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

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

1. \[A B C D\] is a parallelogram having an area of \[60 \mathrm{~cm}^2 . P\] is a point on \[C D\]. Calculate the area of \[\triangle A P B\].

Ans:

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

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

2. \[P Q R S\] is a rectangle in which \[P Q=12 \mathrm{~cm}\] and \[P S=8 \mathrm{~cm}\]. Calculate the area of \[\triangle P R S\].

Ans:

Since \[P Q R S\] is a rectangle, therefore \[P Q=S R\].

\[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, \[P T\] is parallel to \[S R\]. QTSR is a parallelogram and \[P Q S R\] is a rectangle. If the area of \[\triangle\] QTS is \[60 \mathrm{~cm}^2\], find:

(i) the area of \[\| \mathrm{gm}\] QTSR

(ii) the area of the rectangle PQRS

(iii) the area of the triangle PQS.

Ans:

(i) \[\operatorname{ar}(\triangle Q T S)=\dfrac{1}{2} \times \operatorname{ar}(\] parallelogram QTSR \[)\]

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

\[\Rightarrow \operatorname{ar}\] (parallelogram QTSR) \[=2 \times \operatorname{ar}(\triangle\] 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)

\[P Q=Q T \ldots \ldots\]....(iii)

In \[\triangle P S Q\] and \[\triangle Q S T\]

\[Q S=Q S\]

\[P Q=Q T\] (from (iii))

\[\angle P Q S=\angle S Q T=90^{\circ}\]

Therefore, \[\triangle P S Q \cong \triangle Q S T\]

The area of two congruent triangles is equal.

Hence, \[\operatorname{ar}(\triangle P S Q)=\operatorname{ar}(\triangle \mathrm{QTS})=60 \mathrm{~cm}^2\]

4. In the given figure area of \[\| g m P Q R S\] is \[30 \mathrm{~cm}^2\]. Find the height of \[\| g m P Q F E\] if \[P Q=6 \mathrm{~cm}\].

Ans: Area \[(\| g m\] PQRS \[)=\] Area \[(\mid\] gm PQFE \[)\]

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

\[\therefore\] Area \[(\| g m P Q F E)=30 \mathrm{~cm}^2\]

\[\Rightarrow\] Base \[\times\] Height \[=30\]

\[\Rightarrow 6 \times\] Height \[=30\]

\[\Rightarrow \text { Height }=\dfrac{30}{6}=5 \mathrm{~cm}\]

\[\therefore\] The height of a parallelogram PQFE is \[5 \mathrm{~cm}\].

5. In the given figure, \[P Q R S\] is a \[\| g m\]. A straight line through \[P\] cuts \[S R\] at point \[T\] and \[Q R\] produced at \[N\]. Prove that area of triangle QTR is equal to the area of triangle STN.

Ans: Triangle PQT and parallelogram \[P Q R S\] are on the same base \[P Q\] and are between the same parallel lines \[P Q\] and \[\mathrm{SR}\]

\[\therefore\] Area \[(\triangle P Q T)=\dfrac{1}{2}\] Area \[(\] parallelogram \[P Q R S) \ldots\] (i)

\[\triangle P S N\] and parallelogram PQRS are on the same base \[P S\] and between the same parallel lines \[P S\] and \[Q N\].

\[\therefore \operatorname{Area}(\triangle P S N)=\dfrac{1}{2} \operatorname{Area}(\] parallelogram \[P Q R S) \quad\].....(ii)

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

\[\therefore\] Area \[(\triangle P Q T)+\operatorname{Area}(\triangle P S N)=\operatorname{Area}(\] parallelogram \[P Q R S)\]

\[\Rightarrow\] Area (quad. \[P S N Q)-\operatorname{Area}(\triangle Q T N)=\operatorname{Area}(\] parallelogram PQRS)

\[\Rightarrow \operatorname{Area}(\] quad. PSNQ \[)-\operatorname{Area}(\triangle Q T N)=\operatorname{Area}(\] quad. PSNQ \[)-\operatorname{Area}(\triangle S R N)\]

\[\Rightarrow \operatorname{Area}(\triangle Q T N)=\operatorname{Area}(\triangle S R N)\]

Subtracting \[\operatorname{Area}(\triangle R T N)\] from both the sides, we get

\[\operatorname{Area}(\triangle Q T N)-\operatorname{Area}(\Delta R T N)=\operatorname{Area}(\Delta S R N)-\operatorname{Area}(\Delta R T N)\]

\[\Rightarrow \operatorname{Area}(\triangle Q T R)=\operatorname{Area}(\Delta S T N)\]

6. In the given figure, ST \[\|\] PR. Prove that: area of quadrilateral \[P Q R S=\] area of \[\triangle P Q T\].

Ans: We have,

\[\operatorname{Area}(\triangle P S R)=\operatorname{Area}(\triangle P T R)\]

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

parallel lines \[P R\] and \[\mathrm{ST}\] )

Adding \[\operatorname{Area}(\triangle P Q R)\] on both sides, we get

\[\operatorname{Area}(\triangle P S R)+\operatorname{Area}(\triangle P Q R)=\operatorname{Area}(\triangle P T R)+\operatorname{Area}(\triangle P Q R)\]

\[\Rightarrow \operatorname{Area}(\] Quadrilateral \[P Q R S)=\operatorname{Area}(\triangle P Q T)\]

7. In the figure, \[A B C D\] is a parallelogram, and \[A P D\] is an equilateral triangle of side \[80 \mathrm{~cm}\], Calculate the area of the parallelogram \[A B C D\].

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 \[\| g m P Q R S\] is \[84 \mathrm{~cm}^2\]; find the area of

(i) \[\| g m P Q M N\]

(ii) \[\triangle P Q S\]

(iii) \[\triangle P Q N\]

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

Here,

For rectangle \[\mathrm{PQMN}\], base \[=P Q\]

For parallelogram \[\mathrm{PQRS}\], base \[=\mathrm{PQ}\]

So, we get, Area of rectangle \[P Q M N=\] Area of parallelogram \[P Q R S\]

Area of rectangle \[P Q M N=84 \mathrm{~cm}^2\]

(ii) \[\operatorname{ar}(\triangle P Q S)=\dfrac{1}{2} \times \operatorname{ar}(\] parallelogram \[P Q R S)\]

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

\[\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, \[P Q R\] is a straight line. \[S Q\] is parallel to Tp. Prove that the quadrilateral \[P Q S T\] is equal in area to the \[\triangle P S R\].

Ans: In quadrilateral PQST,

\[\operatorname{ar}(\triangle P Q S)=\dfrac{1}{2} \times \operatorname{ar}(\] quadrilateral \[P Q S T)\]

\[\operatorname{ar}(\] quadrilateral \[P Q S T)=2 \operatorname{ar}(\triangle P Q S) \ldots \ldots(i)\]

In \[\triangle P S R\],

\[\operatorname{ar}(\triangle P S R)=\operatorname{ar}(\triangle P Q S)+\operatorname{ar}(\triangle Q S R)\]

but \[\operatorname{ar}(\triangle P Q S)=\operatorname{ar}(\triangle Q S R) \ldots(\] since \[Q S\] is median as \[Q S \| T P)\]

\[\operatorname{ar}(\triangle P S R)=2 \operatorname{ar}(\triangle P Q S) \ldots \ldots .(\mathrm{ii})\]

From (i) and (ii)

\[\operatorname{ar}(\] quadrilateral \[P Q S T)=\operatorname{ar}(\triangle P S R)\]

10. In the given figure, if \[A B\|D C\| F G\] and \[A E\] is a straight line. Also,

\[A D \| F C\]. Prove that: area of \[\| g m A B C D=\] area of \[\| g m\] BFGE.

Ans: Joining \[A C\] and \[F E\], we get

Triangle \[A F C\] and Triangle \[A F E\] are on the same base \[A F\] and both are between the same parallels \[A F\] and \[C E\].

\[\Rightarrow \operatorname{Area}(\triangle A F C)=\operatorname{Area}(\triangle A F E)\]

\[\Rightarrow \operatorname{Area}(\triangle A B F)+\operatorname{Area}(\triangle A B C)=\operatorname{Area}(\triangle A B F)+\operatorname{Area}(\triangle B F E)\]

\[\Rightarrow \operatorname{Area}(\triangle A B C)=\operatorname{Area}(\triangle B F E)\]

\[\Rightarrow \dfrac{1}{2} \operatorname{Area}(\] parallelogram \[A B C D)=\dfrac{1}{2}\] Area (parallelogram \[\left.B F G E\right)\]

\[\Rightarrow A(\] parallelogram \[A B C D)=\] Area (parallelogram \[B F G E)\]

11. In the given figure, the perimeter of parallelogram PQRS is \[42 \mathrm{~cm}\]. Find the lengths of \[P Q\] and PS.

Ans: Area of \[\| g m P Q R S=P Q \times 6\]

Also,

Area of \[\| g m P Q R S=P S \times 8\]

\[\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 \[\| g m P Q R S=P Q+Q R+R S+P S\]

\[\Rightarrow 42=2 P Q+2 P S\] (opposite sides of a parallelogram are equal)

\[\Rightarrow 21=P Q+P S\]

\[\Rightarrow \dfrac{4 P S}{3}+P S=21\] from \[(i)\]

\[\Rightarrow \dfrac{4 P S+3 P S}{3}=21\]

\[\Rightarrow\] 7PS \[=63\]

\[\Rightarrow P S=9 \mathrm{~cm}\]

Now,

\[P Q=\dfrac{4 P S}{3}=\dfrac{4 \times 9}{3}=12 \mathrm{~cm}\]

\[\therefore P Q=12 \mathrm{~cm}\] and \[P S=9 \mathrm{~cm}\].

12. In the given figure, \[P T \| Q R\] and \[Q T \| R S\]. Show that: area of \[\triangle P Q R=\] area of \[\triangle T Q S\].

Ans: By joining TR, we get

Triangle \[P Q R\] and Triangle \[Q T R\] are on the same base \[Q R\] and between the same parallel lines \[Q R\] and \[P T\].

\[\therefore \operatorname{Ar}(\triangle P Q R)=\operatorname{Ar}(\triangle Q T R) \quad \ldots\] (i)

\[\triangle Q T R\] and \[\triangle T Q S\] are on the same base QT and between the same parallel lines

QT and RS.

\[\therefore A r(\triangle Q T R)=\operatorname{Ar}(\triangle T Q S) \ldots(i i)\]

From (i) and (ii), we get

\[\operatorname{Ar}(\triangle P Q R)=\operatorname{Ar}(\triangle T Q S)\]

13. In the given figure, \[\triangle P Q R\] is right-angled at \[P . \mathrm{PABQ}\] and \[Q R S T\] are squares on the side \[P Q\] and hypotenuse \[Q R\]. If \[P N \perp T S\], show that:

(a) \[\triangle Q R B \cong \triangle P Q T\]

(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.

\[\therefore A(\triangle P Q T)=\dfrac{1}{2} A(\] rect. QTNM \[)\]

\[\Rightarrow A(\] rect. QTNM \[)=2 \times A(\triangle P Q T)\]

\[\Rightarrow A(\] rect. QTNM \[)=2 \times \operatorname{ar}(\triangle B Q R)\] From (ii) ... (iii)

Triangle \[B Q R\] and square PABQ are on the same base \[B Q\] and both are between the same parallel lines \[B Q\] and \[A R . \therefore 2 \times A(\triangle B Q R)=A\] (sq. PABQ) .... (iv)

From (iii) and (iv),

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

14. In the figure, \[A B=B E\]. Prove that the area of triangle \[A C E\] is equal in area to the parallelogram \[A B C D\].

Ans: In parallelogram \[A B C D\],

\[\operatorname{ar}(\triangle A B C)=\dfrac{1}{2} \times \operatorname{ar}(\] parallelogram \[A B C D)\]

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

and between the same parallels)

\[\operatorname{ar}(\] parallelogram \[A B C D)=2 \operatorname{ar}(\triangle A B C) \ldots \ldots(i)\]

In \[\triangle A C E\],

\[\operatorname{ar}(\triangle A C E)=\operatorname{ar}(\triangle A B C)+\operatorname{ar}(\triangle B C E)\]

but \[\operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle B C E)\] (since \[B C\] is median)

\[\operatorname{ar}(\triangle A C E)=2 \operatorname{ar}(\triangle A B C)\].........(ii)

From (i) and (ii)

\[\operatorname{ar}(\] parallelogram \[A B C D)=\operatorname{ar}(\triangle A C E)\]

15. The diagonals of a parallelogram \[A B C D\] intersect at \[O\]. A line through \[O\] meets \[A B\] in \[P\] and \[C D\] in \[Q\]. Show that

(a) Area of \[A P Q D=\dfrac{1}{2}\] area of \[\| g m A B C D\]

(b) Area of \[A P Q D=\] Area of \[B P Q C\]

Ans:

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

\[\Rightarrow A(\triangle A D B)=\dfrac{1}{2} A(\| g m A B C D) \quad \ldots\] (i)

In \[\triangle D O Q\] and \[\triangle B O P\],

\[\angle D O Q=\angle B O P\] (Vertically opposite)

\[D O=B O\] (Diagonals of a \[\| \mathrm{gm}\] )

\[\angle O D Q=\angle O B P\] (Alternate angles)

\[\therefore \triangle D O Q \cong \triangle B O P\] (ASA test of congruency)

\[\Rightarrow A r(\triangle D O Q)=\operatorname{Ar}(\triangle B O P)\]

Adding \[\operatorname{Ar}(D O P A)\] on both sides, we get

\[A r(\triangle D O Q)+\operatorname{Ar}(D O P A)=\operatorname{Ar}(\triangle B O P)+\operatorname{Ar}(D O P A)\]

Area of \[A P Q D=\operatorname{Ar}(\triangle A D B)\]

\[\Rightarrow\] Area of \[A P Q D=\dfrac{1}{2} \operatorname{Ar}(\| g m A B C D)(\] From \[(i)) \ldots\] (ii)

(b) \[A r(\triangle A B C)=\dfrac{1}{2} A r(\| g m A B C D)\]

In \[\triangle C O Q\] and \[\triangle A O P\],

\[\angle C O Q=\angle A O P\] (Vertically opposite angles)

\[C O=A O\] (Diagonals of a \[\| \mathrm{gm}\] bisect

\[\angle O C Q=\angle O A P\] (Alternate angles)

\[\therefore \triangle C O Q \cong \triangle A O P\] (ASA test of congruency)

\[\Rightarrow A r(\triangle C O Q)=A r(\triangle A O P)\]

Adding \[\operatorname{Ar}(C O P B)\] on both sides, we get

\[A r(\triangle C O Q)+\operatorname{Ar}(C O P B)=\operatorname{Ar}(\triangle A O P)+\operatorname{Ar}(C O P B)\]

\[\therefore\] Area of \[B P Q C=A(\triangle A B C)\]

\[\Rightarrow\] Area of \[B P Q C=\dfrac{1}{2} A(\| g m A B C D) \quad\]... (iii)

\[\therefore\] Area of \[A P Q D=\] 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 \[B C\].

Since \[A D\] is the median of Triangle \[A B C\]. Therefore, \[D\] is the mid-point line of \[B C\].

\[\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. \[A D\] is a median of a \[\triangle A B C . P\] is any point on \[A D\]. Show that the area of \[\triangle A B P\] is equal to the area of \[\triangle A C P\].

Ans:

\[\mathrm{AD}\] is the median of \[\triangle A B C\]. So, it will divide Triangle \[A B C\] into two triangles of equal areas.

Therefore, Area \[(\triangle A B D)=\operatorname{area}(\triangle A C D) \ldots\] (1)

Now PD is the median of \[\triangle P B C\].

Therefore, Area \[(\triangle P B D)=\] area \[(\triangle P C D)\]...(2)

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

\[\operatorname{Area}(\triangle A B D)-\operatorname{area}(\triangle P B D)=\operatorname{Area}(\triangle A C D)-\operatorname{Area}(\triangle P C D)\]

\[\operatorname{Area}(\triangle A B P)=\operatorname{area}(\triangle A C P)\]

18. In the given figure \[A F=B F\] and \[D C B F\] is a parallelogram. If the area of \[\triangle A B C\] 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 \[A C\] and \[B D\].

\[B O\] is the median in \[\triangle A B C\].

Therefore, it will divide two triangles into equal areas.

\[\therefore \operatorname{ar}(\triangle A O B)=\operatorname{ar}(\triangle B O C) \ldots . .(i)\]

In \[\triangle B C D, C O\] is the median.

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

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

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

20. The diagonals \[A C\] and \[B C\] of a quadrilateral \[A B C D\] intersect at \[O\]. Prove that if \[B O=O D\], then areas of \[\triangle A B C\] and \[\triangle A D C\] are equal.

Ans:

In \[\triangle A B D\],

\[B O=O D\]

\[\Rightarrow O\] is the mid-point of \[B D\]

\[\Rightarrow A O\] is a median.

\[\Rightarrow \operatorname{ar}(\triangle A O B)=\operatorname{ar}(\triangle A O D) \quad \ldots(\mathrm{i})\]

In \[\triangle C B D, O\] is the mid-point of \[B D\]

\[\Rightarrow \mathrm{CO}\] is a median.

\[\Rightarrow \operatorname{ar}(\triangle C O B)=\operatorname{ar}(\triangle C O D) \quad\]...(ii)

Adding (i) and (ii)

\[\operatorname{ar}(\triangle A O B)+\operatorname{ar}(\triangle C O B)=\operatorname{ar}(\triangle A O D)+\operatorname{ar}(\triangle C O D)\]

Therefore, \[\operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle A D C)\]

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, \[O B \perp A C\] and \[O D \perp A C\]

Now, \[\operatorname{ar}(\] rhombus \[A B C D)=\operatorname{ar}(\triangle A B C)+\operatorname{ar}(\triangle A D C)\]

\[=\dfrac{1}{2}(A C \times B O)+\dfrac{1}{2}(A C \times D O)\]

\[=\dfrac{1}{2}\{A C \times(B O+D O)\}\]

\[=\dfrac{1}{2}(A C \times B D)\]

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( \[\triangle P O Q)+\operatorname{area}(\triangle R O S)-\operatorname{area}(\triangle Q O R)+\operatorname{area}(\triangle S O P)=\dfrac{1}{2} \operatorname{area}(\| g m P P R S)\]

Ans:

Let us draw a line \[K L\] which passes through point \[O\] and parallel to line \[P Q\].

In parallelogram \[P Q R S\],

\[P Q \| K L\] (By construction) ... (1)

\[\mathrm{PQRS}\] is a parallelogram.

\[\therefore P S \| Q R\] (Opposite sides of a parallelogram)

\[\Rightarrow P K \| Q L\]...(2)

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

\[P Q \| K L\] and \[P K \| Q L\]

Therefore, quadrilateral PQLK is a parallelogram.

It is clear that \[\triangle P O Q\] and parallelogram PQLK are lying on the same base \[P Q\] and between the same parallel lines \[P K\] and \[Q L\].

\[\therefore\] Area \[(\triangle P O Q)=\dfrac{1}{2}\] Area (parallelogram PQLK) ... (3)

Similarly, for \[\triangle R O S\] and parallelogram KLRS,

Area \[(\triangle R O S)=\dfrac{1}{2}\] Area (parallelogram KLRS) ... (4)

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

Area \[(\triangle P O Q)+\operatorname{Area}(\triangle R O S)=\dfrac{1}{2} \operatorname{Area}(\] parallelogram \[\mathrm{PQLK})+\]

\[\dfrac{1}{2}\] Area (parallelogram KLRS)

Area \[(\triangle P O Q)+\] Area \[(\triangle R O S)=\dfrac{1}{2}\] Area (PQRS) ......(5)

Let us draw a line segment \[M N\] which is passing through point \[O P\] and is parallel to line segment PS.

In parallelogram PQRS,

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

PQRS is a parallelogram.

\[\therefore P Q \| R S\] (Opposite sides of a parallelogram)

\[\Rightarrow\] PN \[\| S N \ldots\] (7)

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

MN || PS and PN || SN

Therefore, quadrilateral PNMS is a parallelogram.

It can be clear that \[\triangle P O S\] and parallelogram PNMS are lying on the same base PS and between the same parallel lines \[P S\] and \[M N\].

\[\therefore\] Area \[(\triangle S O P)=\dfrac{1}{2}\] Area (PNMS) ... (8)

Similarly, for \[\triangle Q O R\] and parallelogram MNQR,

Area \[(\triangle Q O R)=\dfrac{1}{2}\] Area \[(M N Q R) \ldots(9)\]

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

\[\operatorname{Area}(\triangle S O P)+\] Area \[(\triangle Q O R)=\dfrac{1}{2}\] Area \[(P N M S)+\dfrac{1}{2}\] Area \[(M N Q R)\]

\[\operatorname{Area}(\triangle S O P)+\operatorname{Area}(\triangle Q O R)=\dfrac{1}{2}\] Area \[(P Q R S)\]

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

\[\operatorname{Area}(\triangle P O Q)+\operatorname{Area}(\triangle R O S)=\operatorname{Area}(\triangle S O P)+\operatorname{Area}(\triangle Q O R)=\] \[\dfrac{1}{2} \operatorname{Area}(\] || gmPQRS)

23. A quadrilateral \[A B C D\] is such that diagonals \[B D\] divide its area into two equal parts. Prove that \[\mathrm{BD}\] bisects \[\mathrm{AC}\].

Ans:

Join \[A C\]. Suppose \[A C\] and \[B D\] intersect at \[O\]. Draw lines \[A L\] and \[C M\] both perpendicular to \[B D\].

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

Thus \[\triangle A B D\] and \[\triangle A B C\] are on the same base \[A B\] and have equal areas.

Therefore, their corresponding altitudes \[A L=C M\] are equal.

Now, in \[\triangle A L O\] and \[\triangle C M O\],

\[\angle 1=\angle 2\] (vertically opposite angles)

\[\angle A L O=\angle C M O\] (right angles)

\[A L=C M\]

Therefore, \[\triangle A L O \cong \triangle C M O\] (AAS axiom)

\[\Rightarrow A O=O C\]

\[\Rightarrow B D\] bisects \[A C\]

24. In the given figure, \[B C \| D E\].

(a) If the area of \[\triangle A D C\] is \[20 s q\]. units, find the area of \[\triangle A E B\].

(b) If the area of \[\triangle B F D\] is 8 square units, find the area of \[\triangle C E F\]

Ans: (a) Triangles on the same base and between the same parallels are equal in area. \[\therefore A r(\triangle D B C)=\operatorname{Ar}(\triangle E C B)\].... (i)

Now,

\[A r(\triangle A B C)=A r(\triangle A D C)+\operatorname{Ar}(\triangle D B C)=\operatorname{Ar}(\triangle A E B)+\operatorname{Ar}(\triangle E C B)\]

\[\Rightarrow A r(\triangle A D C)+\operatorname{Ar}(\triangle D B C)=\operatorname{Ar}(\triangle A E B)+\operatorname{Ar}(\triangle E C B)\]

\[\Rightarrow A r(\triangle A D C)=\operatorname{Ar}(\triangle A E B)\].... (ii) from (i)

Given, \[\operatorname{Ar}(\triangle A D C)=20\] sq. units

\[\Rightarrow A r(\triangle A E B)=20\] sq. units

(b)

\[\operatorname{Ar}(\triangle A D C)=\operatorname{Ar}(\triangle A E B) \ldots\] From (ii)

\[\Rightarrow A r(\triangle A D C)-A r(A D F E)=A r(\triangle A E B)-A r(A D F E)\]

\[\Rightarrow A r(\triangle C E F)=A r(\triangle B F D)\]

Given, \[A r(\triangle B F D)=8\] sq. units

\[\Rightarrow A r(\triangle C E F)=8 s q\]. units

25. In the given figure, \[A B\|S Q\| D C\] and \[A D\|P R\| B C\]. If the area of

quadrilateral \[A B C D\] is 24 square units, find the area of quadrilateral \[P Q R S\].

Ans: Let \[S Q\] and \[P R\] intersect at point \[O\].

Now,

\[D C \| S Q\] and \[R P \| B C\]

\[\Rightarrow R C \| O Q\] and \[R O \| Q C\]

\[\Rightarrow\] Quadrilateral ROQC is a parallelogram.

Similarly,

Quad rilaterals ROSD, APOS and POQB are parallelograms.

\[\triangle R O Q\] and parallelogram ROQC are on the same base and between the same

parallel lines.

\[\therefore A(\triangle R O Q)=\dfrac{1}{2} \times A\left(\|^{g m} R O Q C\right) \ldots(\] i)

Similarly,

\[A(\triangle P O Q)=\dfrac{1}{2} \times A\left(\|^{g m} P O Q B\right) \ldots\] (ii)

\[A(\triangle P O S)=\dfrac{1}{2} \times A\left(\|^{g m} A P O S\right) \ldots(\] iii)

\[A(\triangle S O R)=\dfrac{1}{2} \times A\left(\|^{g m} R O S D\right) \ldots\] (iv)

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

\[A(\triangle R O Q)+A(\triangle P O Q)+A(\triangle P O S)+A(\triangle S O R)\]

\[=\dfrac{1}{2}\left[A\left(\|^{g m} R O Q C\right)+A\left(\|^{g m} P O Q B\right)+A\left(\|^{g m} A P O S\right)+A\left(\|^{g m} R O S D\right)\right.\]

\[=\dfrac{1}{2} \times 24\]

\[=12 s q\]. units

26. \[\triangle P Q R\] and \[\triangle S Q R\] are on the same base \[Q R\] with \[P\] and \[S\] on opposite sides of line \[Q R\], such that area of \[\triangle P Q R\] is equal to the area of \[\triangle S Q R\]. Show that \[Q R\] bisects \[P S\].

Ans:

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

\[\operatorname{ar}(\triangle P Q R)=\operatorname{ar}(\triangle S Q R)\]

Thus \[\triangle P Q R\] and \[\triangle S Q R\] are on the same base \[Q R\] and have equal areas.

Therefore, their corresponding altitudes \[P M=S N\] are equal.

Now, in \[\triangle P M O\] and \[\triangle S N O\],

\[\angle 1=\angle 2\] (vertically opposite angles)

\[\angle P M O=\angle S N O\] (right angles)

\[P M=S N\]

Therefore, \[\triangle P M O \cong \triangle S N O\] (AAS axiom)

\[\Rightarrow P O=O S\]

\[\Rightarrow Q R\] bisects \[P S\]

27. If the medians of a \[\triangle A B B C\] intersect at \[G\], show that

\[\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C)=\operatorname{ar}(\triangle B G C)=\dfrac{1}{3} \operatorname{ar}(\triangle A B C)\]

Ans:

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

In \[\triangle A B C, A D\] is the median

\[\Rightarrow \operatorname{ar}(\triangle A B D)=\operatorname{ar}(\triangle A C D) \quad \ldots(i)\]

In \[\triangle G B C, G D\] is the median

\[\Rightarrow \operatorname{ar}(\Delta G B D)=\operatorname{ar}(\Delta G C D) \quad\]...(ii)

Subtracting (ii) from (i),

\[\operatorname{ar}(\triangle A B D)-\operatorname{ar}(\triangle G B D)=\operatorname{ar}(\triangle A C D)-\operatorname{ar}(\triangle G C D)\]

\[\Rightarrow \operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C) \ldots \ldots\] (iii)

Similarly, \[\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle B G C) \quad \ldots\] (iv)

From (iii) and (iv),

\[\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle B G C)=\operatorname{ar}(\triangle A G C) \ldots \ldots(v)\]

But \[\operatorname{ar}(\triangle A G B)+\operatorname{ar}(\triangle B G C)+\operatorname{ar}(\triangle A G C)=\operatorname{ar}(\triangle A B C)\]

Therefore, \[3 \operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A B C)\]

\[\Rightarrow \operatorname{ar}(\triangle A G B)=\dfrac{1}{3} \operatorname{ar}(\triangle A B C)\]

Hence, \[\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C)=\operatorname{ar}(\triangle B G C)=\dfrac{1}{3} \operatorname{ar}(\triangle A B C)\].

28. In \[\triangle A B C\], the mid-points of \[A B, B C\] and \[A C\] are \[P, Q\] and \[R\] respectively. Prove that \[B Q R P\] is a parallelogram and that its area is half of \[\triangle A B C\].

Ans:

\[P\] and \[R\] are mid-points of \[A B\] and \[A C\] respectively given in the question.

Therefore, \[P R \| B C\] and \[P R=\dfrac{1}{2} B C\]........(i)

Also, \[Q\] is the mid-point of \[B C\],

\[\Rightarrow Q C=\dfrac{1}{2} B C \text {...(ii) }\]

From (i) and (ii)

\[P R \| B C\] and \[P R=Q C\]

\[\Rightarrow P R \| Q C\] and \[P R=Q C\]

\[Q\] and \[R\] are also mid-points of \[B C\] and \[A C\] respectively as given.

Therefore, \[Q R \| B P\] and \[Q R=B P\]

Hence, \[B Q R P\] is a parallelogram.

\[\Rightarrow P Q\] is a diagonal of \[\| g m B Q R P\]

\[\operatorname{ar}(\triangle P Q R)=\operatorname{ar}(\triangle B Q P) \ldots(v)\] (as diagonal of a \[\|\] gm divides it into two

triangles of equal areas)

Similarly, QCRP and QRAP are \[\| g m\] and

\[\operatorname{ar}(\triangle P Q R)=\operatorname{ar}(\triangle Q C R)=\operatorname{ar}(\triangle A P R) \quad \ldots(\mathrm{vi})\]

From (v) and (vi)

\[\operatorname{ar}(\triangle P Q R)=\operatorname{ar}(\triangle B Q P)=\operatorname{ar}(\triangle Q C R)=\operatorname{ar}(\triangle A P R)\]

Now, \[\operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle B Q P)+\operatorname{ar}(\triangle Q C R)+\operatorname{ar}(\triangle A P R)\]

\[\Rightarrow \operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle P Q R)\]

\[\Rightarrow \operatorname{ar}(\triangle A B C)=4 \operatorname{ar}(\triangle P Q R)\]

\[\Rightarrow \operatorname{ar}(\triangle P Q R)=\dfrac{1}{4} \operatorname{ar}(\triangle A B C) \quad\]...(vii \[)\]

\[\operatorname{ar}(\| g m B Q R P)=\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle B Q P)\]

\[\Rightarrow \operatorname{ar}(\| \lg m B R P)=\operatorname{ar}(\triangle P Q R)+\operatorname{ar}(\triangle P Q R)(\] from \[(v))\]

\[\Rightarrow \operatorname{ar}(\| \lg B Q R P)=2 \operatorname{ar}(\triangle P Q R)\]

\[\Rightarrow \operatorname{ar}(\| l g B Q R P)=2 \times \dfrac{1}{4} \operatorname{ar}(\triangle A B C)\] (from (vii))

\[\Rightarrow \operatorname{ar}(\| g m B Q R P)=\dfrac{1}{2} \operatorname{ar}(\triangle A B C)\]

29. In the given figure, \[P Q\|S R\| M N, P S \| Q M\] and \[S M \| P N\]. 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 \[P Q R S\] 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 \[P Q R S)=A(\] parallelogram \[\mathrm{SMOP})\]....(ii)

From (i) and (ii), we have

\[A(\] parallelogram SMNT \[)=A(\] parallelogram \[P Q R S)\]

31. In \[\triangle P Q R, P S\] is a median. \[T\] is the mid-point of \[S R\] and \[M\] is the mid-point of \[P T\]. Prove that: \[\triangle P M R=\dfrac{1}{8} \triangle P Q R\].

Ans:

Area \[(\triangle P Q R)=\operatorname{area}(\triangle P Q S)+\operatorname{area}(\triangle P S R) \ldots(i)\]

Since \[P S\] is the median of \[\triangle P Q R\] and median divides a triangle into two triangles of equal area.

Therefore, area \[(\triangle P Q S)=\operatorname{area}(\triangle P S R) \quad\]...(ii)

Substituting in (i)

Area \[(\triangle P Q R)=\operatorname{area}(\triangle P S R)+\] area \[(\triangle P S R)\] Area \[(\triangle P Q R)\]

32. In the figure, \[A B C D\] is a parallelogram and \[C P\] is parallel to \[D B\]. Prove that: Area of \[O B P C=\dfrac{3}{4}\] area of \[A B C D\]

Ans: The diagonals of a parallelogram divide it into four triangles of equal area Therefore, the area of \[\triangle A O D=\] area \[\triangle B O C=\] area \[\triangle A B O=\] area \[\triangle C D O\]. \[\Rightarrow\] area \[\triangle B O C=\dfrac{1}{4}\] area \[(\lg m A B C D)\]........(i)

In \[\| \mathrm{gm} A B C D, B D\] is the diagonal

Therefore, area \[(\triangle A B D)=\operatorname{area}(\triangle B C D)\]

\[\Rightarrow\] area \[(\triangle B C D)=\dfrac{1}{2}\] area \[(\mid \operatorname{lgm} A B C D) \ldots .\]. (ii)

In Parallelogram BPCD, BC is the diagonal

Therefore, area \[(\triangle B C D)=\operatorname{area}(\triangle B P C) \ldots .\]. (iii)

From (iii) and (ii)

area \[(\triangle B P C)=\dfrac{1}{2}\] area \[(\| g m A B C D) \ldots \ldots\] (iv)

adding (i) and (iv)

\[\operatorname{area}(\triangle B P C)+\] area \[\triangle B O C=\dfrac{1}{2}\] area \[(\| g m A B C D)+\dfrac{1}{4}\] area \[(\| g m A B C D)\]

Area of \[O B P C=\dfrac{3}{4}\] area of \[A B C D\]

33. The medians \[Q M\] and \[R N\] of \[\triangle P Q R\] intersect at \[O\]. Prove that: area of \[\triangle R O Q=\] area of quadrilateral \[P M O N\].

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, \[M N \| Q R\]

Clearly, \[\triangle Q M N\] and \[\triangle R N M\] are on the same base \[M N\] and between the same parallel lines.

Therefore, \[\operatorname{area}(\triangle Q M N)=\operatorname{area}(\triangle R N M)\]

\[\Rightarrow \operatorname{Area}(\triangle Q M N)-\operatorname{area}(\triangle O N M)=\operatorname{area}(\triangle R N M)-\operatorname{area}(\triangle O N M)\]

\[\Rightarrow \operatorname{area}(\Delta Q O N)=\operatorname{area}(\triangle R O M) \ldots \ldots(i)\]

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

Therefore, area \[(\triangle Q M R)=\operatorname{area}(\triangle P Q M)\]

\[\Rightarrow \operatorname{area}(\triangle R O Q)+\operatorname{area}(\triangle R O M)=\operatorname{area}(\] quad. \[P M O N)+\operatorname{area}(\triangle Q O N)\]

\[\Rightarrow\] area \[(\triangle R O Q)+\operatorname{area}(\triangle R O M)=\] area (quad. PMON \[)+\] area \[(\triangle R O M)\] (from

(i))

\[\Rightarrow \operatorname{area}(\triangle R O Q)=\operatorname{area}(\] quad. \[P M O N)\]

34.

a) In the given figure, \[A B C\] is a triangle, and \[A D\] is the median.

If \[E\] is any point on the median \[A D\]. Show that: Area of \[\triangle A B E=\] Area of \[\triangle A C E\].

b) In the given figure, \[A B C\] is a triangle, and \[A D\] is the median.

If \[E\] is the midpoint of the median \[A D\], prove that: Area of \[\triangle A B C=4 \times\]

Area of \[\triangle A B E\]

Ans: (a) \[A D\] is the median of \[\triangle A B C\]

so it will divide \[\triangle A B C\] into two triangles of equal areas.

\[\therefore \operatorname{Area}(\triangle A B D)=\operatorname{Area}(\triangle A C D) \ldots .\]. (i)

Similarly, ED is the median of \[\triangle E B C\].

\[\therefore \operatorname{Area}(D E B D)=\operatorname{Area}(D E C D) \ldots\] (ii)

Subtracting equation (ii) from (i), we have

\[\operatorname{Area}(\triangle A B D)-\operatorname{Area}(\triangle E B D)=\operatorname{Area}(\triangle A C D)-\operatorname{Area}(\triangle E C D)\]

\[\Rightarrow \operatorname{Area}(\triangle A B E)=\operatorname{Area}(\triangle A C E)\]

(b) \[A D\] is the median of \[\triangle A B C\].

So, it will divide \[\triangle A B C\] into two triangles of equal areas.

\[\therefore \operatorname{Area}(\triangle A B D)=\operatorname{Area}(\triangle A C D) \ldots .\]. (i)

Similarly, ED is the median of \[\triangle E B C\].

\[\therefore \operatorname{Area}(\triangle E B D)=\operatorname{Area}(\triangle E C D) \ldots\] (ii)

Subtracting equation (ii) from (i), we have

\[\operatorname{Area}(\triangle A B D)-\operatorname{Area}(\triangle E B D)=\operatorname{Area}(\triangle A C D)-\operatorname{Area}(\triangle E C D)\]

\[\Rightarrow \operatorname{Area}(\triangle A B E)=\operatorname{Area}(\triangle A C E) \quad \ldots\] (iii)

Since \[E\] is the mid-point of median \[A D\],

\[A E=E D\]

Now,

\[\triangle A B E\] and \[\triangle B E D\] have equal bases and a common vertex \[B\].

\[\therefore \operatorname{Area}(\triangle A B E)=\operatorname{Area}(\triangle B E D)\].....(iv)

From (i), (ii), (iii) and (iv), we get

\[\operatorname{Area}(\triangle A B E)=A(\triangle B E D)=\operatorname{Area}(\triangle A C E)=\operatorname{Area}(\triangle E D C) \ldots(v)\]

Now,

\[\operatorname{Area}(\triangle A B C)=\operatorname{Area}(\triangle A B E)+A(\triangle B E D)+\operatorname{Area}(\triangle A C E)+\operatorname{Area}(\triangle E D C)\]

\[=4 \times \operatorname{Area}(\triangle A B E) \quad\] From \[(v)\]

35. In a parallelogram \[P Q R S, M\] and \[N\] are the midpoints of the sides \[P Q\] and \[P S\] respectively. If the area of \[\triangle P M N\] is 20 square units, find the area of the parallelogram \[P Q R S\].

Ans:

In a parallelogram PQRS, \[\mathrm{SQ}\] is the diagonal.

So, it bisects the parallelogram.

\[\therefore \operatorname{Area}(\mathrm{DPSQ})=\dfrac{1}{2} \times\] Area (parallelogram \[P Q R S\] )

\[S M\] is the median of \[\triangle P S Q\].

\[\therefore \operatorname{Area}(\triangle P S M)=\dfrac{1}{2} \times \operatorname{Area}(\triangle P S Q)=\dfrac{1}{2} \times \dfrac{1}{2} \times\] Area (parallelogram PQRS) \[=\dfrac{1}{4} \times\] Area (parallelogram \[P Q R S\] )

Again, \[\mathrm{MN}\] is the median of \[\triangle P S M\].

\[\therefore \operatorname{Area}(\triangle P M N)=\dfrac{1}{2} \times \operatorname{Area}(\triangle P S M)=\dfrac{1}{2} \times \dfrac{1}{4} \times\] Area (parallelogram PQRS) \[=\dfrac{1}{8} \times\] Area (parallelogram PQRS)

\[\Rightarrow 20=\dfrac{1}{8} \times\] Area(parallelogram PQRS)

\[\Rightarrow\] Area (parallelogram PQRS) \[=160\] square units

36. In a parallelogram \[P Q R S, T\] is any point on the diagonal \[P R\]. If the area of \[\triangle P T Q\] is 18 square units find the area of \[\triangle P T S\].

Ans:

Construction: First, Join QR.

Let the diagonals \[P R\] and \[Q S\] intersect each other at point \[O\].

Since diagonals of a parallelogram bisect each other and \[O\] is the mid-point of both lines \[P R\] and \[Q S\].

Now, medians PR and QS of a triangle divide it into two triangles of equal area.

In \[\triangle P S Q, O P\] is the median.

\[\therefore \operatorname{Area}(\triangle P O S)=\operatorname{Area}(\triangle P O Q) \ldots \ldots\] (i)

Similarly, OT is the median of \[\triangle T S Q\].

\[\therefore \operatorname{Area}(\triangle T O S)=\operatorname{Area}(\triangle T O Q) \ldots \ldots\] (ii)

Subtracting equation (ii) from (i), we have

\[\operatorname{Area}(\triangle \mathrm{POS})-\operatorname{Area}(\triangle \mathrm{TOS})=\operatorname{Area}(\triangle \mathrm{POQ})-\operatorname{Area}(\triangle \mathrm{TOQ})\]

\[\Rightarrow \operatorname{Area}(\triangle P T Q)=\operatorname{Area}(\triangle P T S)\]

\[\Rightarrow \operatorname{Area}(\triangle P T S)=18\] square units

37. \[A B C D\] is a quadrilateral in which diagonals \[A C\] and \[B D\] intersect at a point \[O\].

Prove that: area \[\triangle A O D+\] area \[\triangle B O C+\] area \[\triangle A B O+\] area \[\triangle C D O\].

Ans:

Since the diagonals of a parallelogram bisect each other at the point of intersection.

Therefore, \[O B=O D\] and \[O A=O C\]

In \[\triangle A B C, O B\] is the median and the median divides the triangle into two triangles of equal areas

Therefore, area \[(\triangle B O C)=\operatorname{area}(\triangle A B O) \ldots \ldots \ldots(i)\]

In \[\triangle A D C, O D\] is the median and the median divides the triangle into two triangles of equal areas

Therefore, area \[(\triangle A O D)=\operatorname{area}(\triangle C D O) \ldots \ldots \ldots\] (ii)

Adding (i) and (ii)

area \[(\triangle A O D)+\] area \[(\triangle B O C)=\operatorname{area}(\triangle A B O)+\] area \[(\triangle C D O)\]