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

Answer:

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

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

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

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

\displaystyle \\

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

Answer:

\displaystyle PQRS is a square

\displaystyle \therefore PT = TS = 4 \ cm

\displaystyle QU = UR = 4 \ cm

\displaystyle RS = 4 \ cm

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

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

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

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

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

Answer:

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

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

\displaystyle \therefore ar (PQRS)

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

\displaystyle \\

Question 4: In the adjoining figure, \displaystyle \angle AOB = 90^o, AC = BC, OA = 12 \ cm \text{ and } OC = 6.5 \ cm . Find the area of \displaystyle \triangle AOB .

Answer:

Since mid point of hypotenuse is equidistant from all three vertices

\displaystyle AC = CB = OC = 6.5 \ cm

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

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

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

\displaystyle \\

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

Answer:

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

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

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

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

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

\displaystyle = 6 + 28 + 6 = 40 \ \text{cm}^2

\displaystyle \\

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

Answer:

\displaystyle OCDE is a rectangle.

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

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

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

Answer:

\displaystyle \text{Given } AB \parallel DC

Since \displaystyle \triangle ABD \text{ and } \triangle ABC are between the same parallels and have the same base, therefore

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

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

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

\displaystyle \\

Question 8: In the adjoining figure, \displaystyle ABCD \text{ and } CDEF are parallelograms. Prove that \displaystyle ar( \triangle ADE) = ar( \triangle BCF)

Answer:

\displaystyle ADCB is a parallelogram

\displaystyle \therefore AD = BC

\displaystyle \text{Similarly, } DEFC is a parallelogram

\displaystyle \therefore DE = CF

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

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

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

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

Answer:

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

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

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

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

\displaystyle \\

Question 10: In the adjoining figure, \displaystyle ABC \text{ and } ABD are two triangles on the base \displaystyle AB . If the line segment \displaystyle CD bisected by \displaystyle AB at \displaystyle O , show that \displaystyle ar( \triangle ABC) = ar( \triangle ABD)

Answer:

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

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

Now consider \displaystyle \triangle CLO \text{ and } \triangle DMO

\displaystyle CO = OD (given)

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

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

\displaystyle \therefore \triangle CLO \cong \triangle DMO

\displaystyle \Rightarrow CL = DM

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

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Question 11: If \displaystyle P is any point in the interior of a parallelogram \displaystyle ABCD , then prove that the area of the \displaystyle \triangle APB is less than half the area of the parallelogram.

Answer:

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

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

We know \displaystyle PM < DL

\displaystyle \Rightarrow AB \times PM < AB \times DL

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

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

\displaystyle \\

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

Answer:

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

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

Since \displaystyle BD = DC

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

\displaystyle AG = GD

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

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

Since \displaystyle GD = AG

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

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

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

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

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

\displaystyle \\

Question 13: A point \displaystyle D is taken on the side of \displaystyle BC and of a \displaystyle \triangle ABC such that \displaystyle BD = 2 DC . Prove that \displaystyle ar( \triangle ABD) = 2 \ ar( \triangle ADC)

Answer:

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

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

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

\displaystyle \\

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

\displaystyle \text{i) } ar( \triangle ADO) = ar( \triangle CDO) \text{ii) } ar( \triangle APB) = ar( \triangle CBP)

Answer:

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

In \displaystyle \triangle ACD, \ DO is the median,

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

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

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

In \displaystyle \triangle PAC , since \displaystyle PO is the median

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

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

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

\displaystyle \\

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

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

ii) If the area of \displaystyle \triangle DFB = 3 \ \text{cm}^2 , find the area of \displaystyle {\parallel}^{gm} ABCD

Answer:

\displaystyle \text{Given } BC = CE

Consider \displaystyle \triangle ADF \text{ and } \triangle CEF

\displaystyle AD = CE ( since \displaystyle BC = CE)

\displaystyle \angle AFD = \angle CFE

\displaystyle \angle ADF = \angle FCE

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

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

\displaystyle \Rightarrow DF = FC

Since \displaystyle BF is median in \displaystyle \triangle BCD

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

\displaystyle \Rightarrow ar( \triangle BCD) = 6 \ \text{cm}^2

\displaystyle \therefore ar(ABCD) = 12 \ \text{cm}^2

\displaystyle \\

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

Answer:

Consider \displaystyle \triangle POA \text{ and } \triangle QCO

\displaystyle AO = OC (diagonals bisect each other)

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

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

\displaystyle \therefore \triangle POA \cong \triangle QCO

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

\displaystyle \\

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

Answer:

\displaystyle ABCD is a \displaystyle {\parallel}^{gm}

\displaystyle \therefore AB = DC

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

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

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

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

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

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

\displaystyle \\

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

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

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

Answer:

i) In \displaystyle \triangle APC, CR is the median

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

In \displaystyle \triangle ABC, CP is the median

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

In \displaystyle \triangle BPC, PQ is the median

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

From i), ii) and iii) we get

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

\displaystyle \text{Similarly, } ar( \triangle APC) = 2 \ ar( \triangle ARC) … … … … … v)

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

ii) In \displaystyle \triangle PQA, QR is the median

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

In \displaystyle \triangle APC, CR is the median

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

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

\displaystyle \text{ and } ar( \triangle BPQ) = ar( \triangle APQ)

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

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

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

iii) Since \displaystyle AQ is the median of \displaystyle \triangle ABC

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

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

Since \displaystyle RQ is the median of \displaystyle \triangle RBC

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

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

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

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

\displaystyle \\

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

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

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

Answer:

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

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

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

\displaystyle \therefore ar (AGED) = ar (GBCE)

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

\displaystyle ar (ABCD) = 3x \times h

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

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

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

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

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

\displaystyle \\

Question 20: In the adjoining figure, \displaystyle CD \parallel AE \text{ and } CY \parallel BA

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

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

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

Answer:

\displaystyle \text{Given } CD \parallel AE \text{ and } CY \parallel BA

i) Consider \displaystyle ABCY

\displaystyle \triangle CYB \text{ and } \triangle CYA have the same base are between the same parallels.

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

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

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

\displaystyle \text{ii) } \triangle CDA \text{ and } \triangle CDE are between the same parallels

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

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

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

iii) From i)

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

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

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

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

From \displaystyle \text{ii) } ar( \triangle EDZ) = ar( BYZC )

\displaystyle \\

Question 21: In the adjoining area, \displaystyle PSDA is a parallelogram in which \displaystyle PQ = QR = RS \text{ and } AP \parallel BQ \parallel CR . Prove that \displaystyle ar( \triangle PQE) = ar( \triangle CFD)

Answer:

\displaystyle PSDA is a \displaystyle {\parallel}^{gm}

Also \displaystyle PQ = QR = RS

Since \displaystyle AP \parallel QB \parallel RC \parallel SD

\displaystyle AD = PS \text{ and } AD \parallel PS

\displaystyle PQ = CD

\displaystyle \angle EPQ = \angle FDC

\displaystyle \angle QEP = \angle CFD

\displaystyle \therefore \triangle PQE \cong \triangle CDF

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

\displaystyle \\

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

Answer:

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

Consider \displaystyle \triangle DCY \text{ and } \triangle BPY

\displaystyle CY = YB (given)

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

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

\displaystyle \therefore \triangle DCY \cong \triangle BPY

\displaystyle \Rightarrow DY = YP \text{ and } DC = BP

Since \displaystyle X \text{ and } Y are mid points of \displaystyle DA \text{ and } DP respectively, 

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

ii) From i) we have \displaystyle XY \parallel AB \text{ and } AB \parallel DC (Given)

\displaystyle \therefore XY \parallel DC

\displaystyle \therefore XYCD is a trapezium

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

Let the distance between \displaystyle XY \text{ and } DC be \displaystyle h

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

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

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

\displaystyle \\

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

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

\displaystyle \text{iii) } ar( \triangle BFE) = ar( \triangle AFD)

Answer:

\displaystyle \text{i) } \triangle ABC \text{ and } \triangle BDE are equilateral triangles (given)

\displaystyle BD = DC (given)

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

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

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

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

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

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

\displaystyle \therefore BE \parallel AC

Since \displaystyle DE is median in \displaystyle \triangle BEC

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

Since \displaystyle \triangle BAE \text{ and } \triangle BCE are between the same parallels and have the same base

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

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

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

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

\displaystyle AB \parallel DE

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

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

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

\displaystyle \\

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

Answer:

In \displaystyle \triangle ABE , since \displaystyle BO is the median

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

In \displaystyle \triangle ABD , since \displaystyle AE is the median

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

In \displaystyle \triangle ABC , since \displaystyle AD is the median

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \therefore XY \parallel BC

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

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

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

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

\displaystyle \text{Similarly, } ar (\triangle QAC) = ar (\triangle APB)

\displaystyle \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)

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

\displaystyle \therefore AQ = AP

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

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

\displaystyle \\

Question 26: In the adjoining figure, \displaystyle ABCD \text{ and } AEFD are two parallelogram. Prove that

\displaystyle \text{i) } PE = FQ \text{ii) } ar( \triangle APE) : ar( \triangle PFA) = ar( \triangle QFD) : ar( \triangle PFD)

\displaystyle \text{iii) } ar( \triangle PEA) = ar( \triangle QFD)

Answer:

i) Consider \displaystyle \triangle APE \text{ and } \triangle DFQ

\displaystyle \angle AEP = \angle DFQ (corresponding angles)

\displaystyle EA = FD (opposite sides)

\displaystyle \angle EPA = \angle FQD (corresponding angles)

\displaystyle \therefore \triangle APE \cong \triangle QFD

\displaystyle \Rightarrow EP = FQ

ii) From \displaystyle \text{i) } ar( \triangle APE) = ar( \triangle DFQ) … … … … … i)

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

(Same base and between same two parallels)

Dividing i) by ii)

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

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

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

\displaystyle \\

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

Answer:

Since \displaystyle DC = AB \text{ and } DC \parallel AB

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

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

Since \displaystyle LM \parallel DA \text{ and } PQ \parallel AB

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

\displaystyle \text{Similarly, } ar(\triangle LOC) = ar (\triangle QOC)

Since \displaystyle DC \parallel PQ \text{ and } LM \parallel BC

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

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

\displaystyle \\

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

\displaystyle \text{i) } ar( \triangle LCM) = ar( \triangle LBM) \text{ii) } ar( \triangle LBC) = ar( \triangle MBC)

\displaystyle \text{iii) } ar( \triangle ABM) = ar( \triangle ACL) \text{iv) } ar( \triangle LOB) = ar( \triangle MOC)

Answer:

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

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

\displaystyle \text{ii) } \text{Similarly, } ar( \triangle BCM) = ar( \triangle BCL)

Base is the same and between same parallels

\displaystyle \text{iii) } ar( \triangle ALM) + ar( \triangle LMB) = ar( \triangle ALM) + ar( \triangle LMC)

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

\displaystyle \text{iv) } ar( \triangle LMB) = ar( \triangle LMC)

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

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

\displaystyle \\

Question 29: In the adjoining figure, \displaystyle D \text{ and } E are two points on \displaystyle BC such that \displaystyle BD = DE = EC . Show that \displaystyle ar( \triangle ABD) = ar( \triangle ADE) = ar( \triangle AEC)

Answer:

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

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

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

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

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

\displaystyle \\

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

\displaystyle \text{i) } \triangle MBC \cong \triangle ABD \text{ii) } ar( BYXD) = 2 ar( \triangle MBC)

\displaystyle \text{iii) } ar( BYXD) = ar( ABMN) \text{iv) } \triangle MBC \cong \triangle ABD

\displaystyle \text{v) } ar( CYXE) = 2 ar( \triangle FCB) v\displaystyle \text{i) } ar( CYXE) = ar( ACFG)

v\displaystyle \text{ii) } ar( BCED ) = ar( ABMN) + ar(ACFG)

Answer:

i) Consider \displaystyle \triangle MBC \text{ and } \triangle ABD

\displaystyle MB = AB

\displaystyle BC = BD

\displaystyle \angle MBC = \angle ABC

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

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

ii) Consider \displaystyle \triangle ABD \text{ and } \triangle BDY

Since \displaystyle \triangle ABD \text{ and } \triangle BDY are on the same base an \displaystyle BD and between the same parallels

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

\displaystyle 2 ar( \triangle BYD) = ar( BYXD)

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

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

iii) Consider \displaystyle MB \parallel NC

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

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

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

iv) Consider \displaystyle \triangle FCB \text{ and } \triangle ACE

\displaystyle FC = AC

\displaystyle BC = CE

\displaystyle \angle FCB = \angle ACE

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

\displaystyle \text{v) } ar( \triangle ACE) = ar( \triangle YCE)

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

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

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

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

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

\displaystyle \text{v) } BC^2 = AB^2 + AC^2

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

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

\displaystyle \\

Question 31: In the adjoining figure, \displaystyle PQRS \text{ and } PXYZ are two parallelograms of equal area. Prove that \displaystyle SX is parallel to \displaystyle YR .

Answer:

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

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

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

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

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

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

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

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

\displaystyle \\

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:

Construction: Join \displaystyle AC \text{ and } AR .

\displaystyle AR is the median in \displaystyle \triangle ADC

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

\displaystyle RS is median in \displaystyle \triangle ARD

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

From i) and ii) we get

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

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

Adding iii) and iv) we get

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

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

Similarly we can prove that

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

Adding v) and vi) we get

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

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

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

\displaystyle \\

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

\displaystyle \text{i) } PQ : QC ii) area of \displaystyle \triangle PBC iii) area of \displaystyle {\parallel}^{gm} ABCD

Answer:

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

Consider \displaystyle \triangle CDQ \text{ and } \triangle PBQ

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

\displaystyle \angle DCQ = \angle QPB (alternate angles)

\displaystyle \angle CDQ = \angle QBP (alternate angles)

\displaystyle \therefore \triangle CDQ \sim \triangle PBQ

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

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

\displaystyle \therefore PQ : QC = 1:2

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

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

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

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

\displaystyle \Rightarrow ar( \triangle ACB) = 60 \ \text{cm}^2

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \therefore EBCF is a trapezium.

\displaystyle ar (\triangle ABC) = 100 \ \text{cm}^2

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

Draw altitude from \displaystyle A .

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

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

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

\displaystyle \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)

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

\displaystyle \therefore ar (EBFC) = 100 - 25 = 75 \ \text{cm}^2

\displaystyle \\

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

Answer:

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

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

ii) Since \displaystyle AD is the median

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

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

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

\displaystyle \\

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

\displaystyle \text{i) } ar( \triangle ACD) = ar( \triangle ABE) \text{ii) } ar( \triangle OBD) = ar( \triangle OCE)

Answer:

\displaystyle \text{i) } \text{Given } DE \parallel BC

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

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

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

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

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

ii) from \displaystyle \text{i) } ar( \triangle OCE) = ar( \triangle OBD)

\displaystyle \\

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

Answer:

In \displaystyle \triangle ADB, AO is a median

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

\displaystyle \text{Similarly, } ar( \triangle CDO) = ar( \triangle CBO) … … … … … ii)

Adding i) \displaystyle \text{ and } \text{ii) } ar( \triangle ABC) = ar( \triangle ACD)

\displaystyle \\

Question 38: In the adjoining figure, \displaystyle P is a point on the side \displaystyle BC of \displaystyle \triangle ABC such that \displaystyle PC = 2 BP \text{ and } Q is a pint on \displaystyle AP such that \displaystyle QA = 5 PQ . Find \displaystyle ar( \triangle AQC) : ar( \triangle ABC)

Answer:

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

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

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

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

\displaystyle \\

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

\displaystyle \text{i) } AP : PC \text{ii) } ar( \triangle PDC) : ar( \triangle ABC)

Answer:

\displaystyle \text{Given } AD is median

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

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

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

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

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

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

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

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

ii) from \displaystyle \text{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)}  

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

\displaystyle \\

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

Answer:

Consider \displaystyle \triangle AEG \text{ and } \triangle CGF

\displaystyle \angle AGE = \angle FGC

\displaystyle AE = EB \text{ and } EB = FC \Rightarrow AE = FC

Also since \displaystyle EF \parallel BC

\displaystyle \angle AEG = \angle GFC

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

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

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

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

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

\displaystyle \\

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

Answer:

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

\displaystyle AD \parallel BC . Join \displaystyle DB

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

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

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

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

\displaystyle \\

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

Answer:

Given, \displaystyle ar (\triangle CPQ) = 20 \ \text{cm}^2

We can prove \displaystyle \triangle CDP \sim \triangle BAP

\displaystyle \Rightarrow BP : PC = 1:2

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

\displaystyle \therefore ar (\triangle CPD) = 2^2 \times ar (\triangle CPQ) = 40 \ \text{cm}^2

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

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

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

\displaystyle \\

Question 43: In the adjoining figure, \displaystyle ABCD \text{ and } AEFG are two parallelograms. Prove that \displaystyle ar({\parallel}^{gm} ABCD) = ar({\parallel}^{gm} AEFG)

Answer:

Join \displaystyle BG

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

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

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

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

\displaystyle \\

Question 44: \displaystyle ABCD is a square. \displaystyle E \text{ and } F are mid points of the sides \displaystyle AB \text{ and } AD respectively. Prove that \displaystyle ar( \triangle CEF) = \frac{3}{8} ar( ABCD)

Answer:

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

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

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

Adding i) and ii)

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

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

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

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

\displaystyle \\

Question 45: A point \displaystyle D is taken on the sides \displaystyle BC and of \displaystyle \triangle ABC \text{ and } AD is produced to \displaystyle E such that \displaystyle AD = DE , prove that \displaystyle ar( \triangle BCE) = ar( \triangle ABC)

Answer:

Draw a line \displaystyle \parallel to \displaystyle AD

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

(They have equal bases and between the same parallels)

\displaystyle \text{Similarly, } ar( \triangle CDE) = ar( \triangle CDA) … … … … … ii)

Adding i) and ii) we get

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

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

\displaystyle \\

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

Answer:

Consider \displaystyle \triangle ABG \text{ and } \triangle CDE

\displaystyle DC = AB

\displaystyle \angle BAG = \angle CDE

\displaystyle \angle ABC = \angle DG

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

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

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

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

Now consider, \displaystyle \triangle ADH \text{ and } \triangle EFG

\displaystyle \angle DAH = \angle EFG

\displaystyle \angle ADH = \angle GEF

\displaystyle AD = EG

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

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

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

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

\displaystyle \\

Question 47: \displaystyle ABCD is a rectangle \displaystyle \text{ and } P is mid point of \displaystyle AB . \displaystyle DP is produced to meet \displaystyle CB at \displaystyle Q . Prove that \displaystyle ar (ABCD) = ar (\triangle DQC)

Answer:

Consider \displaystyle \triangle DAP \text{ and } \triangle BQP

\displaystyle \angle DAP = \angle BPQ

\displaystyle AP = PB

\displaystyle \angle DAP = \angle QBP

\displaystyle \therefore \triangle DAP \cong \triangle BQP

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

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

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

\displaystyle \\

Question 48: If each diagonal of a quadrilateral divides it into two triangles of equal areas, then prove that it is a parallelogram.

Answer:

\displaystyle AC divides \displaystyle ABCD into two equal halves

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

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

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

\displaystyle \Rightarrow AB \parallel DC

\displaystyle \text{Similarly, } AD \parallel BC

\displaystyle \therefore ABCD is a parallelogram.

\displaystyle \\