Quadrilaterals

Multiple Choice Questions(MCQs)

1. Three angles of a quadrilateral are $75^{\circ}, 9 0^{\circ}$ and $75^{\circ}$. The fourth angle is

(A) $90^{\circ}$

(B) $95^{\circ}$

(C) $105^{\circ}$

(D) $120^{\circ}$

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Solution

Fourth angle of the quadrilateral $=360^{\circ}-(75^{\circ}+90^{\circ}+75^{\circ})$

$ \begin{aligned} & =360^{\circ}-240^{\circ} \\ & =120^{\circ} \end{aligned} $

Hence, the correct option is (D).

2. A diagonal of a rectangle is inclined to one side of the rectangle at $25^{\circ}$. The acute angle between the diagonals is

(A) $5 5^{\circ}$

(B) $50^{\circ}$

(C) $40^{\circ}$

(D) $25^{\circ}$

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Solution

Let $A B C D$ is a rectangle in which diagonal $A C$ is inclined to one side $A B$ of the rectangle at an angle of $25^{\circ}$.

Now, $AC=BD$ [Diagonal of a rectangle are equal]

$\dfrac{1}{2} A C=\dfrac{1}{2} B D$

$OA=OD$

In triangle $AOB$,

$OA=OD$

Now, $\angle O B A=\angle O A B=25^{\circ}$

And, $\angle A O D=180^{\circ}-130^{\circ}=50^{\circ}$

Hence, the acute angle between the diagonal is $50^{\circ}$.

Therefore, the correct option is (B).

3. $A B C D$ is a rhombus such that $\angle A C B=40^{\circ}$. Then $\angle A D B$ is

(A) $40^{\circ}$

(B) $45^{\circ}$

(C) $50^{\circ}$

(D) $60^{\circ}$

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Solution

Given:

$ABCD$ is a rhombus such that $\angle ACB=40^{\circ}$.

The diagonal of a rhombus bisect each other at right angles.

In right triangle $BOC$,

$ \begin{aligned} \angle O B C & =180^{\circ}-(\angle B O C+\angle B C O) \\ & =180^{\circ}-(90^{\circ}+40^{\circ}) \\ & =50^{\circ} \end{aligned} $

So, $\angle D B C=\angle O B C=50^{\circ}$

Now, $\angle A D B=\angle D B C \quad$ [Alt. int. $\angle s]$

So, $\angle A D B=50^{\circ} \quad[\angle D B C=50^{\circ}]$

Hence, the correct option is (C).

4. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if

(A) PQRS is a rectangle

(B) PQRS is a parallelogram

(C) Diagonals of PQRS are perpendicular

(D) Diagonals of PQRS are equal.

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Solution

If diagonals of PQRS are perpendicular.

Hence, the correct option is (C).

5. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if

(A) PQRS is a rhombus

(B) PQRS is a parallelogram

(C) Diagonals of PQRS are perpendicular

(D) Diagonals of PQRS are equal.

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Solution

If diagonal of PQRS are equal.

Hence, the correct option is (D).

6. If angles $A, B, C$ and $D$ of the quadrilateral $ABCD$, taken in order, are in the ratio 3:7:6:4, then $ABCD$ is a

(A) Rhombus

(B) Parallelogram

(C) Trapezium

(D) Kite

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Solution

Given in the question, ratio of angles of quadrilateral $ABCD$ is $3: 7: 6: 4$.

Let the angles of quadrilateral $ABCD$ be $3 x, 7 x, 6 x$ and $4 x$ respectively. So, $3 x+7 x+6 x+4 x=360^{\circ} \quad$ [Sum of the all angles of a quadrilateral is $360^{\circ}$.

$20 x=360^{\circ}$

$ \begin{aligned} & x=\dfrac{360^{\circ}}{20} \\ & x=18^{\circ} \end{aligned} $

So, angles of the quadrilateral are:

$\angle A=3 \times 18^{\circ}=54^{\circ}$

$\angle B=7 \times 18^{\circ}=126^{\circ}$

$\angle C=6 \times 18^{\circ}=108^{\circ}$

$\angle D=4 \times 18^{\circ}=72^{\circ}$

See the figure, $\angle B C E=180^{\circ}-\angle B C D$

$\angle B C E=180^{\circ}-108^{\circ}=72^{\circ}$

[Linear pair axiom]

$\angle B C E=\angle A D C=72^{\circ}$

Now, $BC || AD$ [The corresponding angles are equal.]

The sum of co interior angles is:

$\angle A+\angle B=126^{\circ}+54^{\circ}=180^{\circ}$

And $\angle C+\angle D=108^{\circ}+72^{\circ}=180^{\circ}$

Hence, $ABCD$ is a trapezium.

7. If bisectors of $\angle A$ and $\angle B$ of a quadrilateral $ABCD$ intersect each other at $P$, of $\angle B$ and $\angle C$ at $Q$, of $\angle C$ and $\angle D$ at $R$ and of $\angle D$ and $\angle A$ at $S$, then PQRS is a

(A) rectangle

(B) rhombus

(C) parallelogram

(D) quadrilateral whose opposite angles are supplementary

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Solution

PQRS is a quadrilateral whose opposite angles are supplementary.

Hence, the correct option is (D).

8. If APB and CQD are two parallel lines, then the bisectors of the angles $A P Q, B P Q, C Q P$ and $PQD$ form

(A) a square

(B) a rhombus

(C) a rectangle

(D) any other parallelogram

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Solution

PNQM is a rectangle.

Hence, the correct option is (C).

9. The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is

(A) a rhombus

(B) a rectangle

(C) a square

(D) any parallelogram

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Solution

The figure will be a rectangle.

Hence, the correct option is (B).

10. $D$ and $E$ are the mid-points of the sides $A B$ and $A C$ of $\triangle A B C$ and $O$ is any point on side $B C$. $O$ is joined to $A$. If $P$ and $Q$ are the mid-points of $O B$ and $O C$ respectively, then $D E Q P$ is

(A) a square

(B) a rectangle

(C) a rhombus

(D) a parallelogram

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Solution

According to the question, the line segment joining the mid-points of any two sides of a triangle of a triangle is parallel to the third side and is half of it. So,

Now,

$D E=\dfrac{1}{2} B C$ and $DE || BC$

Similarly, $D P=\dfrac{1}{2} A O$ and $DP || AO$

And, $E Q=\dfrac{1}{2} A O$ and $EQ || AO$

$DP=EQ[$$ \because$ Each $=\dfrac{1}{2} A O]$

And DP || EQ [Since, DP||AO and EQ||AO]

Now, DEQP is quadrilateral in which one pair of its opposite sides is equal and parallel.

Hence, quadrilateral DEQP is a parallelogram. The correct option is (D).

11. The figure formed by joining the mid-points of the sides of a quadrilateral $A B C D$, taken in order, is a square only if,

(A) $A B C D$ is a rhombus

(B) diagonals of $A B C D$ are equal

(C) diagonals of $ABCD$ are equal and perpendicular

(D) diagonals of $ABCD$ are perpendicular.

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Solution

If diagonals of $ABCD$ are equal and perpendicular.

Hence, the correct option is (C).

12. The diagonals $A C$ and $B D$ of a parallelogram $A B C D$ intersect each other at the point $O$. If $\angle DAC=32^{\circ}$ and $\angle AOB=70^{\circ}$, then $\angle DBC$ is equal to

(A) $24^{\circ}$

(B) $86^{\circ}$

(C) $38^{\circ}$

(D) $32^{\circ}$

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Solution

According to the question,

$AD$ is parallel to $BC$ and $AC$ cuts it. So,

$\angle D A C=\angle A C B[$ Alt. int. $\angle s]$

$\angle D A C=32^{\circ}$ [Given]

So, $\angle A C B=32^{\circ}$

Produce Co to $A$ in triangle $AOB$. So,

Ext. $\angle B O A=\angle O C B+\angle O B C$ [By exterior angle theorem]

$70^{\circ}=32^{\circ}+\angle O B C$

$\angle O B C=70^{\circ}-32^{\circ}=38^{\circ}$

Hence, $\angle D B C=38^{\circ}$. The correct option is (C).

13. Which of the following is not true for a parallelogram?

(A) opposite sides are equal

(B) opposite angles are equal

(C) opposite angles are bisected by the diagonals

(D) diagonals bisect each other.

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Solution

Opposite angles are bisected by the diagonals. That not true for the parallelogram.

Hence, the correct option is (C).

14. $D$ and $E$ are the mid-points of the sides $A B$ and $A C$ respectively of $\triangle A B C$. $D E$ is produced to $F$. To prove that $C F$ is equal and parallel to $D A$, we need an additional information which is

(A) $\angle \angle DAE=\angle EFC$

(B) $A E=E F$

(C) $D E=E F$

(D) $\angle ADE=\angle ECF$

Show Answer

Solution

According to the question, we need $DE=EF$

Hence, the correct option is (C).

Short Answer Questions with Reasoning

1. Diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at $O$. If $O A=3 ~cm$ and $O D=2 ~cm$, determine the lengths of $A C$ and $B D$.

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Solution

As we know that the diagonal of a parallelogram bisect each other. So,

$AC=2 \times OA=2 \times 3 ~cm=6 ~cm$

And, $BD=2 OD=2 \times 2 ~cm=4 ~cm$

Therefore, lengths of $AC$ and $BD$ are $6 ~cm$ and $4 ~cm$ respectively.

2. Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

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Solution

The given statement is not true because diagonal of a parallelogram bisect each other.

3. Can the angles $110^{\circ}, 8 0^{\circ}, 70^{\circ}$ and $95^{\circ}$ be the angles of a quadrilateral? Why or why not?

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Solution

We know that, sum of the angles of a quadrilateral is always $360^{\circ}$.

Sum of these angles $=110^{\circ}+80^{\circ}+70^{\circ}+95^{\circ}=355^{\circ}$ that is not equal to $360^{\circ}$.

Hence, $110^{\circ}, 80^{\circ}, 70^{\circ}$ and $95^{\circ}$ can’t be the angle of a quadrilateral.

4. In quadrilateral $A B C D, \angle A+\angle D=180^{\circ}$. What special name can be given to this quadrilateral?

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Solution

Given:

In quadrilateral $ABCD, \angle A+\angle D=180^{\circ}$.

We know that the sum of the two consecutive angle is $180^{\circ}$. So, pair of opposite side $A B$ and $CD$ are parallel.

Since, the quadrilateral $ABCD$ is trapezium.

Hence, special name can be given to this quadrilateral is trapezium.

5. All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

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Solution

Given:

All the angles of a quadrilateral are equal.

We know that, the sum of angles of a quadrilateral is $360^{\circ}$. Since, each angle is $\dfrac{360^{\circ}}{4}=90^{\circ}$.

Hence, special name is given to this quadrilateral is rectangle.

6. Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give reason for your answer.

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Solution

We know that diagonal of a rectangle need not to be perpendicular.

Hence, the given statement is false.

7. Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.

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Solution

We know that sum of four angles of a quadrilateral is always equal to $360^{\circ}$.

Now, if all the four angles of a quadrilateral be obtuse angles then sum of four angle will be more than $360^{\circ}$.

Hence, the given statement is false.

8. In $\triangle A B C, A B=5 \mathrm{~cm}, B C=8 \mathrm{~cm}$ and $C A=7 \mathrm{~cm}$. If $D$ and $E$ are respectively the mid-points of $A B$ and $B C$, determine the length of $D E$.

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Solution

Given:

In $\triangle ABC$,

$AB=5 ~cm, BC=8 ~cm$ and $CA=7 ~cm$

According to the question, $D$ and $E$ are respectively the mid-points of $AB$ and $BC$. So, using mid-point theorem,

$ \begin{aligned} D E & =\dfrac{1}{2} A C \\ & =\dfrac{1}{2} \times 7 ~cm \\ & =3.5 ~cm \end{aligned} $

9. In Fig., it is given that BDEF and FDCE are parallelograms. Can you say that BD = CD? Why or why not?

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Solution

$BDEF$ is a parallelogram. [Given]

So, $BD=EF$

FDCE is a parallelogram. [Given]

[Opposite side of a parallelogram]

So, $CD=EF$

Now, from equation (I) and (II), get:

$BD=CD$

10. In Fig., $ABCD$ and $AEFG$ are two parallelograms. If $\angle C=55^{\circ}$, determine $\angle F$.

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Solution

We know that opposite angle of parallelogram are equal.

$ABCD$ is a parallelogram. So,

$\angle A=\angle C$

Now, $\angle C=55^{\circ} \quad[$ Given]

In parallelogram $AEFG$,

$\angle F=\angle A=55^{\circ}$

Hence, $\angle F=55^{\circ}$.

11. Can all the angles of a quadrilateral be acute angles? Give reason for your answer.

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Solution

We know that an acute angle is less than $90^{\circ}$ and the sum of angles of quadrilateral is always $360^{\circ}$.

Hence, all the angle of a quadrilateral can’t be acute angle because sum of four angles of a quadrilateral will be less than $360^{\circ}$.

12. Can all the angles of a quadrilateral be right angles? Give reason for your answer.

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Solution

We know that sum of angles of quadrilateral is always $360^{\circ}$. Since, all the angles of a quadrilateral can be right angle, which is true because $90^{\circ} \times 4=360^{\circ}$.

13. Diagonals of a quadrilateral $ABCD$ bisect each other. If $\angle A=35^{\circ}$, determine $\angle B$.

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Solution

Given:

Diagonals of a quadrilateral $ABCD$ bisect each other.

So, $A B C D$ is a parallelogram.

Now, $\angle A+\angle B=180^{\circ}$ [Adjacent angles of a parallelogram are supplementary]

Since,

$ \begin{aligned} 35^{\circ}+\angle B & =180^{\circ} \\ \angle B & =180^{\circ}-35^{\circ} \\ \angle B & =145^{\circ} \end{aligned} $

14. Opposite angles of a quadrilateral $ABCD$ are equal. If $AB=4 ~cm$, determine CD.

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Solution

Given:

Opposite angles of a quadrilateral $ABCD$ are equal.

So, that is a parallelogram.

Now, $ABCD$ is a parallelogram.

So, $AB=CD$. [Opposite of a parallelogram are equal]

$AB=4 ~cm$ [Given]

Therefore, $CD=4 ~cm$.

Short Answer Questions

1. One angle of a quadrilateral is of $108^{\circ}$ and the remaining three angles are equal. Find each of the three equal angles.

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Solution

We know that the sum of all the angles in a quadrilateral is $360^{\circ}$.

According to the question, the remaining three angles are equal. So, let it is $x$.

Now,

$ \begin{aligned} 108^{\circ}+x+x+x & =360^{\circ} \\ 3 x & =360^{\circ}-108^{\circ} \\ 3 x & =252^{\circ} \\ x & =84^{\circ} \end{aligned} $

Hence, each of the three equal angles is $84^{\circ}$.

2. $A B C D$ is a trapezium in which $A B || D C$ and $\angle A=\angle B=45^{\circ}$. Find angles $C$ and $D$ of the trapezium.

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Solution

Given:

$ABCD$ is a trapezium in which $AB || DC$ and $\angle A=\angle B=45^{\circ}$.

Now, $AB || DC$ and $AD$ is transversal.

So, $\angle A+\angle D=180^{\circ}$ [sum of interior angles on the side of the transversal is $180^{\circ}$ ] $45^{\circ}+\angle D=180^{\circ}$

$ \begin{aligned} & \angle D=180^{\circ}-45^{\circ} \\ & \angle D=135^{\circ} \end{aligned} $

Similarly, $\angle B+\angle C=180^{\circ}$

$45^{\circ}+\angle C=180^{\circ}$

$ \begin{aligned} & \angle C=180^{\circ}-45^{\circ} \\ & \angle C=135^{\circ} \end{aligned} $

Therefore, $\angle A=\angle B=45^{\circ}$ and $\angle C=\angle D=135^{\circ}$.

3. The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60^{\circ}$. Find the angles of the parallelogram.

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Solution

In quadrilateral DPBQ:

$ \begin{aligned} \angle 1+\angle 2+\angle B+\angle 3 & .=360^{\circ} \quad \text{ [Angle sum property of quadrilateral }] \\ 60^{\circ}+90^{\circ}+\angle B+90^{\circ} & =360^{\circ} \\ \angle B+240^{\circ} & =360^{\circ} \\ \angle B & =360^{\circ}-240^{\circ} \\ \angle B & =120^{\circ} \end{aligned} $

Since, $\angle A D C=\angle B=120^{\circ} \quad$ [Opposite angles of a parallelogram are equal]

$\angle A+\angle B=180^{\circ} \quad$ [Sum of consecutive interior angle is $180^{\circ}$ ]

$\angle A+120^{\circ}=180^{\circ}$

$ \angle A=180^{\circ}-120^{\circ} $

$ \angle A=60^{\circ} $

So, $\angle C=\angle A=60^{\circ}$ [Opposite angle of a parallelogram are equal]

4. $A B C D$ is a rhombus in which altitude from $D$ to side $A B$ bisects $A B$. Find the angles of the rhombus.

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Solution

See the below figure, in triangle APD and triangle BPD,

$ \begin{aligned} & AP=BP \quad[\text{ Given }] \\ & \angle 1=\angle 2 \quad[\text{ Each equal to } 90^{\circ}] \\ & PD=PD \quad[\text{ Common side }] \end{aligned} $

So, by SAS criterion of congruence,

$\triangle A P D \cong \triangle B P D$

$\angle A=\angle 3$

[CPCT]

$\angle 3=\angle 4$

$\angle A=\angle 3=\angle 4$

[Diagonal bisect opposite angles of a rhombus]

Now, $AD || BC$

So, $\angle A+\angle A B C=180^{\circ}$

$\angle A+\angle 3+\angle 4=180^{\circ}$

[Sum of consecutive interior angles is $180^{\circ}$ ]

$\angle A+\angle A+\angle A=180^{\circ}$

$3 \angle A=180^{\circ}$

$\angle A=\dfrac{180^{\circ}}{3}$

$\angle A=60^{\circ}$

Now,

$\angle A B C=\angle 3+\angle 4$

$=60^{\circ}+60^{\circ}$

$\angle A B C=120^{\circ}$

[Opposite angles of a rhombus are equal]

$\angle A D C=\angle A B C=120^{\circ}$

[Opposite angles of a rhombus are equal]

5. $E$ and $F$ are points on diagonal $A C$ of a parallelogram $A B C D$ such that $AE=CF$. Show that BFDE is a parallelogram.

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Solution

Given:

$E$ and $F$ are points on diagonal $AC$ of a parallelogram $ABCD$ such that $AE=CF$.

To prove that BFDE is parallelogram,

Proof: $ABCD$ is a parallelogram.

$OD=OB \quad \ldots$ (I) [Diagonals of parallelogram bisect each other]

$OA=OC \quad \ldots$ (II) [Diagonals of parallelogram bisect each other]

$AE=EF \quad \ldots$ (III)[Given]

Subtracting equation (III) from equation (II), get:

$OA-AE=OC-CF$

$OE=OF \quad \ldots$ (IV)

Now, BFDE is parallelogram. [Since, $OD=OB$ and $OE=OF$ ]

Hence, proved.

6. $E$ is the mid-point of the side $A D$ of the trapezium $A B C D$ with $A B$ || DC. A line through $E$ drawn parallel to $A B$ intersect $B C$ at $F$. Show that $F$ is the mid-point of BC. [Hint: Join AC]

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Solution

Given

$E$ is the mid-point of the side $AD$ of the trapezium $ABCD$ with $AB || DC$. Also, $EF || AB$.

To prove that $F$ is the mid-point of $BC$.

Construction: Join $AC$ which intersect $EF$ at $O$.

Proof: In triangle $ADC, E$ is the midpoint of $AD$ and $EF || DC$. [$ \because EF || AB$ and $DC || AB$. So, $AB||EF|| DC]$

$O$ is the mid-point of $AC$ and $OF || AB$.

Now, OF bisect BC. [Converse of mid-point theorem]

Or $F$ is the mid-point of $BC$.

Hence, proved.

7. Through $A, B$ and $C$, lines $R Q, P R$ and $Q P$ have been drawn, respectively parallel to sides $BC, CA$ and $AB$ of a $\triangle ABC$ as shown in Fig. Show that $BC=\dfrac{1}{2} QR$.

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Solution

Given in the question, Triangle $A B C$ and $P Q R$ in which $A B||Q P, B C|| R Q$ and $C A || P R$.

To prove that $B C=\dfrac{1}{2} Q R$

Proof: In quadrilateral BCAR, $BR || CA$ and $BC || RA$

So, quadrilateral, BCAR is a parallelogram.

$BC=AR \ldots$ (I)

Now, in quadrilateral BCQA, $BC || AQ$ and $AB || QC$

So, quadrilateral BCQA is a parallelogram,

$BC=AQ$

Now, adding equation (I) and (II), get:

$2 BC=AR+AQ$

$2 BC=RQ$

$BC=\dfrac{1}{2} QR$

Now, $BEDF$ is a quadrilateral, in which $\angle BED=\angle BFD=90^{\circ}$

$\angle FSE=360^{\circ}-(\angle FDE+\angle BED+\angle BFD)=360^{\circ}-(60^{\circ}+90^{\circ}+90^{\circ})$

$=360^{\circ}-240^{\circ}$

$=120^{\circ}$

8. $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$, respectively of an equilateral triangle $ABC$. Show that $\triangle DEF$ is also an equilateral triangle.

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Solution

Given in the question, $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$, respectively of an equilateral $\triangle ABC$.

To proof that $\triangle DEF$ is an equilateral triangle.

Proof: In $\triangle ABC, E$ and $F$ are the mid-points of $AC$ and $AB$ respectively, then $EF || BC$. So, $E F=\dfrac{1}{2} B C$

$DF||AC, DE|| AB$

$DE=\dfrac{1}{2} AB$ and $FD=\dfrac{1}{2} AC$ [By mid-point theorem $]$.

Now, $\triangle ABC$ is an equilateral triangle.

$AB=BC=CA$ $\dfrac{1}{2} AB=\dfrac{1}{2} BC=\dfrac{1}{2} CA$ [Dividing by 2 in the above equation]

So, $DE=EF=FD \quad$ [From Equation. (I) and (II)]

Since, all sides of ADEF are equal.

Hence, $\triangle DEF$ is an equilateral triangle.

Hence proved.

9. Points $P$ and $Q$ have been taken on opposite sides $A B$ and $C D$, respectively of a parallelogram $A B C D$ such that $A P=C Q$. Show that $A C$ and $P Q$ bisect each other.

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Solution

Given in the question, points $P$ and $Q$ have been taken on opposite sides $AB$ and $CD$, respectively of a parallelogram $ABCD$ such that $AP=CQ$.

In triangle $AOP$ and triangle COQ:

$AP=CQ \quad[$ Given]

$\angle 1=\angle 2 \quad$ [Alternate interior angles]

$\angle 3=\angle 4 \quad$ [Vertically opposite angles]

$\triangle A O P \cong \triangle C O Q$ [By AAS Congruence rule]

So, $OA=OC$ and $OP=OQ \quad[CPCT]$

Hence, $A C$ and $P Q$ bisect each other.

10. In Fig., $P$ is the mid-point of side $B C$ of a parallelogram $A B C D$ such that $\angle B A P=\angle D A P$. Prove that $A D=2 C D$.

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Solution

Given in the question, in a parallelogram $ABCD, P$ is a mid-point of $BC$ such that $\angle B A P=\angle D A P$.

To prove that $AD=2 CD$

Proof: $ABCD$ is a parallelogram.

So, $AD || BC$ and $AB$ is transversal, then:

$\angle A+\angle B=180^{\circ}$

$\angle B=180^{\circ}-\angle A$

[Sum of cointerior angles is $180^{\circ}$ ]

Now, in triangle $A B P$,

$\angle P A B+\angle B+\angle B P A=180^{\circ} \quad$ [By angle sum property of a triangle]

$\dfrac{1}{2} \angle A+180^{\circ}-\angle A+\angle B P A=180^{\circ} \quad$ [From equation (I)]

$\angle B P A-\dfrac{\angle A}{2}=0$

$\angle B P A=\dfrac{\angle A}{2}$

$\angle B P A=\angle B P A$

$AB=BP \quad$ [Opposite sides of equal angles are equal]

In above equation multiplying both side by 2 , get:

$2 AB=2 BP$

$2 AB=BC \quad[P$ is the mid-point of $BC]$

$2 CD=AD \quad[ABCD$ is a parallelogram, then $AB=CD$ and $BC=AD]$

Long Answer Questions

1. A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

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Solution

$A B C$ is an isosceles right triangle with $\mathrm{AB}=\mathrm{BC}$. A square $\mathrm{BFED}$ is inscribed in triangle $\mathrm{ABC}$ so that $\angle \mathrm{B}=$ common $=90^{\circ}$. In $\triangle \mathrm{ADE}$ and $\triangle \mathrm{EFC}$, we have

$$\mathrm{DE}=\mathrm{EF}\quad …(1)$$

$$[\because \text{Sides of a square are equal} ]$$

$$ \angle 1+\angle 2=180^{\circ}\quad \text{[Linear pair axiom]} $$

$$ \left.\begin{array}{rlrl} \Rightarrow & 90^{\circ}+\angle 2 & =180^{\circ} & {\left[\because \text { Each angle of a square }=90^{\circ}\right.} \end{array}\right] $$

$$\Rightarrow \quad \angle 2=90^\circ$$

Similarly, $\quad \angle 4=90^{\circ}$

$\therefore \quad \angle 2=\angle 4 \quad …(2)\left[\because \quad\right.$ Each $\left.=90^{\circ}\right]$

Now, $\mathrm{AB}=\mathrm{BC}\quad \text{[Given]}$

$\therefore \quad \angle \mathrm{C}=\angle \mathrm{A} \quad …(3)$

$[\because$ Angles opp. to equal sides are equal]

From (1), (2) and (3), we get

$$ \begin{aligned} \Delta \mathrm{ADE} & \cong \Delta \mathrm{EFC} \quad \text{[By AAS Congruence rule]} \\ \text{Hence,} \quad \mathrm{AE} & =\mathrm{EC} \quad \text{[CPCT]} \end{aligned} $$

2. ABCD is a parallelogram, in which AB=10 cm and AD=6b cm.The bisector of $\angle A$ meets DC in E. AE and BC produced meet at F.

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Solution

ABCD is a parallelogram, in which AB = 10 cm and AD = 6 cm. The bisector of $\angle A$ meets DC in E. AE and BC produced meet at F.

$\begin{aligned} & \angle \mathrm{BAE}=\angle \mathrm{EAD}\quad \text {…(1) }[\because \text { AE bisects } \angle \mathrm{A}] \\ & \angle \mathrm{EAD}=\angle \mathrm{EFB} \quad \text {…(2) }[\text { Alt. } \angle s] \\ & \Rightarrow \quad \angle \mathrm{BAE}=\angle \mathrm{EFB} \quad[\text { From (1) and (2)] } \\ & \therefore \quad \mathrm{BF}=\mathrm{AB} \quad[\because \text { Sides opposite to equal } \angle s \text { are equal }] \\ & \Rightarrow \quad \mathrm{BF}=10 \mathrm{~cm} \quad {[\because \mathrm{AB}=10 \mathrm{~cm}]} \\ & \Rightarrow \quad \mathrm{BC}+\mathrm{CF}=10 \mathrm{~cm} \Rightarrow 6 \mathrm{~cm}+\mathrm{CF}=10 \mathrm{~cm} \\ & {[\because \mathrm{BC}=\mathrm{AD}=6 \mathrm{~cm} \text {, opposite sides of a } || \mathrm{gm}]} \\ & \Rightarrow \quad \mathrm{CF}=10 \mathrm{~cm}-6 \mathrm{~cm}=4 \mathrm{~cm} \\ & \end{aligned}$

3. P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

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Solution

Given: A quadrilateral ABCD in which AC = BD and P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of quadrilateral ABCD.

To prove: PQRS is a rhombus.

Proof: In $\triangle \mathrm{ABC}, \mathrm{P}$ and $\mathrm{Q}$ are the mid-points of sides $\mathrm{AB}$ and $\mathrm{BC}$ respectively.

That is, PQ joins mid-points of sides $A B$ and $B C$.

$$\therefore \quad \mathrm{PQ} || \mathrm{AC} \quad …(1)$$

$$ \text{and} \quad \mathrm{PQ}=\dfrac{1}{2} \mathrm{AC} \quad …(2) \text{[Mid-point theorem]} $$

In $\triangle \mathrm{ADC}, \mathrm{R}$ and $\mathrm{S}$ are the mid-points of $\mathrm{CD}$ and $\mathrm{AD}$ respectively.

$$\therefore \quad \mathrm{SR} || \mathrm{AC} \quad …(3)$$

$$ \text{and} \quad \mathrm{SR}=\dfrac{1}{2} \mathrm{AC} \text{…(4) [Mid-point theorem]} $$

From (1) and (3), we get

$$ \mathrm{PQ}||\mathrm{SR} $$

$\Rightarrow $ PQRS is a parallelogram.

From (2) and (4), we get

$$ \mathrm{PQ}=\mathrm{SR} $$

In $\triangle \mathrm{DAB}, \mathrm{SP}$ joins mid-points of sides $\mathrm{DA}$ and $\mathrm{AB}$ respectively.

$$ \begin{array}{rlrl} \therefore & \mathrm{SP} =\dfrac{1}{2} \mathrm{BD} \quad \text{…(5) [Mid-point theorem]} \\ \mathrm{AC} & =\mathrm{BD} \quad …(6)\text{[Given]} \end{array} $$

From equations (2), (5) and (6), we get (6) [Given] $$ \mathrm{SP}=\mathrm{PQ} $$

$\therefore$ Parallelogram $\mathrm{PQRS}$ is a rhombus.

Hence, proved.

4. P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that $AC \perp BD$. Prove that PQRS is a rectangle.

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Solution

Given: A quadrilateral ABCD in which AC ⊥ BD and P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of quadrilateral ABCD.

To prove: PQRS is a rectangle.

Proof: In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.

That is, PQ joins mid-points of sides AB and BC.

$$ \begin{array}{ll} \therefore & \mathrm{PQ} || \mathrm{AC} \quad …(1)\\ \text { and } & \mathrm{PQ}=\dfrac{1}{2} \mathrm{AC} \quad …(2) \text{[Mid-point theorem]} \end{array} $$

In $\triangle \mathrm{ADC}, \mathrm{R}$ and $\mathrm{S}$ are the mid-points of $\mathrm{CD}$ and $\mathrm{AD}$ respectively.

$$ \therefore \quad \mathrm{SR} || \mathrm{AC} \quad …(3) $$

$$ \text { and }\mathrm{SR}=\dfrac{1}{2} \mathrm{AC} \quad …(4) \text{[Mid-point theorem]} $$

From (1) and (3), we get

PQ $||$ SR

From (2) and (4), we get

PQ=SR

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

$$\mathrm{PQ} || \mathrm{AC}$$

$$\Rightarrow \quad \mathrm{PE} || \mathrm{GF}$$

In $\triangle A B D, P S$ joins mid-points of sides $A B$ and $A D$ respectively.

$$\therefore \text { PS } || \text { BD }\quad \text{[Proved above]}$$

$$\Rightarrow \text { PG } || \text { EF }$$

$\Rightarrow \quad PEFG~ \text{is a parallelogram}\quad \therefore [PE||GF \text{ and } PG|| EF] $

$\Rightarrow \quad \angle 1=\angle 2 \quad [\therefore \text{opposite angles of a parallelogram are equal}]$

But, $\quad \angle 1=90^{\circ} \quad [\because AC \perp BD]$

$\therefore \quad \angle 2=90^{\circ}$

$\Rightarrow$ Parallelogram PQRS is a rectangle.

5. P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and $AC \perp BD$. Prove that PQRS is a square.

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Solution

Given : A quadrilateral ABCD is which AC = BD and AC ⊥ BD. P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD.

To prove : PQRS is a square.

Proof : Parallelogram PQRS is a rectangle.

[Same as in Q4]

$$ \mathrm{PQ}=\dfrac{1}{2} \mathrm{AC}\quad … (1) \text{[Proved as in Q4]} $$

PS joins mid-points of sides $A B$ and $A D$ respectively.

$$ \begin{aligned} & \mathrm{PS}=\dfrac{1}{2} \mathrm{BD}\quad … (2) \text{[Mid-point theorem]} \\ & \mathrm{AC}=\mathrm{BD} \quad …(3) \text{[Given]} \\ & \end{aligned} $$

From (1), (2) and (3), we get

$$ \mathrm{PS}=\mathrm{PQ} $$

$\Rightarrow$ Rectangle $PQRS$ is a square.

6. A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

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Solution

ABCD is a parallelogram and diagonal AC bisects $\angle A$. We have to show that ABCD is a rhombus.

$\angle 1 = \angle 2 \quad…(1) [\because \text{ AC bisects $\angle A$]}$

$\angle 2 = \angle 4\quad …(2) \text{[∵ Alt. interior angles]}$

From (1) and (2), we get

$\angle 1 = \angle 4$

Now, in ∆ABC, we have

$\angle 1 = \angle 4 \quad \text{[Proved above]}$

$BC = ABq\quad$ [∵ Sides opp. to equal ∠s are equal]

Also, $AB = DC$ and $AD = BC\quad$ [∵ Opposite sides of a parallelogram are equal]

So, ABCD is a parallelogram in which its sides AB = BC = CD = AD.

Hence, ABCD is a rhombus.

7. P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PQRS is a parallelogram.

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Solution

Given : A parallelogram ABCD in which P and Q are the mid-points of the sides AB and CD respectively. AQ intersects DP at S and BQ intersects CP at R.

To prove : PRQS is a parallelogram

Proof: $\quad \mathrm{DC} || \mathrm{AB}[\because$ Opposite sides of a parallelogram are parallel $]$

$\Rightarrow \quad \mathrm{AP} || \mathrm{QC}$

$\mathrm{DC}=\mathrm{AB} \quad[\because$ Opposite sides of a parallelogram are equal $]$

$\Rightarrow \quad \dfrac{1}{2} \mathrm{DC}=\dfrac{1}{2} \mathrm{AB}$

$\Rightarrow \quad \mathrm{QC}=\mathrm{AP} \quad[\because \mathrm{P}$ is mid-point of $\mathrm{AB}$ and $\mathrm{Q}$ is mid-point of CD]

$\Rightarrow \mathrm{APCQ}$ is a parallelogram.$\quad [\because \mathrm{AP} | \mathrm{QC}$ and $\mathrm{QC}=\mathrm{AP}]$

$\Rightarrow \mathrm{AQ} || \mathrm{PC}\quad $$\quad [\because$ Opposite sides of a $|$ gm are parallel $]$

$\Rightarrow \mathrm{SQ} || \mathrm{PR}$

Similarly, SP $|| QR$

$\therefore$ Quadrilateral PRQS is a parallelogram.

Hence, proved.

8. ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that $\angle A= \angle B$ and $\angle C = \angle D$.

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Solution

Given : A quadrilateral ABCD in which AB || DC and AD = BC.

To prove : $\angle A= \angle B$ and $\angle C = \angle D$

Construction : Draw $DP \perp AB$ and $CQ \perp AB$.

Proof : In $\triangle APD$ and $\triangle BQC$, we have

$\angle 1 = \angle 2\qquad $[∵Each equal to 90°]

AD = BC [Given]

DP = CQ

[Distance between parallel lines]

So, by RHS criterion of congruence, we have

$$ \begin{array}{rlrl} & & \triangle \mathrm{APD} & \cong \triangle \mathrm{BQC} \\ & \angle \mathrm{A} =\angle \mathrm{B}\quad [CPCT] \end{array} $$

Now, $\mathrm{DC} || \mathrm{AB}$

$$ \begin{aligned} & \angle \mathrm{A}+\angle 3=180^{\circ}\quad …(1)\text{[$\because$ Sum of consective interior angles is $180^{\circ}$]} \\ & \angle \mathrm{B}+\angle 4=180^{\circ} \quad …(2)\text{[Same reason]} \end{aligned} $$

From (1) and (2), we get

$$ \begin{aligned} & & \angle \mathrm{A}+\angle 3 & =\angle \mathrm{B}+\angle 4 \\ \Rightarrow & & \angle 3 & =\angle 4 \qquad [\because \angle \mathrm{A}=\angle \mathrm{B}]\\ \Rightarrow & & \angle \mathrm{C} & =\angle \mathrm{D} \end{aligned} $$

Hence, proved.

9. In the given figure, AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

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Solution

Given : $\mathrm{AB}||\mathrm{DE}, \mathrm{AB}=\mathrm{DE}, \mathrm{AC}|| \mathrm{DF}$ and $\mathrm{AC}=\mathrm{DF}$

To prove : $\mathrm{BC} || \mathrm{EF}$ and $\mathrm{BC}=\mathrm{EF}$

Proof: $\quad \mathrm{AC} || \mathrm{DF}$ [Given]

and $\quad \mathrm{AC}=\mathrm{DF}$ [Given]

$\therefore$ ACFD is a parallelogram.

$\Rightarrow$ $\mathrm{AD} || \mathrm{CF}$ $\ldots(1)[\because$ Opposite sides of a $||$ gm are parallel $]$

and $\mathrm{AD}=\mathrm{CF}$ $\ldots(2)[\because$ Opposite sides of a $||$ gm are equal]

Now, $\mathrm{AB} || \mathrm{DE}\quad$ [Given]

and $\mathrm{AB}=\mathrm{DE}\quad$ [Given]

$\therefore \mathrm{ABED}$ is a parallelogram.

$\Rightarrow \quad \mathrm{AD} || \mathrm{BE} \quad$…(3) $[\because$ Opposite sides of a $||$ gm are parallel $]$

and $\quad \mathrm{AD}=\mathrm{BE} \quad . .(4)[\because$ Opposite sides of $\mathrm{a} || \mathrm{gm}$ are equal $]$

From (1) and (3), we get

$$ \mathrm{CF} || \mathrm{BE} $$

And, from (2) and (4), we get

$$ \mathrm{CF}=\mathrm{BE} $$

$\therefore$ BCFE is a parallelogram.

$$ \begin{array}{lrr} \Rightarrow & \mathrm{BC} || \mathrm{EF} & {[\because \text { Opposite sides of a } || \text { gm are parallel }]} \\ \text { and } & \mathrm{BC}=\mathrm{EF} & {[\because \text { Opposite sides of a } || \text { gm are equal }]} \end{array} $$

Hence, proved.



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