Class 9: Areas of Parallelograms and Triangles – Exercise 14.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

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 $PQ = 8 \ cm$ .

Answer:

$PQRS$ 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.

Answer:

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

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

$\therefore ar (PQRS)$

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

$\\$

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

Answer:

Since mid point of hypotenuse is equidistant from all three vertices

$AC = CB = OC = 6.5 \ cm$

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

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

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

$\\$

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:

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

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

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

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

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

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

$\\$

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.

Answer:

$OCDE$ is a rectangle.

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

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

$\\$

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

Answer:

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

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

$\\$

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:

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

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

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

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

$\\$

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

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)

$\therefore \triangle CLO \cong \triangle DMO$

$\Rightarrow CL = DM$

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

$\\$

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.

Answer:

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

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

We know $PM < DL$

$\Rightarrow AB \times PM < AB \times DL$

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

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

$\\$

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:

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

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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) $ar( \triangle APB) = ar( \triangle CBP)$

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

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

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

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

$\\$

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 ${\parallel}^{gm} ABCD$

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 $\triangle BCD$

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

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

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

$\\$

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:

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 $ABCD$ .

Answer:

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

Answer:

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

Answer:

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 $ar( BCZY) = ar( \triangle EDZ)$

Answer:

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

Answer:

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

Also $PQ = QR = RS$

Since $AP \parallel QB \parallel RC \parallel SD$

$AD = PS$ and $AD \parallel PS$

$PQ = CD$

$\angle EPQ = \angle FDC$

$\angle QEP = \angle CFD$

$\therefore \triangle PQE \cong \triangle CDF$

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

$\\$

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:

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

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)

$AB \parallel DE$

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

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

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

$\\$

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:

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

Answer:

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

Answer:

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

Answer:

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

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

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

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

$\\$

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

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.

$\\$

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) $ar( BCED ) = ar( ABMN) + ar(ACFG)$

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

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

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

$\\$

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

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, $SX \parallel RY$

$\\$

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 $AC$ 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 ${\parallel}^{gm} ABCD$

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

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

$\\$

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:

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

Answer:

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) $ar( \triangle OBD) = ar( \triangle OCE)$

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) $ar( \triangle OCE) = ar( \triangle OBD)$

$\\$

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

Answer:

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 $ar( \triangle AQC) : ar( \triangle ABC)$

Answer:

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

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

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

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

$\\$

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) $ar( \triangle PDC) : ar( \triangle ABC)$

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) $\frac{ar( \triangle ADC)}{ar( \triangle ADP)}$ $=$ $\frac{ar( \triangle ADP) + ar(\triangle PDC) }{ar( \triangle ADP)}$ $= 1 +$ $\frac{ar( \triangle PDC)}{ar( \triangle ADP)}$

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

$\\$

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 $ar({\parallel}^{gm} EBCF)$

Answer:

Consider $\triangle AEG$ and $\triangle CGF$

$\angle AGE = \angle FGC$

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

Also since $EF \parallel BC$

$\angle AEG = \angle GFC$

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

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

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

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

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

$\\$

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:

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

$AD \parallel BC$ . 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 $ar({\parallel}^{gm} ABCD)$ .

Answer:

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

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

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

$\\$

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

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)

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

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

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

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

$\\$

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

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

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

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

$\\$

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