ADP)} = 1 + \frac{ar( \triangle PDC)}{ar( \triangle ADP)} 

\Rightarrow \frac{ar( \triangle PDC)}{ar( \triangle ADP)} = \frac{3}{2} -1 = \frac{1}{2} 


Question 40: In the adjoining figure, E is the midpoint of the side AB of \triangle ABC and EBCF is a parallelogram. If  ar( \triangle ABC) = 25 \ cm^2 , find ar({\parallel}^{gm} EBCF) 2019-02-09_22-56-16


Consider \triangle AEG and \triangle CGF

\angle AGE = \angle FGC

AE = EB and EB = FC \Rightarrow AE = FC

Also since EF \parallel BC

\angle AEG = \angle GFC

\therefore \triangle AEG \cong \triangle CGF ( By SAA criterion)

\therefore ar( {\parallel}^{gm} EFCG) = ar( {\parallel}^{gm} EGCB) + ar( \triangle FGC)

\Rightarrow ar( {\parallel}^{gm} EFCG) = ar( {\parallel}^{gm} EGCB) + ar( \triangle AEG)

\Rightarrow ar( {\parallel}^{gm} EFCG) = ar( \triangle ABC)

\Rightarrow ar( {\parallel}^{gm} EFCG) = 25 \ cm^2


Question 41: In the adjoining figure, E and F are mid points of sides AB and CD respectively of parallelogram ABCD . If ar({\parallel}^{gm} ABCD) = 40 \ cm^2 find ar( \triangle APD) . Name the parallelogram whose area is equal to the area of \triangle APD .


ar( {\parallel}^{gm} ABCD) = 40 \ cm^2

AD \parallel BC . Join DB

ar( \triangle ADB) = ar( \triangle ADP)

\Rightarrow  \frac{1}{2} ar( {\parallel}^{gm} ABCD)= ar( \triangle ADP)

\Rightarrow ar( \triangle ADP) = \frac{40}{2} = 20 \ cm^2

\therefore ar( {\parallel}^{gm} AEFD) = 20 \ cm^2


Question 42: In the adjoining figure, P is a point on side BC of a parallelogram ABCD such that BP: PC = 1:2 . If DP produced meets AB produced at Q and ar(\triangle CPQ) = 20 \ cm^2 , find  ar(\triangle CDP) and ar({\parallel}^{gm} ABCD) .


Given, ar (\triangle CPQ) = 20 \ cm^2

We can prove \triangle CDP \sim \triangle BAP

\Rightarrow BP : PC = 1:2

ar (\triangle BQP) = \frac{1}{2} ar (\triangle CPQ) = 10 \ cm^2

\therefore ar (\triangle CPD) = 2^2 \times ar (\triangle CPQ) = 40 \ cm^2

\therefore ar ({\parallel}^{gm} DCPM) = 80 \ cm^2

Also ar ({\parallel}^{gm} ABPM) = \frac{1}{2} \times 80 \ cm^2 = 40 \ cm^2

\therefore ar ({\parallel}^{gm} ABCD) = 80 + 40 = 120 \ cm^2


Question 43: In the adjoining figure, ABCD and AEFG are two parallelograms. Prove that ar({\parallel}^{gm} ABCD) = ar({\parallel}^{gm} AEFG) 2019-02-09_22-46-43


Join BG

\triangle ABG and {\parallel}^{gm} ABCD are on the same base and between the same parallel

\therefore ar( \triangle ABG) = \frac{1}{2} ar({\parallel}^{gm} ABCD)

Similarly, ar( \triangle ABG) = \frac{1}{2} ar({\parallel}^{gm} AEFG)

\therefore ar({\parallel}^{gm} ABCD) = ar({\parallel}^{gm} AEFG)


Question 44: ABCD is a square. E and F are mid points of the sides AB and AD respectively. Prove that ar( \triangle CEF) = \frac{3}{8} ar( ABCD) 2019-02-09_22-43-05


ar( \triangle CFA) = \frac{1}{2} ar( \triangle CDA)

\Rightarrow ar( \triangle CFM) + ar( \triangle FMA) = \frac{1}{2} ar( \triangle CDA)    … … … … … i)

Similarly, \Rightarrow ar( \triangle CEM) + ar( \triangle EMA) = \frac{1}{2} ar( \triangle CAB)    … … … … … ii)

Adding i) and ii)

ar( \triangle CFM) + ar( \triangle FMA) + ar( \triangle CEM) + ar( \triangle EAM) = \frac{1}{2} ar( ABCD)

ar(\triangle CEF) = \frac{1}{2} ar(ABCD) - ar(\triangle FAE)

\Rightarrow ar(\triangle CEF) = \frac{1}{2} ar(ABCD) - \frac{1}{8} ar(\triangle ABCD)

\Rightarrow ar(\triangle CEF) = \frac{3}{8} ar(ABCD)


Question 45: A point D is taken on the sides BC and of \triangle ABC and AD is produced to E such that AD = DE , prove that ar( \triangle BCE) =  ar( \triangle ABC) 2019-02-09_22-40-18


Draw a line \parallel to AD

\therefore  ar( \triangle BDE) =  ar( \triangle BDA)  … … … … … i)

(They have equal bases and between the same parallels)

Similarly, ar( \triangle CDE) =  ar( \triangle CDA)  … … … … … ii)

Adding i) and ii) we get

ar( \triangle BDE) + ar( \triangle CDE) =  ar( \triangle BDA) + ar( \triangle CDA)

\Rightarrow ar( \triangle BCE) =  ar( \triangle ABC)


Question 46: In the adjoining figure,  if AB \parallel DC \parallel EF and AD \parallel BE and DE \parallel AF . Prove that ar( DEFH) =  ar( ABCD)


Consider \triangle ABG and \triangle CDE


\angle BAG = \angle CDE

\angle ABC = \angle DG

\therefore \triangle ABG \cong \triangle CDE (By ASA criterion)

\Rightarrow ar (\triangle ABG) = ar (\triangle CDE)

\Rightarrow ar (\triangle ABG) - ar (\triangle HCG) = ar (\triangle CDE) - ar (\triangle HCG)

\Rightarrow ar (ABCH) = ar (\triangle DHGE)

Now consider, \triangle ADH and \triangle EFG

\angle DAH = \angle EFG

\angle ADH = \angle GEF


\therefore \triangle ADH \cong \triangle EFG (By ASA criterion)

\therefore ar (\triangle ADH) = ar (\triangle EFG)

\Rightarrow ar (\triangle ADH) + ar (ABCH) = ar (\triangle EFG) + ar (\triangle DHGE)

\Rightarrow ar (\triangle ABCD) = ar (\triangle DEFG)


Question 47: ABCD is a rectangle and P is mid point of AB . DP is produced to meet CB at Q . Prove that ar (ABCD) = ar (\triangle DQC) 2019-02-09_22-35-02


Consider \triangle DAP and \triangle BQP

\angle DAP = \angle BPQ


\angle DAP = \angle QBP

\therefore \triangle DAP \cong \triangle BQP

\therefore ar( \triangle DAP) =  ar( \triangle BQP)

\Rightarrow ar( \triangle DAP) + ar (BCDP) =  ar( \triangle BQP)+ ar (BCDP)

\Rightarrow ar (ABCD) = ar (\triangle DQC)


Question 48: If each diagonal of a quadrilateral divides it into two triangles of equal areas, then prove that it is a parallelogram.2019-02-09_22-32-31


AC divides ABCD into two equal halves

\therefore ar(\triangle ABC) = \frac{1}{2} ar(ABCD)

Similarly, ar(\triangle ABD) = \frac{1}{2} ar(ABCD)

\therefore ar(\triangle ABC) = ar(\triangle ABD)

\Rightarrow AB \parallel DC

Similarly, AD \parallel BC

\therefore ABCD is a parallelogram.