RD Sharma 2020 solution class 9 chapter 13 Quadrilaterals FBQS

FBQS

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Question 1:

Three angles of a quadrilateral are ${75}^{\circ },{90}^{\circ }$ and ${75}^{\circ }$. The measure of the fourth angle is _____________ .

Answer 1:

Let the measure of the fourth angle be x.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ 75° + 90° + 75° + x = 360°
⇒ 240° + x = 360°
​⇒ x = 360° − 240°
​⇒ x = 120°


Hence, the measure of the fourth angle is 120°.

Question 2:

Diagonals of a parallelogram ABCD intersect at O. If ∠AOB = 90° and ∠BDC = 40°, then ∠OAB is __________.

Answer 2:

Let ∠OAB be x.

Since, AB || DC
Therefore, ∠DBA = ∠BDC = 40°

In ∆AOB,
∠OAB  + ∠ABD  + ∠AOB = 180°
⇒  x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° − 130°
⇒ x = 50°

Hence, ∠OAB is 50°.

Question 3:

The angle A, B, C and D of a quadrilateral are in the ratio 1 : 2 : 4 : 5. If bisectors of ∠C and ∠D meet of O, the ∠COD = __________.

Answer 3:

Let the angle A, B, C and D of a quadrilateral be x, 2x, 4x and 5x, respectively.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ x + 2x + 4x + 5x = 360°
⇒ 12x = 360°
​⇒ x = 30°

Thus, ∠A = 30°
∠B = 60°
∠C = 120°
∠D = 150°

Now, the bisectors of ∠C and ∠D meet of O
Thus, in ∆COD,
$\frac{1}{2}$∠DCO  + ∠COD  + $\frac{1}{2}$∠ODC = 180°
⇒ $\frac{1}{2}$ × 120° + ∠COD  + $\frac{1}{2}$ × 150° = 180°
⇒ 60° + ∠COD  + 75° = 180°
⇒ 135° + ∠COD = 180°
⇒ ∠COD = 180° − 135°
⇒ ∠COD = 45°

Hence, ∠COD = 45°.

Question 4:

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is __________.

Answer 4:

In ∆AOB,
∠OAB  + ∠ABO  + ∠AOB = 180°
⇒ 25° + 25° + ∠AOB = 180°
⇒ ∠AOB + 50° = 180°
⇒ ∠AOB = 180° − 50°
⇒ ∠AOB = 130° which is an obtuse angle

Now,  ∠AOB + ∠AOD = 180°    (angles on a straight line)
⇒ 130° + ∠AOD = 180°
⇒ ∠AOD = 180° − 130°
⇒ ∠AOD = 50° which is an acute angle

Hence, the acute angle between the diagonals is 50°.

Question 5:

ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB = _________.

Answer 5:

Let ∠ADB be x.

As we know, angles formed by the diagonals of a rhombus is 90°



In ∆BOC,
∠OBC  + ∠BCO  + ∠COB = 180°
⇒ ∠OBC  + 40° + 90° = 180°
⇒ ∠OBC + 130° = 180°
⇒ ∠OBC = 180° − 130°
⇒ ∠OBC = 50°

Since, AD || BC
Thus, ∠ADB = ∠DBC     (alternate interior angles)
⇒ x = 50°

Hence, ∠ADB = 50°.

Question 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 _________.

Answer 6:

Let the angle A, B, C and D of a quadrilateral be 3x, 7x, 6x and 4x, respectively.

We have,
Sum of all the angles of a quadrilateral = 360°
⇒ 3x + 7x + 6x + 4x = 360°
⇒ 20x = 360°
​⇒ x = 18°

Thus, ∠A = 54°
∠B = 126°
∠C = 108°
∠D = 72°



Now,
∠BCE = 180° − 108° = 72° = ∠ADE
⇒ AD || BC

Hence, ABCD is a trapezium.

Question 7:

If the diagonals of a parallelogram ABCD are equal then ∠ABC = ___________.

Answer 7:

The diagonals of a parallelogram ABCD are equal
⇒ ABCD is a rectangle
⇒ ∠ABC = 90°

Hence, ∠ABC = 90°.

Question 8:

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, then AC = _________ and BD = __________.

Answer 8:

Given:
ABCD is a parallelogram
Diagonals AC and BD intersect each other at O.
OA = 3 cm and OD = 2 cm

As we know, diagonals of a parallelogram bisects each other.
Thus, AC = 2OA = 2× 3 = 6 cm
and BD = 2OD = 2× 2 = 4 cm

Hence, AC = 6 cm and BD = 4 cm.

Question 9:

ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°. Then, ∠C = ______ and ∠D = ______.

Answer 9:

Given:
ABCD is a trapezium
AB || DC
∠A = ∠B = 45°

Since, AB || DC
Thus, AD and BC are the transversals.
Therefore, sum of the interior angles must be equal to 180°.

∠A + ∠D = 180°
⇒ 45° + ∠D = 180°
⇒ ∠D = 180° − 45°
⇒ ∠D = 135°

∠B + ∠C = 180°
⇒ 45° + ∠C = 180°
⇒ ∠C = 180° − 45°
⇒ ∠C = 135°

Hence, ∠C = 135° and ∠D = 135°.

Question 10:

ABCD is a rhombus in which altitude from vertex D to side AB bisects AB. The measures of angles of the rhombus are ________.

Answer 10:

Given:
ABCD is a rhombus
Altitude from vertex D to side AB bisects AB.

Let DE biescts AB.
⇒ AE = EB                                     ...(1)

In ∆ADE and  ∆BDE

DE = DE (Common side)
∠AED = ∠BED (Right angle)
AE = EB (from (1))

By SAS property,  ∆ADE ≅ ∆BDE

Thus, AD = BD (By C.P.C.T)          ...(2)

Since, sides of a rhombus are equal
Thus, AD = AB                                ...(3)

From (2) and (3)
AD = BD = AB 
Therefore, ∆ADB is an equilateral triangle.

Hence, ∠A = 60°
∠A = ∠C = 60° (opposite angles of rhombus are equal)

Sum of adjacent angles of a rhombus is 180°
⇒ ∠A + ∠D = 180° and ∠C + ∠B = 180°
⇒ 60° + ∠D = 180° and 60° + ∠B = 180°
⇒ ∠D = 180° − 60° and ∠B = 180° − 60°
⇒ ∠D = 120° and ∠B = 120°

Hence, the measures of angles of the rhombus are 60°, 120°, 60° and 120°.

Question 11:

Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35°, then ∠B = _________.

Answer 11:

Given: Diagonals of a quadrilateral ABCD bisect each other

Since, the diagonals of a quadrilateral ABCD bisect each other
Therefore, It is a parallelogram

Sum of interior angles of a parallelogram is 180°
⇒ ∠A + ∠B = 180°
⇒ 35° + ∠B = 180°
⇒ ∠B = 180° − 35°
⇒ ∠B = 145°

Hence, ∠B =  145°.

Question 12:

In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, then DE = __________.

Answer 12:

Given:
In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm
D and E are respectively the mid-points of AB and BC


Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

Therefore,
$$\begin{gathered} DE=\frac{1}{2}\left(AC\right) \\ =\frac{1}{2}\times 7 \\ =3.5 \mathrm{cm} \end{gathered}$$


Hence, DE = 3.5 cm.

Question 13:

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, then CD = ___________.

Answer 13:

Given:
Opposite angles of a quadrilateral ABCD are equal
AB = 4 cm

If Opposite angles of a quadrilateral ABCD are equal, then ABCD is a parallelogram.
Thus, CD = AB = 4 cm

Hence, CD = 4 cm.

Question 14:

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC = _______.

Answer 14:

Given:
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O
∠DAC = 32°
∠AOB = 70°




Let ∠DBC be x.

Since, AB || DC
Therefore, ∠DAC = ∠ACB = 32°   (alternate angles)


Also, ∠AOB  + ∠BOC = 180° (angles on a straight line)
⇒ 70° + ∠BOC = 180°
⇒ ∠BOC = 180° − 70°
⇒ ∠BOC = 110°


In ∆COB,
∠OCB  + ∠CBO  + ∠BOC = 180°
⇒  32° + x + 110° = 180°
⇒ x + 142° = 180°
⇒ x = 180° − 142°
⇒ x = 38°

Hence, ∠DBC = 38°.

Question 15:

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

Answer 15:

Given:
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rectangle.
Let the rectangle be ABCD.



A is the mid-point of PQ, B is the mid-point of QR, C is the mid-point of RS and D is the mid-point of PS.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB || PR

and In ∆QRS,
BC || QS

Since, AB$\perp$BC
Therefore, PR$\perp$QS


Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken  in order, is a rectangle, if diagonals of PQRS are perpendicular.

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Question 16:

The quadrilateral formed joining the mid-points of the sides of quadrilaterals PQRS, taken in order, is a rhombus, if diagonals of PQRS are _____________.

Answer 16:

Given: 
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rhombus.
Let the rhombus be ABCD.



A is the mid-point of PQ, B is the mid-point of QR, C is the mid-point of RS and D is the mid-point of PS.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB = $\frac{1}{2}$PR

and In ∆QRS,
BC = $\frac{1}{2}$QS

Since, AB = BC (sides of a rhombus are equal)
Therefore, $\frac{1}{2}$PR = $\frac{1}{2}$QS
⇒ PR = QS
​

Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken  in order, is rhombus, if diagonals of PQRS are equal.

Question 17:

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, diagonals of ABCD are ___________ and _________.

Answer 17:

Given: 
ABCD is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral ABCD, is a square.
Let the square be PQRS.



P is the mid-point of AB, Q is the mid-point of BC, R is the mid-point of CD and S is the mid-point of AD.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
PQ = $\frac{1}{2}$AC

and In ∆BCD,
QR = $\frac{1}{2}$BD

Since, PQ = QR (sides of a square are equal)
Therefore, $\frac{1}{2}$AC = $\frac{1}{2}$BD
⇒ AC = BC

Hence, the diagonals are equal.
​
In ∆ABC,
PQ || AC

and In ∆BCD,
QR || BD

Since, PQ$\perp$QR
Therefore, AC$\perp$BD

Hence, the diagonals are perpendicular.

Hence, the figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, diagonals of ABCD are perpendicular and equal.

Question 18:

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a quadrilateral whose opposite angles are __________.

Answer 18:

Given: 
ABCD is a quadrilateral
bisectors of ∠A and ∠B intersect each other at P
bisectors of ∠B and ∠C intersect each other at Q
bisectors of ∠C and ∠D intersect each other at R
bisectors of ∠D and ∠A intersect each other at S

In ∆DAS,
∠ASD + ∠SDA + ∠DAS = 180° (angle sum property)
⇒ ∠ASD + $\frac{1}{2}$∠D + $\frac{1}{2}$∠A = 180°
⇒ ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A

Also, ∠PSR = ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A
⇒ ∠PSR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A      ...(1)

Similarly,
In ∆BQC,
∠BQC + ∠QCB + ∠CBQ = 180° (angle sum property)
⇒ ∠BQC + $\frac{1}{2}$∠C + $\frac{1}{2}$∠B = 180°
⇒ ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B

Also, ∠PQR = ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
⇒ ∠PQR = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B      ...(2)

Adding (1) and (2), we get
∠PSR + ∠PQR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A + 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
                          = 360° − $\frac{1}{2}$(∠D + ∠A + ∠C + ∠B)
                          = 360° − $\frac{1}{2}$(360°)
                          = 360° − 180°​
                          = 180°

Since, sum of angles of a quadrilateral is 360°
Therefore, ∠PSR + ∠PQR + ∠QRS + ∠QPS = 360°
⇒ ∠QRS + ∠QPS = 180°

Hence, the sum of the opposite angles of the quadrilateral PQRS is 180°.

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Question 19:

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form a ____________.

Answer 19:

Given: 
APB and CQD are two parallel lines

Let the bisectors of the angle APQ and CQP intersects at the point R and the bisectors of the angle BPQ and PQD intersects at the point S.

Join PR, RQ, QS and SP as shown in the figure.



APB || CQD
⇒ ∠APQ = ∠PQD (alternate angles)
⇒ 2∠RPQ = 2∠PQS
⇒ ∠RPQ = ∠PQS
⇒  RP || SQ      ...(1)

Similarly,
APB || CQD
⇒ ∠BPQ = ∠PQC (alternate angles)
⇒ 2∠SPQ = 2∠PQR
⇒ ∠SPQ = ∠PQR
​⇒  RQ || SP      ...(2)

From (1) and (2),
PSQR is a parallelogram,

Also, ∠CQP + ∠PQD = 180° (angles on a straight line)
⇒ 2∠RQP + 2∠PQS = 180°
⇒ ∠RQP + ∠PQS = 90°
⇒ ∠RQS = 90°

Therefore, PSQR is a rectangle.

Hence, if APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form a rectangle.

Question 20:

D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is a ____________.

Answer 20:

Given: 
D and E are the mid-points of the sides AB and AC of ∆ABC.
O is any point on side BC
P and Q are the mid-points of OB and OC respectively.



Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
DE || BC || PQ        ...(1) 
DE = $\frac{1}{2}$BC
⇒ DE = $\frac{1}{2}$(BP + PO + OQ + QC)
⇒ DE = $\frac{1}{2}$(2OP + 2OQ)
⇒ DE = OP + OQ
⇒ DE = PQ             ...(2)

In ∆AOB,
DP || AO                  ...(3) 
DP = $\frac{1}{2}$AO             ...(4)


In ∆AOC,
EQ || AO                  ...(5) 
EQ = $\frac{1}{2}$AO             ...(6)

From (3) and (5),
DP || EQ

From (4) and (6),
DP = EQ

Thus, DE || PQ and DP || EQ
DP = EQ and DE = PQ

Hence, DEQP is a parallelogram.

Question 21:

When all the sides of a quadrilateral are equal, then it is either a ____________or a _____________.

Answer 21:

All the sides of a square or a rhombus are equal.

Hence, when all the sides of a quadrilateral are equal, then it is either a square or a rhombus.

Question 22:

The bisectors of two adjacent sides of a parallelogram ABCD meet at a point P inside the parallelogram. The angle made by these bisectors at point P is ____________.

Answer 22:

Given: 
ABCD is a parallelogram
Bisectors of ∠A and ∠B intersect each other at P


∠A + ∠B = 180°      (interior angles)
⇒ $\frac{1}{2}$∠A + $\frac{1}{2}$∠B = $\frac{1}{2}$(180°)
⇒ ∠PAB + ∠PBA = 90°      ....(1)

Now, in ∆APB
∠PAB + ∠PBA + ∠APB = 180° (angle sum property)
⇒ 90° + ∠APB = 180°         (From (1))
⇒ ∠APB = 180° − 90°
⇒ ∠APB = 90°


Hence, the angle made by these bisectors at point P is 90°.

Question 23:

In the given figure, PQRS is a rhombus. If ∠OPQ = 35°, then ∠ORS + ∠OQP = ____________.

Answer 23:

Given: 
PQRS is a rhombus
∠OPQ = 35°


PQRS is a rhombus
PQ || RS
∠OPQ = ∠ORS       (alternate angles)
⇒ ∠ORS = 35°      ....(1)

We know, diagonals of a rhombus intersect at right angle
∴ ∠POQ = 90°      ....(2)

Now, in ∆OPQ
∠OPQ + ∠POQ + ∠OQP = 180° (angle sum property)
⇒ 35° + 90° + ∠OQP = 180°
⇒ 125° + ∠OQP = 180°
⇒ ∠OQP = 180° − 125°
⇒ ∠OQP = 55°      ....(3)

Adding (1) and (3), we get
∠ORS + ∠OQP = 35° + 55° = 90°


Hence, ∠ORS + ∠OQP =  90°.

Question 24:

In the given figure, ABC is an isosceles triangle in which AB = AC. AEDC is a parallelogram. If ∠CDF = 70° and ∠BFE = 100°, then ∠FBA = ____________.

Answer 24:

Given: 
ABC is an isosceles triangle
AB = AC
AEDC is a parallelogram
∠CDF = 70°
∠BFE = 100°


AEDC is a parallelogram
∠ACD + ∠CDE = 180°     (interior angles)
⇒ ∠ACD + 70° = 180°
⇒ ∠ACD = 180° − 70°
⇒ ∠ACD = 110°      ....(1)


Now, ∠ACD + ∠ACB = 180° (angles on a straight line)
⇒ 110° + ∠ACB = 180°
⇒ ∠ACB = 180° − 110°
⇒ ∠ACB = 70°       ....(2)

Also, ∠BFE + ∠BFD = 180° (angles on a straight line)
⇒ 100° + ∠BFD = 180°
⇒ ∠BFD = 180° − 100°
⇒ ∠BFD = 80°       ....(3)


Now, in ∆BFD
∠FBD + ∠BDF + ∠BFD = 180° (angle sum property)
⇒ ∠FBD + 70° + 80° = 180°
⇒ ∠FBD + 150° = 180°
⇒ ∠FBD = 180° − 150°
⇒ ∠FBD = 30°      ....(4)

Since, ABC is an isosceles triangle with AB = AC
Thus, ∠ABC = ∠ACB = 70° (From (2))

∠ABC = ∠ABF + ∠FBD
⇒  70° = ∠ABF + 30°
⇒  ∠ABF = 70° − 30°
⇒  ∠ABF = 40°


Hence, ∠FBA = 40°.