JEE Main 2022 June 24th Shift 1 Maths Question Paper with Answer Keys and Solutions

JEE Main Question Papers

Content-Type:

Text, Images, Videos and PDF

Exam:

JEE Main

Academic Session:

2024

Medium:

English Medium

Subject:

Maths

Year:

2022

Date and Month:

24th- June

Shift:

Shift-1 Morning

Available Material:

JEE Main Previous Year Question Papers with PDF

Access Frank Solutions for Maths Class 9 Chapter 21 Areas Theorems on Parallelograms

1. $$ABCD$$ is a parallelogram having an area of $$60{\mathrm{cm}}^2.P$$ is a point on $$CD$$. Calculate the area of $$\mathrm{△}APB$$.

Ans:

$$\mathrm{ar}(\mathrm{△}APB)={\displaystyle \frac{1}{2}}\times \mathrm{ar}($$ parallelogram $$ABCD)$$

(The area of a triangle is half of a parallelogram on the same base and between the same parallel lines)

\[\Rightarrow \operatorname{ar}(\triangle A P B)=\dfrac{1}{2} \times 60 \mathrm{~cm}^2 \\

\Rightarrow \operatorname{ar}(\triangle A P B)=30 \mathrm{~cm}^2\]

2. $$PQRS$$ is a rectangle in which $$PQ=12\mathrm{cm}$$ and $$PS=8\mathrm{cm}$$. Calculate the area of $$\mathrm{△}PRS$$.

Ans:

Since $$PQRS$$ is a rectangle, therefore $$PQ=SR$$.

\[S R=12 \mathrm{~cm} \\

P S=8 \mathrm{~cm} \\

\Rightarrow \operatorname{ar}(\triangle P R S)=\dfrac{1}{2} \times \text { base } x \text { height } \\

\Rightarrow \operatorname{ar}(\triangle P R S)=\dfrac{1}{2} \times S R \times P S \\

\Rightarrow \operatorname{ar}(\triangle P R S)=\dfrac{1}{2} \times 12 \times 8 \\

\Rightarrow \operatorname{ar}(\triangle P R S)=48 \mathrm{~cm}^2\]

3. In the figure, $$PT$$ is parallel to $$SR$$. QTSR is a parallelogram and $$PQSR$$ is a rectangle. If the area of $$\mathrm{△}$$ QTS is $$60{\mathrm{cm}}^2$$, find:

(i) the area of $$\Vert \mathrm{gm}$$ QTSR

(ii) the area of the rectangle PQRS

(iii) the area of the triangle PQS.

Ans:

(i) $$\mathrm{ar}(\mathrm{△}QTS)={\displaystyle \frac{1}{2}}\times \mathrm{ar}($$ parallelogram QTSR $$)$$

(The area of a triangle is half of a parallelogram on the same base and between the same parallel lines)

$$\Rightarrow \mathrm{ar}$$ (parallelogram QTSR) $$=2\times \mathrm{ar}(\mathrm{△}$$ QTS $$)$$

\[\Rightarrow \operatorname{ar}(\text { parallelogram QTSR })=2 \times 60 \mathrm{~cm}^2 \\

\Rightarrow \operatorname{ar}(\text { parallelogram QTSR })=120 \mathrm{~cm}^2 \\

\text { (ii) } \operatorname{ar}(\triangle Q T S)=\dfrac{1}{2} \times \operatorname{ar}(\text { parallelogram QTSR) } \\

\operatorname{ar}(\triangle Q T S)=\operatorname{ar}(\triangle R S Q)=60 \mathrm{~cm}^2\]

Now,

\[\operatorname{ar}(\triangle R S Q)=\dfrac{1}{2} \times \operatorname{ar}(\text { rectangle } P Q R S) \\

\Rightarrow \operatorname{ar}(\text { rectangle } P Q R S)=2 \times \operatorname{ar}(\triangle R S Q) \\

\Rightarrow \operatorname{ar}(\text { rectangle } P Q R S)=2 \times 60 \mathrm{~cm}^2 \\

\Rightarrow \operatorname{ar}(\text { rectangle } P Q R S)=120 \mathrm{~cm}^2\]

(iii) Since PQRS is a rectangle,

Therefore RS = PQ ......(i)

QTSR is a parallelogram,

Therefore, RS = QT ......(ii)

From (i) and (ii)

$$PQ=QT\dots \dots$$....(iii)

In $$\mathrm{△}PSQ$$ and $$\mathrm{△}QST$$

$$QS=QS$$

$$PQ=QT$$ (from (iii))

$$\mathrm{\angle }PQS=\mathrm{\angle }SQT={90}^{\circ }$$

Therefore, $$\mathrm{△}PSQ\cong \mathrm{△}QST$$

The area of two congruent triangles is equal.

Hence, $$\mathrm{ar}(\mathrm{△}PSQ)=\mathrm{ar}(\mathrm{△}\mathrm{QTS})=60{\mathrm{cm}}^2$$

4. In the given figure area of $$\Vert gmPQRS$$ is $$30{\mathrm{cm}}^2$$. Find the height of $$\Vert gmPQFE$$ if $$PQ=6\mathrm{cm}$$.

Ans: Area $$(\Vert gm$$ PQRS $$)=$$ Area $$(\mid$$ gm PQFE $$)$$

( IIgm on same base PQ and between same parallel lines)

$$\therefore$$ Area $$(\Vert gmPQFE)=30{\mathrm{cm}}^2$$

$$\Rightarrow$$ Base $$\times$$ Height $$=30$$

$$\Rightarrow 6\times$$ Height $$=30$$

$$\Rightarrow \text{ Height }={\displaystyle \frac{30}{6}}=5\mathrm{cm}$$

$$\therefore$$ The height of a parallelogram PQFE is $$5\mathrm{cm}$$.

5. In the given figure, $$PQRS$$ is a $$\Vert gm$$. A straight line through $$P$$ cuts $$SR$$ at point $$T$$ and $$QR$$ produced at $$N$$. Prove that area of triangle QTR is equal to the area of triangle STN.

Ans: Triangle PQT and parallelogram $$PQRS$$ are on the same base $$PQ$$ and are between the same parallel lines $$PQ$$ and $$\mathrm{SR}$$

$$\therefore$$ Area $$(\mathrm{△}PQT)={\displaystyle \frac{1}{2}}$$ Area $$($$ parallelogram $$PQRS)\dots$$ (i)

$$\mathrm{△}PSN$$ and parallelogram PQRS are on the same base $$PS$$ and between the same parallel lines $$PS$$ and $$QN$$.

$$\therefore \mathrm{Area}(\mathrm{△}PSN)={\displaystyle \frac{1}{2}}\mathrm{Area}($$ parallelogram $$PQRS)$$.....(ii)

Adding equations (i) and (ii), we get

$$\therefore$$ Area $$(\mathrm{△}PQT)+\mathrm{Area}(\mathrm{△}PSN)=\mathrm{Area}($$ parallelogram $$PQRS)$$

$$\Rightarrow$$ Area (quad. $$PSNQ)-\mathrm{Area}(\mathrm{△}QTN)=\mathrm{Area}($$ parallelogram PQRS)

$$\Rightarrow \mathrm{Area}($$ quad. PSNQ $$)-\mathrm{Area}(\mathrm{△}QTN)=\mathrm{Area}($$ quad. PSNQ $$)-\mathrm{Area}(\mathrm{△}SRN)$$

$$\Rightarrow \mathrm{Area}(\mathrm{△}QTN)=\mathrm{Area}(\mathrm{△}SRN)$$

Subtracting $$\mathrm{Area}(\mathrm{△}RTN)$$ from both the sides, we get

$$\mathrm{Area}(\mathrm{△}QTN)-\mathrm{Area}(\mathrm{\Delta }RTN)=\mathrm{Area}(\mathrm{\Delta }SRN)-\mathrm{Area}(\mathrm{\Delta }RTN)$$

$$\Rightarrow \mathrm{Area}(\mathrm{△}QTR)=\mathrm{Area}(\mathrm{\Delta }STN)$$

6. In the given figure, ST $$\Vert$$ PR. Prove that: area of quadrilateral $$PQRS=$$ area of $$\mathrm{△}PQT$$.

Ans: We have,

$$\mathrm{Area}(\mathrm{△}PSR)=\mathrm{Area}(\mathrm{△}PTR)$$

(Both the triangles on the same base PR and between the same

parallel lines $$PR$$ and $$\mathrm{ST}$$ )

Adding $$\mathrm{Area}(\mathrm{△}PQR)$$ on both sides, we get

$$\mathrm{Area}(\mathrm{△}PSR)+\mathrm{Area}(\mathrm{△}PQR)=\mathrm{Area}(\mathrm{△}PTR)+\mathrm{Area}(\mathrm{△}PQR)$$

$$\Rightarrow \mathrm{Area}($$ Quadrilateral $$PQRS)=\mathrm{Area}(\mathrm{△}PQT)$$

7. In the figure, $$ABCD$$ is a parallelogram, and $$APD$$ is an equilateral triangle of side $$80\mathrm{cm}$$, Calculate the area of the parallelogram $$ABCD$$.

Ans: \[\Rightarrow \operatorname{ar}(\triangle A P D)=\dfrac{\sqrt{3} s^2}{4} \\

\Rightarrow \operatorname{ar}(\triangle A P D)=\dfrac{\sqrt{3} \times 8^2}{4} \\

\Rightarrow \operatorname{ar}(\triangle A P D)=\dfrac{\sqrt{3} \times 64}{4} \\

\Rightarrow \operatorname{ar}(\triangle A P D)=\sqrt{3} \times 16 \\

\Rightarrow 16 \sqrt{3} \mathrm{~cm}^2 \operatorname{ar}(\triangle A P D) \\

\Rightarrow \dfrac{1}{2} \times \operatorname{ar}(\text { parallelogram } A B C D)\]

(The area of a triangle is half of a parallelogram on the same base and between the same parallel lines)

\[\Rightarrow \operatorname{ar}(\text { parallelogram } A B C D)=2 \times \operatorname{ar}(\triangle A P D) \\

\Rightarrow \operatorname{ar}(\text { parallelogram } A B C D)=2 \times 16 \sqrt{3} \mathrm{~cm}^2 \\

\Rightarrow \operatorname{ar}(\text { parallelogram } A B C D)=32 \sqrt{3} \mathrm{~cm}^2\]

8. In the figure, if the area of $$\Vert gmPQRS$$ is $$84{\mathrm{cm}}^2$$; find the area of

(i) $$\Vert gmPQMN$$

(ii) $$\mathrm{△}PQS$$

(iii) $$\mathrm{△}PQN$$

Ans: (i) Area of both rectangle and parallelogram on the same base is equal.

Here,

For rectangle $$\mathrm{PQMN}$$, base $$=PQ$$

For parallelogram $$\mathrm{PQRS}$$, base $$=\mathrm{PQ}$$

So, we get, Area of rectangle $$PQMN=$$ Area of parallelogram $$PQRS$$

Area of rectangle $$PQMN=84{\mathrm{cm}}^2$$

(ii) $$\mathrm{ar}(\mathrm{△}PQS)={\displaystyle \frac{1}{2}}\times \mathrm{ar}($$ parallelogram $$PQRS)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}PQS)={\displaystyle \frac{1}{2}}\times 84{\mathrm{cm}}^2$$

\[\Rightarrow \operatorname{ar}(\triangle P Q S)=42 \mathrm{~cm}^2 \\

\text { (iii) } \operatorname{ar}(\triangle P Q N)=\dfrac{1}{2} \times \operatorname{ar}(\text { rectangle } P Q M N) \\

\Rightarrow \operatorname{ar}(\triangle P Q N)=\dfrac{1}{2} \times 84 \mathrm{~cm}^2 \\

\Rightarrow \operatorname{ar}(\triangle P Q N)=42 \mathrm{~cm}^2\]

9. In the figure, $$PQR$$ is a straight line. $$SQ$$ is parallel to Tp. Prove that the quadrilateral $$PQST$$ is equal in area to the $$\mathrm{△}PSR$$.

Ans: In quadrilateral PQST,

$$\mathrm{ar}(\mathrm{△}PQS)={\displaystyle \frac{1}{2}}\times \mathrm{ar}($$ quadrilateral $$PQST)$$

$$\mathrm{ar}($$ quadrilateral $$PQST)=2\mathrm{ar}(\mathrm{△}PQS)\dots \dots (i)$$

In $$\mathrm{△}PSR$$,

$$\mathrm{ar}(\mathrm{△}PSR)=\mathrm{ar}(\mathrm{△}PQS)+\mathrm{ar}(\mathrm{△}QSR)$$

but $$\mathrm{ar}(\mathrm{△}PQS)=\mathrm{ar}(\mathrm{△}QSR)\dots ($$ since $$QS$$ is median as $$QS\Vert TP)$$

$$\mathrm{ar}(\mathrm{△}PSR)=2\mathrm{ar}(\mathrm{△}PQS)\dots \dots .(\mathrm{ii})$$

From (i) and (ii)

$$\mathrm{ar}($$ quadrilateral $$PQST)=\mathrm{ar}(\mathrm{△}PSR)$$

10. In the given figure, if $$AB\Vert DC\Vert FG$$ and $$AE$$ is a straight line. Also,

$$AD\Vert FC$$. Prove that: area of $$\Vert gmABCD=$$ area of $$\Vert gm$$ BFGE.

Ans: Joining $$AC$$ and $$FE$$, we get

Triangle $$AFC$$ and Triangle $$AFE$$ are on the same base $$AF$$ and both are between the same parallels $$AF$$ and $$CE$$.

$$\Rightarrow \mathrm{Area}(\mathrm{△}AFC)=\mathrm{Area}(\mathrm{△}AFE)$$

$$\Rightarrow \mathrm{Area}(\mathrm{△}ABF)+\mathrm{Area}(\mathrm{△}ABC)=\mathrm{Area}(\mathrm{△}ABF)+\mathrm{Area}(\mathrm{△}BFE)$$

$$\Rightarrow \mathrm{Area}(\mathrm{△}ABC)=\mathrm{Area}(\mathrm{△}BFE)$$

$$\Rightarrow {\displaystyle \frac{1}{2}}\mathrm{Area}($$ parallelogram $$ABCD)={\displaystyle \frac{1}{2}}$$ Area (parallelogram $$BFGE)$$

$$\Rightarrow A($$ parallelogram $$ABCD)=$$ Area (parallelogram $$BFGE)$$

11. In the given figure, the perimeter of parallelogram PQRS is $$42\mathrm{cm}$$. Find the lengths of $$PQ$$ and PS.

Ans: Area of $$\Vert gmPQRS=PQ\times 6$$

Also,

Area of $$\Vert gmPQRS=PS\times 8$$

\[\therefore P Q \times 6=P S \times 8 \\

\Rightarrow P Q=\dfrac{8 P S}{6} \\

\Rightarrow P Q=\dfrac{4 P S}{3} \ldots(i)\]

The perimeter of $$\Vert gmPQRS=PQ+QR+RS+PS$$

$$\Rightarrow 42=2PQ+2PS$$ (opposite sides of a parallelogram are equal)

$$\Rightarrow 21=PQ+PS$$

$$\Rightarrow {\displaystyle \frac{4PS}{3}}+PS=21$$ from $$(i)$$

$$\Rightarrow {\displaystyle \frac{4PS+3PS}{3}}=21$$

$$\Rightarrow$$ 7PS $$=63$$

$$\Rightarrow PS=9\mathrm{cm}$$

Now,

$$PQ={\displaystyle \frac{4PS}{3}}={\displaystyle \frac{4\times 9}{3}}=12\mathrm{cm}$$

$$\therefore PQ=12\mathrm{cm}$$ and $$PS=9\mathrm{cm}$$.

12. In the given figure, $$PT\Vert QR$$ and $$QT\Vert RS$$. Show that: area of $$\mathrm{△}PQR=$$ area of $$\mathrm{△}TQS$$.

Ans: By joining TR, we get

Triangle $$PQR$$ and Triangle $$QTR$$ are on the same base $$QR$$ and between the same parallel lines $$QR$$ and $$PT$$.

$$\therefore \mathrm{Ar}(\mathrm{△}PQR)=\mathrm{Ar}(\mathrm{△}QTR)\dots$$ (i)

$$\mathrm{△}QTR$$ and $$\mathrm{△}TQS$$ are on the same base QT and between the same parallel lines

QT and RS.

$$\therefore Ar(\mathrm{△}QTR)=\mathrm{Ar}(\mathrm{△}TQS)\dots (ii)$$

From (i) and (ii), we get

$$\mathrm{Ar}(\mathrm{△}PQR)=\mathrm{Ar}(\mathrm{△}TQS)$$

13. In the given figure, $$\mathrm{△}PQR$$ is right-angled at $$P.\mathrm{PABQ}$$ and $$QRST$$ are squares on the side $$PQ$$ and hypotenuse $$QR$$. If $$PN\perp TS$$, show that:

(a) $$\mathrm{△}QRB\cong \mathrm{△}PQT$$

(b) Area of square $$\mathrm{PABQ}=$$ area of rectangle QTNM.

Ans: \[\angle B Q R=\angle B Q P+\angle P Q R \\

\Rightarrow \angle B Q R=90^{\circ}+\angle P Q R \\

\angle P Q T=\angle T Q R+\angle P Q R \\

\Rightarrow \angle P Q T=90^{\circ}+\angle P Q R \\

\Rightarrow \angle B Q R=\angle P Q T \ldots .(i) \\

\text { (a) In } \triangle Q R B \text { and } \triangle P Q T, \\

B Q=P Q \ldots .(\text { sides of a square } P A B Q \text { ) } \\

Q R=Q T \ldots \text { (sides of a square QRST) } \\

\angle B Q R=\angle P Q T \text { From (i) } \\

\therefore \triangle Q R B \cong \triangle P Q T \text { (by SAS congruence criterion) } \\

\Rightarrow A(\triangle B Q R)=\text { A( } \triangle P Q T) \ldots \text { (ii) }\]

(b) Triangle PQT and rectangle QTNM are on the same base QT and both are between the same parallel lines QT and PN.

$$\therefore A(\mathrm{△}PQT)={\displaystyle \frac{1}{2}}A($$ rect. QTNM $$)$$

$$\Rightarrow A($$ rect. QTNM $$)=2\times A(\mathrm{△}PQT)$$

$$\Rightarrow A($$ rect. QTNM $$)=2\times \mathrm{ar}(\mathrm{△}BQR)$$ From (ii) ... (iii)

Triangle $$BQR$$ and square PABQ are on the same base $$BQ$$ and both are between the same parallel lines $$BQ$$ and $$AR.\therefore 2\times A(\mathrm{△}BQR)=A$$ (sq. PABQ) .... (iv)

From (iii) and (iv),

$$A($$ sq. $$\mathrm{PABQ})=A($$ rect. QTNM $$)$$

14. In the figure, $$AB=BE$$. Prove that the area of triangle $$ACE$$ is equal in area to the parallelogram $$ABCD$$.

Ans: In parallelogram $$ABCD$$,

$$\mathrm{ar}(\mathrm{△}ABC)={\displaystyle \frac{1}{2}}\times \mathrm{ar}($$ parallelogram $$ABCD)$$

(The area of a triangle is half that of a parallelogram that lies on the same base

and between the same parallels)

$$\mathrm{ar}($$ parallelogram $$ABCD)=2\mathrm{ar}(\mathrm{△}ABC)\dots \dots (i)$$

In $$\mathrm{△}ACE$$,

$$\mathrm{ar}(\mathrm{△}ACE)=\mathrm{ar}(\mathrm{△}ABC)+\mathrm{ar}(\mathrm{△}BCE)$$

but $$\mathrm{ar}(\mathrm{△}ABC)=\mathrm{ar}(\mathrm{△}BCE)$$ (since $$BC$$ is median)

$$\mathrm{ar}(\mathrm{△}ACE)=2\mathrm{ar}(\mathrm{△}ABC)$$.........(ii)

From (i) and (ii)

$$\mathrm{ar}($$ parallelogram $$ABCD)=\mathrm{ar}(\mathrm{△}ACE)$$

15. The diagonals of a parallelogram $$ABCD$$ intersect at $$O$$. A line through $$O$$ meets $$AB$$ in $$P$$ and $$CD$$ in $$Q$$. Show that

(a) Area of $$APQD={\displaystyle \frac{1}{2}}$$ area of $$\Vert gmABCD$$

(b) Area of $$APQD=$$ Area of $$BPQC$$

Ans:

(a) A diagonal divides a parallelogram into two triangles of equal areas.

$$\Rightarrow A(\mathrm{△}ADB)={\displaystyle \frac{1}{2}}A(\Vert gmABCD)\dots$$ (i)

In $$\mathrm{△}DOQ$$ and $$\mathrm{△}BOP$$,

$$\mathrm{\angle }DOQ=\mathrm{\angle }BOP$$ (Vertically opposite)

$$DO=BO$$ (Diagonals of a $$\Vert \mathrm{gm}$$ )

$$\mathrm{\angle }ODQ=\mathrm{\angle }OBP$$ (Alternate angles)

$$\therefore \mathrm{△}DOQ\cong \mathrm{△}BOP$$ (ASA test of congruency)

$$\Rightarrow Ar(\mathrm{△}DOQ)=\mathrm{Ar}(\mathrm{△}BOP)$$

Adding $$\mathrm{Ar}(DOPA)$$ on both sides, we get

$$Ar(\mathrm{△}DOQ)+\mathrm{Ar}(DOPA)=\mathrm{Ar}(\mathrm{△}BOP)+\mathrm{Ar}(DOPA)$$

Area of $$APQD=\mathrm{Ar}(\mathrm{△}ADB)$$

$$\Rightarrow$$ Area of $$APQD={\displaystyle \frac{1}{2}}\mathrm{Ar}(\Vert gmABCD)($$ From $$(i))\dots$$ (ii)

(b) $$Ar(\mathrm{△}ABC)={\displaystyle \frac{1}{2}}Ar(\Vert gmABCD)$$

In $$\mathrm{△}COQ$$ and $$\mathrm{△}AOP$$,

$$\mathrm{\angle }COQ=\mathrm{\angle }AOP$$ (Vertically opposite angles)

$$CO=AO$$ (Diagonals of a $$\Vert \mathrm{gm}$$ bisect

$$\mathrm{\angle }OCQ=\mathrm{\angle }OAP$$ (Alternate angles)

$$\therefore \mathrm{△}COQ\cong \mathrm{△}AOP$$ (ASA test of congruency)

$$\Rightarrow Ar(\mathrm{△}COQ)=Ar(\mathrm{△}AOP)$$

Adding $$\mathrm{Ar}(COPB)$$ on both sides, we get

$$Ar(\mathrm{△}COQ)+\mathrm{Ar}(COPB)=\mathrm{Ar}(\mathrm{△}AOP)+\mathrm{Ar}(COPB)$$

$$\therefore$$ Area of $$BPQC=A(\mathrm{△}ABC)$$

$$\Rightarrow$$ Area of $$BPQC={\displaystyle \frac{1}{2}}A(\Vert gmABCD)$$... (iii)

$$\therefore$$ Area of $$APQD=$$ Area of BPQC From (ii) and (iii)

16. Prove that the median of a triangle divides it into two triangles of equal area.

Ans:

Draw AL perpendicular to $$BC$$.

Since $$AD$$ is the median of Triangle $$ABC$$. Therefore, $$D$$ is the mid-point line of $$BC$$.

\[\Rightarrow B D=D C \\

\Rightarrow B D \times A L=D C \times A L(\text { multiplying by } A L) \\

\Rightarrow \dfrac{1}{2}(B D \times A L)=\dfrac{1}{2}(D C \times A L) \\

\Rightarrow \operatorname{ar}(\triangle A B D)=\operatorname{ar}(\triangle A D C)\]

17. $$AD$$ is a median of a $$\mathrm{△}ABC.P$$ is any point on $$AD$$. Show that the area of $$\mathrm{△}ABP$$ is equal to the area of $$\mathrm{△}ACP$$.

Ans:

$$\mathrm{AD}$$ is the median of $$\mathrm{△}ABC$$. So, it will divide Triangle $$ABC$$ into two triangles of equal areas.

Therefore, Area $$(\mathrm{△}ABD)=\mathrm{area}(\mathrm{△}ACD)\dots$$ (1)

Now PD is the median of $$\mathrm{△}PBC$$.

Therefore, Area $$(\mathrm{△}PBD)=$$ area $$(\mathrm{△}PCD)$$...(2)

Subtracting equation (2) from equation (1), we get

$$\mathrm{Area}(\mathrm{△}ABD)-\mathrm{area}(\mathrm{△}PBD)=\mathrm{Area}(\mathrm{△}ACD)-\mathrm{Area}(\mathrm{△}PCD)$$

$$\mathrm{Area}(\mathrm{△}ABP)=\mathrm{area}(\mathrm{△}ACP)$$

18. In the given figure $$AF=BF$$ and $$DCBF$$ is a parallelogram. If the area of $$\mathrm{△}ABC$$ is 30 square units, find the area of the parallelogram DCBF.

Ans: \[\text { In } \triangle A B C \\

A F=F B \text { and } E F \| B C \ldots \text { (given) } \\

\therefore A E=E C \ldots \text { (Converse of Midpoint theorem) ...(i) } \\

\text { In } \triangle A E F \text { and } \triangle C E D, \\

\angle F E A=\angle D E C \ldots \text { (Vertically opposite angles) } \\

C E=A E \text {...From (i) } \\

\angle F A E=\angle D C E \ldots \text { (Alternate angles) } \\

\therefore \triangle F A E \cong \triangle C E D \ldots(\text { ASA test of congruency) } \\

\Rightarrow A r(\triangle A E F)=A r(\triangle C E D) \ldots . \text { (ii) } \\

A r(\triangle A B C)=A r(\triangle A E F)+A r(E F B C) \\

=A r(\triangle C E D)+A r(E F B C) \ldots . \text { (From (ii)) } \\

\therefore A(\triangle A B C)=A\left(\|^{g m} D C B F\right) \\

\Rightarrow A\left(\|^{g m} D C B F\right)=30 \text { sq. units. }\]

19. Prove that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans:

The diagonals of the parallelogram bisect each other.

So, $$O$$ is the mid-point of lines $$AC$$ and $$BD$$.

$$BO$$ is the median in $$\mathrm{△}ABC$$.

Therefore, it will divide two triangles into equal areas.

$$\therefore \mathrm{ar}(\mathrm{△}AOB)=\mathrm{ar}(\mathrm{△}BOC)\dots ..(i)$$

In $$\mathrm{△}BCD,CO$$ is the median.

From (i), (ii), and (iii)

$$\mathrm{ar}(\mathrm{△}AOB)=\mathrm{ar}(\mathrm{△}BOC)=\mathrm{ar}(\mathrm{△}COD)=\mathrm{ar}(\mathrm{△}AOD)$$

Hence, the diagonals of a parallelogram divide it into four triangles of equal areas.

20. The diagonals $$AC$$ and $$BC$$ of a quadrilateral $$ABCD$$ intersect at $$O$$. Prove that if $$BO=OD$$, then areas of $$\mathrm{△}ABC$$ and $$\mathrm{△}ADC$$ are equal.

Ans:

In $$\mathrm{△}ABD$$,

$$BO=OD$$

$$\Rightarrow O$$ is the mid-point of $$BD$$

$$\Rightarrow AO$$ is a median.

$$\Rightarrow \mathrm{ar}(\mathrm{△}AOB)=\mathrm{ar}(\mathrm{△}AOD)\dots (\mathrm{i})$$

In $$\mathrm{△}CBD,O$$ is the mid-point of $$BD$$

$$\Rightarrow \mathrm{CO}$$ is a median.

$$\Rightarrow \mathrm{ar}(\mathrm{△}COB)=\mathrm{ar}(\mathrm{△}COD)$$...(ii)

Adding (i) and (ii)

$$\mathrm{ar}(\mathrm{△}AOB)+\mathrm{ar}(\mathrm{△}COB)=\mathrm{ar}(\mathrm{△}AOD)+\mathrm{ar}(\mathrm{△}COD)$$

Therefore, $$\mathrm{ar}(\mathrm{△}ABC)=\mathrm{ar}(\mathrm{△}ADC)$$

21. Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.

Ans:

The diagonals of a rhombus intersect each other at right angles,

Therefore, $$OB\perp AC$$ and $$OD\perp AC$$

Now, $$\mathrm{ar}($$ rhombus $$ABCD)=\mathrm{ar}(\mathrm{△}ABC)+\mathrm{ar}(\mathrm{△}ADC)$$

$$={\displaystyle \frac{1}{2}}(AC\times BO)+{\displaystyle \frac{1}{2}}(AC\times DO)$$

$$={\displaystyle \frac{1}{2}}\{AC\times (BO+DO)\}$$

$$={\displaystyle \frac{1}{2}}(AC\times BD)$$

Therefore, the area of a rhombus is equal to half the rectangle contained by its diagonals.

22. PQRS is a parallelogram and $$O$$ is any point in its interior. Prove that: area( $$\mathrm{△}POQ)+\mathrm{area}(\mathrm{△}ROS)-\mathrm{area}(\mathrm{△}QOR)+\mathrm{area}(\mathrm{△}SOP)={\displaystyle \frac{1}{2}}\mathrm{area}(\Vert gmPPRS)$$

Ans:

Let us draw a line $$KL$$ which passes through point $$O$$ and parallel to line $$PQ$$.

In parallelogram $$PQRS$$,

$$PQ\Vert KL$$ (By construction) ... (1)

$$\mathrm{PQRS}$$ is a parallelogram.

$$\therefore PS\Vert QR$$ (Opposite sides of a parallelogram)

$$\Rightarrow PK\Vert QL$$...(2)

From equations (1) and (2), we obtain

$$PQ\Vert KL$$ and $$PK\Vert QL$$

Therefore, quadrilateral PQLK is a parallelogram.

It is clear that $$\mathrm{△}POQ$$ and parallelogram PQLK are lying on the same base $$PQ$$ and between the same parallel lines $$PK$$ and $$QL$$.

$$\therefore$$ Area $$(\mathrm{△}POQ)={\displaystyle \frac{1}{2}}$$ Area (parallelogram PQLK) ... (3)

Similarly, for $$\mathrm{△}ROS$$ and parallelogram KLRS,

Area $$(\mathrm{△}ROS)={\displaystyle \frac{1}{2}}$$ Area (parallelogram KLRS) ... (4)

Adding equations (3) and (4), we obtain

Area $$(\mathrm{△}POQ)+\mathrm{Area}(\mathrm{△}ROS)={\displaystyle \frac{1}{2}}\mathrm{Area}($$ parallelogram $$\mathrm{PQLK})+$$

$${\displaystyle \frac{1}{2}}$$ Area (parallelogram KLRS)

Area $$(\mathrm{△}POQ)+$$ Area $$(\mathrm{△}ROS)={\displaystyle \frac{1}{2}}$$ Area (PQRS) ......(5)

Let us draw a line segment $$MN$$ which is passing through point $$OP$$ and is parallel to line segment PS.

In parallelogram PQRS,

MN || PS (By construction) ... (6)

PQRS is a parallelogram.

$$\therefore PQ\Vert RS$$ (Opposite sides of a parallelogram)

$$\Rightarrow$$ PN $$\Vert SN\dots$$ (7)

From equations (6) and (7), we obtain

MN || PS and PN || SN

Therefore, quadrilateral PNMS is a parallelogram.

It can be clear that $$\mathrm{△}POS$$ and parallelogram PNMS are lying on the same base PS and between the same parallel lines $$PS$$ and $$MN$$.

$$\therefore$$ Area $$(\mathrm{△}SOP)={\displaystyle \frac{1}{2}}$$ Area (PNMS) ... (8)

Similarly, for $$\mathrm{△}QOR$$ and parallelogram MNQR,

Area $$(\mathrm{△}QOR)={\displaystyle \frac{1}{2}}$$ Area $$(MNQR)\dots (9)$$

Adding equations (8) and (9), we obtain

$$\mathrm{Area}(\mathrm{△}SOP)+$$ Area $$(\mathrm{△}QOR)={\displaystyle \frac{1}{2}}$$ Area $$(PNMS)+{\displaystyle \frac{1}{2}}$$ Area $$(MNQR)$$

$$\mathrm{Area}(\mathrm{△}SOP)+\mathrm{Area}(\mathrm{△}QOR)={\displaystyle \frac{1}{2}}$$ Area $$(PQRS)$$

By comparing (5) and (10), we obtain

$$\mathrm{Area}(\mathrm{△}POQ)+\mathrm{Area}(\mathrm{△}ROS)=\mathrm{Area}(\mathrm{△}SOP)+\mathrm{Area}(\mathrm{△}QOR)=$$ $${\displaystyle \frac{1}{2}}\mathrm{Area}($$ || gmPQRS)

23. A quadrilateral $$ABCD$$ is such that diagonals $$BD$$ divide its area into two equal parts. Prove that $$\mathrm{BD}$$ bisects $$\mathrm{AC}$$.

Ans:

Join $$AC$$. Suppose $$AC$$ and $$BD$$ intersect at $$O$$. Draw lines $$AL$$ and $$CM$$ both perpendicular to $$BD$$.

$$\mathrm{ar}(\mathrm{△}ABD)=\mathrm{ar}(\mathrm{△}BDC)$$

Thus $$\mathrm{△}ABD$$ and $$\mathrm{△}ABC$$ are on the same base $$AB$$ and have equal areas.

Therefore, their corresponding altitudes $$AL=CM$$ are equal.

Now, in $$\mathrm{△}ALO$$ and $$\mathrm{△}CMO$$,

$$\mathrm{\angle }1=\mathrm{\angle }2$$ (vertically opposite angles)

$$\mathrm{\angle }ALO=\mathrm{\angle }CMO$$ (right angles)

$$AL=CM$$

Therefore, $$\mathrm{△}ALO\cong \mathrm{△}CMO$$ (AAS axiom)

$$\Rightarrow AO=OC$$

$$\Rightarrow BD$$ bisects $$AC$$

24. In the given figure, $$BC\Vert DE$$.

(a) If the area of $$\mathrm{△}ADC$$ is $$20sq$$. units, find the area of $$\mathrm{△}AEB$$.

(b) If the area of $$\mathrm{△}BFD$$ is 8 square units, find the area of $$\mathrm{△}CEF$$

Ans: (a) Triangles on the same base and between the same parallels are equal in area. $$\therefore Ar(\mathrm{△}DBC)=\mathrm{Ar}(\mathrm{△}ECB)$$.... (i)

Now,

$$Ar(\mathrm{△}ABC)=Ar(\mathrm{△}ADC)+\mathrm{Ar}(\mathrm{△}DBC)=\mathrm{Ar}(\mathrm{△}AEB)+\mathrm{Ar}(\mathrm{△}ECB)$$

$$\Rightarrow Ar(\mathrm{△}ADC)+\mathrm{Ar}(\mathrm{△}DBC)=\mathrm{Ar}(\mathrm{△}AEB)+\mathrm{Ar}(\mathrm{△}ECB)$$

$$\Rightarrow Ar(\mathrm{△}ADC)=\mathrm{Ar}(\mathrm{△}AEB)$$.... (ii) from (i)

Given, $$\mathrm{Ar}(\mathrm{△}ADC)=20$$ sq. units

$$\Rightarrow Ar(\mathrm{△}AEB)=20$$ sq. units

(b)

$$\mathrm{Ar}(\mathrm{△}ADC)=\mathrm{Ar}(\mathrm{△}AEB)\dots$$ From (ii)

$$\Rightarrow Ar(\mathrm{△}ADC)-Ar(ADFE)=Ar(\mathrm{△}AEB)-Ar(ADFE)$$

$$\Rightarrow Ar(\mathrm{△}CEF)=Ar(\mathrm{△}BFD)$$

Given, $$Ar(\mathrm{△}BFD)=8$$ sq. units

$$\Rightarrow Ar(\mathrm{△}CEF)=8sq$$. units

25. In the given figure, $$AB\Vert SQ\Vert DC$$ and $$AD\Vert PR\Vert BC$$. If the area of

quadrilateral $$ABCD$$ is 24 square units, find the area of quadrilateral $$PQRS$$.

Ans: Let $$SQ$$ and $$PR$$ intersect at point $$O$$.

Now,

$$DC\Vert SQ$$ and $$RP\Vert BC$$

$$\Rightarrow RC\Vert OQ$$ and $$RO\Vert QC$$

$$\Rightarrow$$ Quadrilateral ROQC is a parallelogram.

Similarly,

Quad rilaterals ROSD, APOS and POQB are parallelograms.

$$\mathrm{△}ROQ$$ and parallelogram ROQC are on the same base and between the same

parallel lines.

$$\therefore A(\mathrm{△}ROQ)={\displaystyle \frac{1}{2}}\times A\left({\Vert }^{gm}ROQC\right)\dots ($$ i)

Similarly,

$$A(\mathrm{△}POQ)={\displaystyle \frac{1}{2}}\times A\left({\Vert }^{gm}POQB\right)\dots$$ (ii)

$$A(\mathrm{△}POS)={\displaystyle \frac{1}{2}}\times A\left({\Vert }^{gm}APOS\right)\dots ($$ iii)

$$A(\mathrm{△}SOR)={\displaystyle \frac{1}{2}}\times A\left({\Vert }^{gm}ROSD\right)\dots$$ (iv)

Adding equations (i), (ii), (iii) and (iv), we get

$$A(\mathrm{△}ROQ)+A(\mathrm{△}POQ)+A(\mathrm{△}POS)+A(\mathrm{△}SOR)$$

$$={\displaystyle \frac{1}{2}}[A\left({\Vert }^{gm}ROQC\right)+A\left({\Vert }^{gm}POQB\right)+A\left({\Vert }^{gm}APOS\right)+A\left({\Vert }^{gm}ROSD\right)$$

$$={\displaystyle \frac{1}{2}}\times 24$$

$$=12sq$$. units

26. $$\mathrm{△}PQR$$ and $$\mathrm{△}SQR$$ are on the same base $$QR$$ with $$P$$ and $$S$$ on opposite sides of line $$QR$$, such that area of $$\mathrm{△}PQR$$ is equal to the area of $$\mathrm{△}SQR$$. Show that $$QR$$ bisects $$PS$$.

Ans:

Join PS. Suppose PS and QR intersect at O. Draw PM and SN both perpendicular to $$\mathrm{QR}$$.

$$\mathrm{ar}(\mathrm{△}PQR)=\mathrm{ar}(\mathrm{△}SQR)$$

Thus $$\mathrm{△}PQR$$ and $$\mathrm{△}SQR$$ are on the same base $$QR$$ and have equal areas.

Therefore, their corresponding altitudes $$PM=SN$$ are equal.

Now, in $$\mathrm{△}PMO$$ and $$\mathrm{△}SNO$$,

$$\mathrm{\angle }1=\mathrm{\angle }2$$ (vertically opposite angles)

$$\mathrm{\angle }PMO=\mathrm{\angle }SNO$$ (right angles)

$$PM=SN$$

Therefore, $$\mathrm{△}PMO\cong \mathrm{△}SNO$$ (AAS axiom)

$$\Rightarrow PO=OS$$

$$\Rightarrow QR$$ bisects $$PS$$

27. If the medians of a $$\mathrm{△}ABBC$$ intersect at $$G$$, show that

$$\mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}AGC)=\mathrm{ar}(\mathrm{△}BGC)={\displaystyle \frac{1}{3}}\mathrm{ar}(\mathrm{△}ABC)$$

Ans:

Property: The median of a triangle divides it into two triangles of equal areas.

In $$\mathrm{△}ABC,AD$$ is the median

$$\Rightarrow \mathrm{ar}(\mathrm{△}ABD)=\mathrm{ar}(\mathrm{△}ACD)\dots (i)$$

In $$\mathrm{△}GBC,GD$$ is the median

$$\Rightarrow \mathrm{ar}(\mathrm{\Delta }GBD)=\mathrm{ar}(\mathrm{\Delta }GCD)$$...(ii)

Subtracting (ii) from (i),

$$\mathrm{ar}(\mathrm{△}ABD)-\mathrm{ar}(\mathrm{△}GBD)=\mathrm{ar}(\mathrm{△}ACD)-\mathrm{ar}(\mathrm{△}GCD)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}AGC)\dots \dots$$ (iii)

Similarly, $$\mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}BGC)\dots$$ (iv)

From (iii) and (iv),

$$\mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}BGC)=\mathrm{ar}(\mathrm{△}AGC)\dots \dots (v)$$

But $$\mathrm{ar}(\mathrm{△}AGB)+\mathrm{ar}(\mathrm{△}BGC)+\mathrm{ar}(\mathrm{△}AGC)=\mathrm{ar}(\mathrm{△}ABC)$$

Therefore, $$3\mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}ABC)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}AGB)={\displaystyle \frac{1}{3}}\mathrm{ar}(\mathrm{△}ABC)$$

Hence, $$\mathrm{ar}(\mathrm{△}AGB)=\mathrm{ar}(\mathrm{△}AGC)=\mathrm{ar}(\mathrm{△}BGC)={\displaystyle \frac{1}{3}}\mathrm{ar}(\mathrm{△}ABC)$$.

28. In $$\mathrm{△}ABC$$, the mid-points of $$AB,BC$$ and $$AC$$ are $$P,Q$$ and $$R$$ respectively. Prove that $$BQRP$$ is a parallelogram and that its area is half of $$\mathrm{△}ABC$$.

Ans:

$$P$$ and $$R$$ are mid-points of $$AB$$ and $$AC$$ respectively given in the question.

Therefore, $$PR\Vert BC$$ and $$PR={\displaystyle \frac{1}{2}}BC$$........(i)

Also, $$Q$$ is the mid-point of $$BC$$,

$$\Rightarrow QC={\displaystyle \frac{1}{2}}BC\text{...(ii) }$$

From (i) and (ii)

$$PR\Vert BC$$ and $$PR=QC$$

$$\Rightarrow PR\Vert QC$$ and $$PR=QC$$

$$Q$$ and $$R$$ are also mid-points of $$BC$$ and $$AC$$ respectively as given.

Therefore, $$QR\Vert BP$$ and $$QR=BP$$

Hence, $$BQRP$$ is a parallelogram.

$$\Rightarrow PQ$$ is a diagonal of $$\Vert gmBQRP$$

$$\mathrm{ar}(\mathrm{△}PQR)=\mathrm{ar}(\mathrm{△}BQP)\dots (v)$$ (as diagonal of a $$\Vert$$ gm divides it into two

triangles of equal areas)

Similarly, QCRP and QRAP are $$\Vert gm$$ and

$$\mathrm{ar}(\mathrm{△}PQR)=\mathrm{ar}(\mathrm{△}QCR)=\mathrm{ar}(\mathrm{△}APR)\dots (\mathrm{vi})$$

From (v) and (vi)

$$\mathrm{ar}(\mathrm{△}PQR)=\mathrm{ar}(\mathrm{△}BQP)=\mathrm{ar}(\mathrm{△}QCR)=\mathrm{ar}(\mathrm{△}APR)$$

Now, $$\mathrm{ar}(\mathrm{△}ABC)=\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}BQP)+\mathrm{ar}(\mathrm{△}QCR)+\mathrm{ar}(\mathrm{△}APR)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}ABC)=\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}PQR)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}ABC)=4\mathrm{ar}(\mathrm{△}PQR)$$

$$\Rightarrow \mathrm{ar}(\mathrm{△}PQR)={\displaystyle \frac{1}{4}}\mathrm{ar}(\mathrm{△}ABC)$$...(vii $$)$$

$$\mathrm{ar}(\Vert gmBQRP)=\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}BQP)$$

$$\Rightarrow \mathrm{ar}(\Vert \lg mBRP)=\mathrm{ar}(\mathrm{△}PQR)+\mathrm{ar}(\mathrm{△}PQR)($$ from $$(v))$$

$$\Rightarrow \mathrm{ar}(\Vert \lg BQRP)=2\mathrm{ar}(\mathrm{△}PQR)$$

$$\Rightarrow \mathrm{ar}(\Vert lgBQRP)=2\times {\displaystyle \frac{1}{4}}\mathrm{ar}(\mathrm{△}ABC)$$ (from (vii))

$$\Rightarrow \mathrm{ar}(\Vert gmBQRP)={\displaystyle \frac{1}{2}}\mathrm{ar}(\mathrm{△}ABC)$$

29. In the given figure, $$PQ\Vert SR\Vert MN,PS\Vert QM$$ and $$SM\Vert PN$$. Prove that: ar. $$(\mathrm{SMNT})=$$ ar. $$(\mathrm{PQRS})$$.

Ans: \[S M \| P N \\

\Rightarrow S M \| T N \\

\text { Also, SR | MN } \\

\Rightarrow S T \| M N\]

Hence, SMNT is a parallelogram.

\[S M \| P N \\

\Rightarrow S M \| P O \\

\text { Also, PS | QM } \\

\Rightarrow P S \| O M\]

Hence, SMOP is a parallelogram.

Now, parallelograms SMNT and SMOP are on the same base SM and both are between the same parallel lines SM and PN.

$$\therefore$$ Area $$($$ parallelogram SMNT) $$=$$ Area (parallelogram SMOP) ....(i)

Similarly, we can prove that quadrilaterals $$PQRS$$ is a parallelograms.

Now, parallelograms PQRS and SMOP are on the same base PS, and both are between the same parallel lines $$\mathrm{PS}$$ and $$\mathrm{QM}$$.

$$\therefore A($$ parallelogram $$PQRS)=A($$ parallelogram $$\mathrm{SMOP})$$....(ii)

From (i) and (ii), we have

$$A($$ parallelogram SMNT $$)=A($$ parallelogram $$PQRS)$$

31. In $$\mathrm{△}PQR,PS$$ is a median. $$T$$ is the mid-point of $$SR$$ and $$M$$ is the mid-point of $$PT$$. Prove that: $$\mathrm{△}PMR={\displaystyle \frac{1}{8}}\mathrm{△}PQR$$.

Ans:

Area $$(\mathrm{△}PQR)=\mathrm{area}(\mathrm{△}PQS)+\mathrm{area}(\mathrm{△}PSR)\dots (i)$$

Since $$PS$$ is the median of $$\mathrm{△}PQR$$ and median divides a triangle into two triangles of equal area.

Therefore, area $$(\mathrm{△}PQS)=\mathrm{area}(\mathrm{△}PSR)$$...(ii)

Substituting in (i)

Area $$(\mathrm{△}PQR)=\mathrm{area}(\mathrm{△}PSR)+$$ area $$(\mathrm{△}PSR)$$ Area $$(\mathrm{△}PQR)$$

32. In the figure, $$ABCD$$ is a parallelogram and $$CP$$ is parallel to $$DB$$. Prove that: Area of $$OBPC={\displaystyle \frac{3}{4}}$$ area of $$ABCD$$

Ans: The diagonals of a parallelogram divide it into four triangles of equal area Therefore, the area of $$\mathrm{△}AOD=$$ area $$\mathrm{△}BOC=$$ area $$\mathrm{△}ABO=$$ area $$\mathrm{△}CDO$$. $$\Rightarrow$$ area $$\mathrm{△}BOC={\displaystyle \frac{1}{4}}$$ area $$(\lg mABCD)$$........(i)

In $$\Vert \mathrm{gm}ABCD,BD$$ is the diagonal

Therefore, area $$(\mathrm{△}ABD)=\mathrm{area}(\mathrm{△}BCD)$$

$$\Rightarrow$$ area $$(\mathrm{△}BCD)={\displaystyle \frac{1}{2}}$$ area $$(\mid \mathrm{lgm}ABCD)\dots .$$. (ii)

In Parallelogram BPCD, BC is the diagonal

Therefore, area $$(\mathrm{△}BCD)=\mathrm{area}(\mathrm{△}BPC)\dots .$$. (iii)

From (iii) and (ii)

area $$(\mathrm{△}BPC)={\displaystyle \frac{1}{2}}$$ area $$(\Vert gmABCD)\dots \dots$$ (iv)

adding (i) and (iv)

$$\mathrm{area}(\mathrm{△}BPC)+$$ area $$\mathrm{△}BOC={\displaystyle \frac{1}{2}}$$ area $$(\Vert gmABCD)+{\displaystyle \frac{1}{4}}$$ area $$(\Vert gmABCD)$$

Area of $$OBPC={\displaystyle \frac{3}{4}}$$ area of $$ABCD$$

33. The medians $$QM$$ and $$RN$$ of $$\mathrm{△}PQR$$ intersect at $$O$$. Prove that: area of $$\mathrm{△}ROQ=$$ area of quadrilateral $$PMON$$.

Ans:

Join two points MN.

Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Therefore, $$MN\Vert QR$$

Clearly, $$\mathrm{△}QMN$$ and $$\mathrm{△}RNM$$ are on the same base $$MN$$ and between the same parallel lines.

Therefore, $$\mathrm{area}(\mathrm{△}QMN)=\mathrm{area}(\mathrm{△}RNM)$$

$$\Rightarrow \mathrm{Area}(\mathrm{△}QMN)-\mathrm{area}(\mathrm{△}ONM)=\mathrm{area}(\mathrm{△}RNM)-\mathrm{area}(\mathrm{△}ONM)$$

$$\Rightarrow \mathrm{area}(\mathrm{\Delta }QON)=\mathrm{area}(\mathrm{△}ROM)\dots \dots (i)$$

We know that a median of a triangle divides it into two triangles of equal areas.

Therefore, area $$(\mathrm{△}QMR)=\mathrm{area}(\mathrm{△}PQM)$$

$$\Rightarrow \mathrm{area}(\mathrm{△}ROQ)+\mathrm{area}(\mathrm{△}ROM)=\mathrm{area}($$ quad. $$PMON)+\mathrm{area}(\mathrm{△}QON)$$

$$\Rightarrow$$ area $$(\mathrm{△}ROQ)+\mathrm{area}(\mathrm{△}ROM)=$$ area (quad. PMON $$)+$$ area $$(\mathrm{△}ROM)$$ (from

(i))

$$\Rightarrow \mathrm{area}(\mathrm{△}ROQ)=\mathrm{area}($$ quad. $$PMON)$$