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 $\displaystyle ABCD$ .

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, $\displaystyle TO \parallel P$

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

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

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

$\displaystyle \\$

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.

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

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

$\displaystyle \\$

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

$\displaystyle \\$

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)

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

$\displaystyle \Rightarrow CL = DM$

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

$\displaystyle \\$

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