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# JEE Main 2022 Maths Question Paper with Solutions (24 June Morning Shift)

## Download JEE Main 2022 Maths 24 June Question Paper with Solutions PDF (Morning Session)

Last updated date: 21st Mar 2023
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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 $(\triangle C D O)$

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## JEE Main 2022 24 June morning shift Maths Question Paper Analysis

Mentioned below is the summarised analysis of the mathematics question paper.

• The mathematics section was deemed moderately challenging.

• The mathematics section was comparatively more difficult than physics and chemistry.

• The questions had lengthy calculations.

• Calculus and algebra were the important topics that had the most weightage in the question paper.

• Apart from it vector and induction was given importance.

• Binomial Theorem, Permutations and Combinations, Coordinate Geometry were some of the other important topics covered in the examination.

• Regular practice of JEE Main 2022 mock test and JEE Main previous year question paper were the most important aspects of preparation which helped students to perform well in the examination.

Maths is the most scoring part of JEE Main exams so solve and practice the JEE Main 2022 Maths question paper 24 June Morning shift to get an added advantage in the exam. By going through the exam, you will be able to understand the types of questions asked and the level of difficulty of the paper. So download our JEE Main 24 June morning 2022 Maths question paper with solutions to be completely exam ready! The solved JEE Main 2022 Maths Question Paper for the other days of exam are also provided on Vedantu.

For more information on JEE Main 2022 Maths, visit our website and check below-given the JEE Main 2022 Maths question paper FAQs.

## FAQs on JEE Main 2022 Maths Question Paper with Solutions (24 June Morning Shift)

1. In how many sessions will the JEE Main 2022 exam be held?

The National Testing Agency (NTA) has scheduled to conduct the JEE Main 2022 exams in two sessions this year.

Session 1: 23, 24, 25, 26, 27, 28, and 29 June 2022

Session 2: 21, 22, 23, 24, 25, 26, 27, 28, 29 and 30 July 2022

2. How should I prepare for JEE Main 2022 Maths exam?

Maths is the most scoring section in the JEE Main exam and you can ace it by following the right preparation strategy and study resources, the plethora of which you can find on Vedantu. The main technique is to analyse your weaknesses and strengths and prepare accordingly. You can do so with our JEE Main Maths solutions 2022 24 June shift 1 and solving JEE Main Previous Year Question Papers.

3. What are some of the most important topics covered in JEE Main 2022 24 June morning shift Maths question paper?

Some of the most important topics that were covered in the JEE main mathematics question paper of 2022 included the following:

• Calculus

• Coordinate Geometry

• Algebra

• Vector

• 3D Geometry

• Calculus

• Vectors

4. What was the overall difficulty level of the JEE main 2022 maths question paper for the 24 June morning shift?

According to the question paper analysis the paper can be considered moderately difficult. The mathematics section was deemed to be the most challenging as compared to physics and chemistry. The paper constituted lengthy calculative questions, the weightage was primarily given to calculus, algebra and vector. The students who regularly practised the mock paper and questions would have performed well in the examination.