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

Answer:2019-02-04_21-39-17

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

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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 .

Answer:2019-02-04_21-44-51

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

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

Answer:2019-02-04_22-09-56

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

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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

Answer:

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

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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 .

Answer:2019-02-04_22-31-20

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

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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

Answer:

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

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

Answer:2019-02-04_22-51-09

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)

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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

Answer:

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)

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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)

Answer:2019-02-05_20-58-53

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)

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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

Answer:

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)

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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

Answer:

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)

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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)

Answer:2019-02-05_21-13-53.jpg

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)

AG = GD

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)

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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

Answer:

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)

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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

Answer:

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)

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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

Answer:

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

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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)

Answer:2019-02-07_17-52-41

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)

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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 .

Answer:2019-02-07_17-57-22

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)

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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)

Answer:2019-02-07_18-02-16

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)

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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)    

Answer:2019-02-10_8-13-28

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)

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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)

Answer:2019-02-07_18-40-38

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 )

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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

Answer:

PSDA is a {\parallel}^{gm}

Also PQ = QR = RS

Since AP \parallel QB \parallel RC \parallel SD

AD = PS and AD \parallel PS

PQ = CD

\angle EPQ = \angle FDC

\angle QEP = \angle CFD

\therefore \triangle PQE \cong \triangle CDF

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

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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)

Answer:2019-02-09_7-22-31

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}

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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

Answer:

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)

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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)

Answer:2019-02-09_7-36-26

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)

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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)

Answer:2019-02-09_7-41-37

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)

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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)

Answer:2019-02-09_7-46-11

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)

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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)

Answer:2019-02-09_7-52-17

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)

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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

Answer:

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)

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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

Answer:

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.

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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

Answer:

i) Consider \triangle MBC and \triangle ABD

MB = AB

BC = BD

\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

FC = AC

BC = CE

\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)

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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

Answer:

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

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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.

Answer:2019-02-09_11-32-58

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)

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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

Answer:

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

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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 .

Answer:2019-02-09_11-38-12

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

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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)

Answer:2019-02-09_11-41-13

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)

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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

Answer:

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)

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

Answer:2019-02-09_23-05-27

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)

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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

Answer:

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)

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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

Answer:

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} 

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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

Answer:

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

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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 .

Answer:2019-02-09_22-53-51

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

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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) .

Answer:2019-02-09_22-49-01

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

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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

Answer:

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)

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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

Answer:

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)

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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

Answer:

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)

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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)

Answer:2019-02-10_8-39-56

Consider \triangle ABG and \triangle CDE

DC = AB

\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

AD = EG

\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)

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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

Answer:

Consider \triangle DAP and \triangle BQP

\angle DAP = \angle BPQ

AP = PB

\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)

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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

Answer:

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.

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