RS AGGARWAL CLASS 9 CHAPTER 10 QUADRILATERALS MCQ

 MULTIPLE CHOICE QUESTIONS 

PAGE NO-349

Question 1:

Three angles of a quadrilateral are 80°, 95° and 112°. Its fourth angle is
(a) 78°
(b) 73°
(c) 85°
(d) 100°

Answer 1:

(b) 73°​

Explanation: 
Let the measure of the fourth angle be x^o.
Since the sum of the angles of a quadrilateral is 360^o, we have:
80^o + 95^o + 112^o + x = 360^o  
⇒ 287^o + x = 360^o ​
⇒ x = 73^o
Hence, the measure of the fourth angle is 73^o.

Question 2:

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is
(a) 45°
(b) 60°
(c) 36°
(d) 48°

Answer 2:

(b) 60°​

Explanation:
Let ∠A = 3x​, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360^o, we have:
3x + 4x + 5x + 6x = 360^o   
⇒ 18x = 360^o ​
⇒ x = 20^o
∴ ∠A = 60^o​, ∠B = 80^o, ∠C = 100^o and ∠D = 120^{o
​Hence, the smallest angle is 60$°$.}

Question 3:

In the given figure, ABCD is a parallelogram in which ∠BAD = 75° and ∠CBD = 60°. Then, ∠BDC = ?
(a) 60°
(b) 75°
(c) 45°
(d) 50°

Answer 3:

(c) 45°

Explanation: 
∠B = 180^o − ∠A 
⇒ ∠B = 180^o − 75^o = 105^o
​Now, ∠B =​ ∠ABD + ∠CBD 
⇒​​ 105^o​ = ∠ABD + 60^o
⇒ ∠ABD​ = 105^o − 60^o = 45^o
⇒ ∠ABD = ​∠BDC​ = 45^o          (Alternate angles)

Question 4:

ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?
(a) 40°
(b) 25°
(c) 65°
(d) 130°

Answer 4:

We know that diagonals of rhombus bisect each other at 90°.

Then, in ΔBOC,

90° + 50° + ∠OBC = 180°        (Angle sum property of triangle)

$\Rightarrow$ ∠OBC = 180° $-$ 140°

$\Rightarrow$ ∠OBC = 40°

But ∠OBC = ∠ADB     (Alternate interior angles)

Thus, ∠ADB = 40°

Hence, the corerct option is (a).

Question 5:

In which of the following figures are the diagonals equal?
(a) Parallelogram
(b) Rhombus
(c) Trapezium
(d) Rectangle

Answer 5:

(d) Rectangle.

The diagonals of a rectangle are equal.

Question 6:

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a
(a) trapezium
(b) parallelogram
(c) rectangle
(d) rhombus

Answer 6:

(d) rhombus

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

PAGE NO-350

Question 7:

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is
(a) 10 cm
(b) 12 cm
(c) 9 cm
(d) 8 cm

Answer 7:

(a) 10 cm

Explanation:

Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆​AOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.
Now, AB² = OA² + OB²              [Pythagoras theorem]
⇒ ​AB² = (8)² + (6)²
⇒​ AB² =​ 64 + 36 = 100
⇒​ AB =​ 10 cm

Hence, the side of the rhombus is 10 cm.

Question 8:

The length of each side of a rhombus is 10 cm and one if its diagonals is of length 16 cm. The length of the other diagonal is
(a) 13 cm
(b) 12 cm
(c) $2\sqrt{39} \mathrm{cm}$
(d) 6 cm

Answer 8:

(b) 12 cm

Explanation: 


Let ABCD be the rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of the rhombus.  
Let AC be x and BD be 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle in which OA = AC ^$÷$2 = x ^$÷$÷ 2  and OB = BD ^_$÷$÷2 = 16 ^$÷$÷ 2 = 8 cm.

Now, AB²= OA² + OB²              [Pythagoras theorem]
$$\begin{gathered} \Rightarrow {10}^2 = {\left(\frac{x}{2}\right)}^2 + 8^2 \\ \Rightarrow {\left(\frac{x}{2}\right)}^2 = 36 = 6^2 \\ \Rightarrow x = 2\times 6 =12 \mathrm{cm} \end{gathered}$$
⇒102 = (x2)2 + 82⇒100 − 64 = x24⇒36 ×4 = 

Question 9:

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is
(a) 55°
(b) 70°
(c) 45°
(d) 50°

Answer 9:

 

Given: In rectangle ABCD, ∠OAD = 35°.

Since, ∠BAD = 90°

$\Rightarrow$ ∠OAB = 90° $-$ 35° = 55°

In ΔOAB,

Since, OA = OB      (Diagonals of a rectangle are equal and bisect each other)

$\Rightarrow$ ∠OAB = ∠OBA = 55°     (Angles opposite to equal sides are equal)

Now, in ΔODA,
 
55° + 55° + ∠DOA = 180°       (Angle sum property of a triangle)

$\Rightarrow$ ∠DOA = 180° $-$ 110°

$\Rightarrow$ ∠DOA = 70°

Thus, the acute angle between the diagonals is 70°.

Hence, the correct option is (b).

Question 10:

If ABCD is a parallelogram with two adjacent angles ∠A = ∠B, then the parallelogram is a
(a) rhombus
(b) trapezium
(c) rectangle
(d) none of these

Answer 10:

(c) Rectangle

Explanation:
∠A = ∠B
​Then ∠A + ∠B = 180^o
⇒ 2∠A = 180^o
⇒ ∠A​ = 90^o
⇒ ∠A​ =​ ∠B​ =​∠C​ =​​∠D = 90^o
∴​ The parallelogram is a rectangle.

Question 11:

In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A and ∠B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB = ?
(a) 40°
(b) 50°
(c) 80°
(d) 100°

Answer 11:

(b) 50^o

​​Explanation:

∠C = 70^o and ∠D = 30^o
​Then ∠A + ∠B = 360^o - (70 +30)^o = 260^o
∴ $\frac{1}{2}$(∠A +∠B) =$\frac{1}{2}$ (260^o) = 130^o
In ∆​ AOB, we have:
∠AOB​ = 180^o - [$\frac{1}{2}$(∠A +∠B)​]
⇒ ∠AOB​ = ​180 - 130 = 50^o

Question 12:

The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 40°
(b) 45°
(c) 60°
(d) 90°

Answer 12:

(d) 90° 

Explanation:
Sum of two adjacent angles = 180^o
Now, sum of angle bisectors of two adjacent angles = $\frac{1}{2}\times \left({180}^{\mathrm{o}}\right) = {90}^{\mathrm{o}}$
∴ Intersection angle of bisectors of two adjacent angles =  180^o − 90^o =  90^o

Question 13:

The bisectors of the angles of a parallelogram enclose a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 13:

(c) Rectangle

The bisectors of the angles of a parallelogram encloses a rectangle.

Question 14:

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
(a) rectangle
(b) parallelogram
(c) rhombus
(d) quadrilateral whose opposite angles are supplementary

Answer 14:

Given: In quadrilateral ABCD, AS, BQ, CQ and DS are angle bisectors of angles A, B, C and D, respectively.

∠QPS = ∠APB        (Vertically opposite angles)          ...(1)

In $∆$APB,

∠APB + ∠PAB + ∠ABP = 180°        (Angle sum property of triangle.)

$\Rightarrow$ ∠APB + $\frac{1}{2}$∠A + $\frac{1}{2}$∠B = 180°

$\Rightarrow$ ∠APB  = 180° – $\frac{1}{2}$(∠A + ∠B)                  ...(2)

From (1) and (2), we get

∠QPS = 180° – $\frac{1}{2}$(∠A + ∠B)                          ...(3)

Similarly, ∠QRS = 180° – $\frac{1}{2}$(∠C + ∠D)        ...(4)

From (3) and (4), we get

∠QPS + ∠QRS = 360° – $\frac{1}{2}$(∠A + ∠B + ∠C + ∠D)

                          = 360° – $\frac{1}{2}$(360°)

                          = 360° – 180°

                          = 180°

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

Hence, the correct option is (d).

Question 15:

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 15:

(d) parallelogram

The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.

Question 16:

The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 16:

(b) Square

The figure formed by joining the mid points of the adjacent sides of a square is a square.

Question 17:

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 17:

(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is  a parallelogram.

Question 18:

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 18:

(a) rhombus

The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.

PAGE NO-351

Question 19:

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 19:

(c) Rectangle

The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.

Question 20:

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if
(a) ABCD is a Parallelogram
(b) ABCD is rectangle
(c) diagonals of ABCD are equal
(d) diagonals of ABCD are perpendicular to each other.

Answer 20:

Since,

The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram ,

The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular to each other is rectangle.

Hence, the correct option is (d).

Question 21:

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if
(a) ABCD is a Parallelogram
(b) ABCD is rhombus
(c) diagonals of ABCD are equal
(4) diagonals of ABCD are perpendicular to each other.

Answer 21:

Given:

The quadrilateral ABCD is a rhombus.

So, the sides AB, BC, CD and AD are equal.

Now, in $∆PQS,$ we have

$DC=\frac{1}{2}QS$     (Using mid-point theorem)          ...(1)

Similarly, in $∆PSR,$

$BC=\frac{1}{2}PR$                     ...(2)

As, BC = DC

$\Rightarrow \frac{1}{2}QS = \frac{1}{2}PR$        [From (1) and (2)]

So, QS = PR

Thus, the diagonals of PQRS are equal.

Hence, the correct option is (c).

Question 22:

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if
(a) ABCD is a rhombus
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are perpendicular
(d) diagonals of ABCD are equal and perpendicular

Answer 22:

Since,

The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular is rectangle, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal and perpendicular is square.

Hence, the correct option is (d).

Question 23:

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 72°
(d) 81°

Answer 23:

(c) 72°​

Explanation: 
 Let ABCD be a parallelogram. 
  ∴ ∠A = ∠C and ∠B = ∠D      (Opposite angles)
   Let ∠A = x and ∠​B = $\frac{2}{3}x$(4x5)
  ∴ ∠​A + ∠B = 180^o                (Adjacent angles are supplementary)
⇒ x + $\frac{2}{3}$x = 180^o
⇒ $\frac{5}{3}x=180°$
⇒ x = 108^o
∴ ∠B = $\frac{2}{3}\times$ (108^o) = 72^o
Hence, ∠A = ∠C  = 108^o  and ∠B = ∠D = 72^o
⇒9x5 = 180o⇒x = 100o∠A = 100o an

Question 24:

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is
(a) 68°
(b) 102°
(c) 112°
(d) 136°

Answer 24:

(c)112°​

Explanation:
Let ABCD is a parallelogram. 
  ∴ ∠​A = ∠C and ∠​B = ∠D          (Opposite angles)
 Let ∠A be the smallest angle whose measure is x.
 ∴​∠B  = (2x − 24)^o
 Now, ∠​A + ∠B = 180^o                (Adjacent angles are supplementary) 
   ⇒ x + 2x − 24^o = 180^o
   ⇒ 3x =  204^o
  ⇒ x = 68^o
∴​∠​B = 2 ⨯ 68^o − 24^o = 112^o
 Hence, ∠A = ∠C = 68^o and ∠B = ∠D = 112^o

Question 25:

If ∠A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3 : 7 : 6 : 4 then ABCD is a
(a) rhombus
(b) kite
(c) trapezium
(d) parallelogram

Answer 25:

(c) Trapezium

Explanation:

Let the angles be (3x), (7x), (6x) and (4x)​.
Then 3x + 7x + 6x + 4x = 360^o
∴ x = 18^o
​Thus, the angles are 3 ⨯18^o = 54^o, ​7 ⨯ 18^o = 126^o, ​6 ⨯ 18^o = 108^o and ​4 ⨯18^o = 72^o.
But 54^o + 126^o = 180^o and 72^o + 108^o  = 180^o
∴ ABCD is a trapezium.

Question 26:

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.

Answer 26:

(c) Opposite angles are bisected by the diagonals.

Question 27:

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a
(a) square
(b) rhombus
(c) rectangle
(d) kite

Answer 27:

(c) Rectangle

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a rectangle.

PAGE NO-352

Question 28:

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD = ?
(a) 45°
(b) 55°
(c) 60°
(d) 75°

Answer 28:

(c)​ 60°

Explanation:
∠BAD = ∠BCD = 75^o       [Opposite angles are equal]
In ∆ BCD, ∠ C = 75^o            
∴ ​∠CBD = 180^o − (75^o + 45^o) = 60^o

Question 29:

If area of a || gm with sides a and b is A and that of a rectangle with sides a and b is B, then
(a) A > B
(b) A = B
(c) A < B
(d) A ≥ B

Answer 29:

(c) A < B

Explanation:
Let h be the height of parallelogram.
Then clearly, h < b
∴ ​A = a ⨯​ h < a ⨯ b = B
Hence, A < B

Question 30:

In the given figure, ABCD is a || gm and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then,
(a) $AF=\frac{3}{2}AB$
(b) AF = 2AB
(c) AF = 3AB
(d) AF² = 2AB²

Answer 30:

(b) AF = 2 AB

Explanation:
I​n parallelogram ABCD, we have:
AB || DC
∠DCE = ∠​ EBF            (Alternate interior angles)
In ∆ DCE and ​ ∆ BFE, we have:
∠DCE = ∠ EBF              (Proved above)

∠DEC = ∠ BEF              (Vertically opposite angles)
BE = CE           ( Given)
i.e., ∆ DCE ≅​ ∆ BFE     (By ASA congruence rule)
∴  DC = BF         (CPCT)

But DC= AB, as ABCD is a parallelogram.
∴ DC = AB =  BF                 ...(i)

Now, AF = AB + BF             ...(ii)
   From (i), we get:
∴ AF = AB + AB = 2AB

Question 31:

P is any point on the side BC of a ∆ABC. P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is
(a) a trapezium
(b) a parallelogram
(c) a rectangle
(d) a rhombus

Answer 31:

Given: In ∆ABC, M, N, D and E are the mid-points of BP, CP, AB and AC, respectively.

In ∆ABP,

$\because$ D and M are the mid-points of AB,and BP, respectively.      (Given)

$\therefore$ BM = $\frac{1}{2}$AP and BM || AP          (Mid-point theorem)       ...(i)

Again, in ∆ACP,

$\because$ E and N are the mid-points of AC,and CP, respectively.      (Given)

$\therefore$ EN = $\frac{1}{2}$AP and EN || AP          (Mid-point theorem)       ...(ii)

From (i) and (ii), we get

BM = EN and BM || EN

But this a pair of opposite sides of the quadrilateral DENM.

So, DENM is a parallelogram.

Hence, the correct option is (b).

Question 32:

The parallel sides of a trapezium are a and b respectively. The line joining the mid-points of its non-parallel sides will be
(a) $\frac{1}{2}(a-b)$
(b) $\frac{1}{2}(a+b)$
(c) $\frac{2ab}{(a+b)}$
(d) $\sqrt{ab}$

Answer 32:

(b) $\frac{1}{2}\left(a+b\right)$

Explanation:

Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in ∆ DAB, E is the midpoint of AD and EM || AB.
∴ M is the mid point of BD and EM = $\frac{1}{2}\left(a\right)$
Similarly, M is the mid point of BD and MF || DC.
i.e., F is the midpoint of BC and MF = $\frac{1}{2}\left(b\right)$

∴ EF =  EM + MF = $\frac{1}{2}\left(a+b\right)$
 

Question 33:

In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?
(a) $\frac{1}{2}AB$
(b) $\frac{1}{2}CD$
(c) $\frac{1}{2}(AB+CD)$
(d) $\frac{1}{2}(AB-CD)$

Answer 33:

(d) $\frac{1}{2}\left(AB - CD\right)$

Explanation:

Join CF and produce it to cut AB at G.
Then ∆CDF  ≅ ∆GBF                [∵ DF = BF, ∠​DCF = ∠​BGF and ∠​CDF = ∠​GBF]
∴ CD = GB
Thus, in ∆​CAG, the points E and F are the mid points of AC and CG, respectively. 
∴ EF = $\frac{1}{2}\left(AG\right)$ = $\frac{1}{2}\left(AB - GB\right) =\frac{1}{2}\left(AB - CD\right)$

Question 34:

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB = ?
(a) 45°
(b) 60°
(c) 90°
(d) 30°

Answer 34:

(c) ​90°

Explanation:
∠B = ∠D  
⇒ $\frac{1}{2}$∠B = $\frac{1}{2}$∠D​
⇒​ ∠ADB = ∠​ABD
∴ ∆ABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of ∆ABD.
∴ AM ⊥ BD and, hence, ​∠AMB =​​ 90°

Question 35:

In the given figure, ABCD is a rhombus. Then,
(a) AC² + BD² = AB²
(b) AC² + BD² = 2AB²
(c) AC² + BD² = 4AB²
(d) 2(AC² + BD²) = 3AB²

Answer 35:

(c) ​AC² + BD² = 4AB²

Explanation:
We know that the diagonals of a rhombus bisect each other at right angles.
Here, OA = $\frac{1}{2}$AC, OB = $\frac{1}{2}$BD and ∠​AOB ​= 90°
Now, AB²= OA² + OB² = $\frac{1}{4}$(AC)² + $\frac{1}{4}$(BD)²
∴ 4AB² = (AC² + BD²)

PAGE NO-353

Question 36:

In a trapezium ABCD, if AB || CD, then (AC² + BD²) = ?
(a) BC² + AD² + 2BC ⋅ AD
(b) AB² + CD² + 2AB ⋅ CD
(c) AB² + CD² + 2AD ⋅ BC
(d) BC² + AD² + 2AB ⋅ CD

Answer 36:

(c) BC² + AD² + 2AB.CD


Explanation:

Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
 ∴​ DEFC is a parallelogram and EF = CD.

In ∆ABC, ∠B is acute.
∴ AC²= BC² + AB² - 2AB.AE
In ∆ABD, ∠A is acute.​
∴ ​BD² = AD² + AB² - 2AB.AF
∴ ​AC² + BD² = (BC² + AD²) + (AB² + AB² ) - 2AB(AE + BF)
                     = (BC² + AD²) + 2AB(AB - AE - BF)                [∵ AB = AE + EF + FB and AB - AE =  BE]
                     = (BC² + AD²) + 2AB(BE - BF)
                     = (BC² + AD²) + 2AB.EF 
∴ AC² + BD² = ​(BC² + AD²) + 2AB.CD          

Question 37:

Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 3
(d) 1 : 1

Answer 37:

(d) 1:1

Area of a parallelogram = base ⨯ height
If both parallelograms stands on the same base and between the same parallels, then their heights are the same.
So, their areas will also be the same.

Question 38:

In the given figure, AD is a median of ∆ABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?
(a) $\frac{1}{2}AC$
(b) $\frac{1}{3}AC$
(3) $\frac{2}{3}AC$
(4) $\frac{3}{4}AC$

Answer 38:

(b) ⅓ AC

Explanation:

Let G be the mid point of FC. Join DG.
​In ∆BCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF 
⇒ DG || EF

​In ∆ ADG, E is the mid point of AD and EF || DG.
i.e., F is the mid point of AG.
Now, AF = FG = GC       [∵ G is the mid point of FC] 
∴ AF =⅓ AC​
4AC

Question 39:

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30° and ∠AOB = 70°. Then, ∠DBC = ?
(a) 40°
(b) 35°
(c) 45°
(d) 50°

Answer 39:

(a) 40°

Explanation:
∠OAD = ​∠OCB = 30^o              (Alternate interior angles)
∠AOB + ∠BOC = 180^o              (Linear pair of angles)
∴ ∠BOC = 180^o − 70^o = 110^o       (∠ AOB = 70^o)
In ∆BOC, we have:
∠OBC = 180^o − (110^o + 30^o) = 40^o
∴ ​∠DBC = 40^o

Question 40:

Three statements are given below:
I. In a || gm, the angle bisectors of two adjacent angles enclose a right angle.
II. The angle bisectors of a || gm form a rectangle.
III. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer 40:

(c) I and II

Statements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.

Question 41:

Three statements are given below:
I. In a rectangle ABCD, the diagonal AC bisects ∠A as well as ∠C.
II. In a square ABCD, the diagonal AC bisects ∠A as well as ∠C.
III. In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.
(a) I only
(b) II and III
(c) I and III
(d) I and II

Answer 41:

(b) II and III

Clearly, statements II and III are true. Statement I is false, as diagonal of a rectangle does not bisect ∠A and ∠C (∴ adjacent sides are not equal).

Question 42:

In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ.

Answer 42:

Since, the opposite angles of the quadrilateral PQRS are equal.

$\Rightarrow$ Quadrilateral PQRS is a parallelogram.

$\Rightarrow SR = PQ$  (Opposite sides of a parallelogram are equal.)

$\therefore SR = PQ = 2\mathit{ }\mathrm{cm}$

PAGE NO-354

Question 43:

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

Answer 43:

No, the statement is incorrect because the diagonals of a parallelogram bisect each other.

Question 44:

What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 180°?

Answer 44:

Since, in quadrilateral PQRS, ∠P + ∠S = 180°.

i.e. the sum of adjacent angles is 180$°$.

So, PQRS is a parallelogram.

Question 45:

All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.

Answer 45:

No, the statement is false because if all the four angles of a quadrilateral are less than 90$°$, then the sum of all four angles will be less than 360$°$.

Question 46:

All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.

Answer 46:

Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles.

Question 47:

All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.

Answer 47:

No, the statement is false because if all angles are greater than 90$°$, then the sum of four obtuse angles will be greater than 360$°$.

Question 48:

Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.

Answer 48:

Since, the sum of all angles (i.e. 70° + 115° + 60° + 120° = 365$°$).

So, we cannot form a quadrilateral with these angles.

Question 49:

What special name can be given to a quadrilateral whose all angles are equal?

Answer 49:

We know that, sum of all angles in a quadrilateral is 360°.

Let each angle of the quadrilateral be x.

x + x + x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

⇒ All angles of the quadrilateral are 90°.

Hence, given quadrilateral is a rectangle.

Question 50:

If D and E are respectively the midpoints of the sides AB and BC of ∆ABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then determine the length of DE.

Answer 50:

In $∆$ABC,

Since, D and E are respectively the mid-points of sides AB and BC.     (Given)

So, DE = $\frac{1}{2}$AC       (Uing mid-point theorem)

But AC = 3.6 cm      (Given)

DE = $\frac{1}{2}$(3.6)

or, DE = 1.8 cm

Hence, the length of DE is 1.8 cm.

Question 51:

In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.

Answer 51:

Since, the diagonals PR and QS of quadrilateral PQRS bisect each other.        (Given)

So, PQRS is a parallelogram.

Now, $\angle Q+\angle R=180°$(Adjacent angles are supplementary.)

$$\begin{gathered} \Rightarrow 56°+\angle R=180° \\ \Rightarrow \angle R=180°-56° \\ \Rightarrow \angle R=124° \end{gathered}$$

Question 52:

In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not

Answer 52:

Given: Parallelograms BDEF and AFDE.

Since, BF = DE     (Opposite sides of parallelogram BDEF)       ...(i)

And, AF = DE     (Opposite sides of parallelogram AFDE)       ...(ii)

From (i) and (ii), we get

AF = FB
 

Question 53:

Is quadrilateral ABCD a || gm?
I. Diagonals AC and BD bisect each other.
II. Diagonals AC and BD are equal.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

Answer 53:

We know that if the diagonals of a ​quadrilateral bisects each other, then it is a parallelogram.
∴ I gives the answer.
If the diagonals of a quadrilateral are equal, then it is not necessarily a ​parallelogram.
∴ II does not give the answer.

Hence, the correct answer is (a).

PAGE NO-355

Question 54:

Is quadrilateral ABCD a  rhombus?
I. Quad. ABCD is a || gm.
II. Diagonals AC and BD are perpendicular to each other.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

Answer 54:

Clearly, I alone is not sufficient to answer the given question.
Also, II alone is not sufficient to answer the given question.​
However, both I and II together will give the answer.
∴ Hence, the correct answer is (c).

Question 55:

Is || gm ABCD a square?
I. Diagonals of || gm ABCD are equal.
II. Diagonals of || gm ABCD intersect at right angles.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

Answer 55:

When the diagonals of a parallelogram are equal, it is either a rectangle or a square.
Also, if the diagonals intersects at a right angle, then it is a square.
∴ Both I and II together will give the answer.
Hence, the correct answer is (c).

Question 56:

Is quad. ABCD a parallelogram?
I. Its opposite sides are equal.
II. Its opposite angles are equal.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

Answer 56:

We know that a quadrilateral is a parallelogram when either I or II holds true.

​Hence, the correct answer is (b).

Question 57:

Assertion: If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angles is 100°.
Reason: The sum of all the angle of a quadrilateral is 360°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 57:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Explanation:
Fourth angle = 360^o − (130^o + 70^o + 60^o) = 100^o
Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.
Hence, the correct answer is (a).

Question 58:

Assertion: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then, PQRS is a parallelogram.
Reason: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 58:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true and reason (R) gives assertion (A).
Hence, the correct answer is (a).

Question 59:

Assertion: In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.
Reason: The diagonals of a rhombus bisect each other at right angles.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 59:

(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true.

But reason (R) does not give assertion (A).

Hence, the correct answer is (b).

PAGE NO-356

Question 60:

Assertion: Every parallelogram is a rectangle.
Reason: The angle bisectors of a parallelogram form a rectangle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 60:

(d) Assertion is false and Reason is true.

​Explanation:
We can easily prove reason (R). So, reason (R) is true.

Clearly,  assertion (A) is false (as every parallelogram is not necessarily a rectangle).

Hence, the correct answer is (d).

Question 61:

Assertion: The diagonals of a || gm bisect each other.
Reason: If the diagonals of a || gm are equal and intersect at right angles, then the parallelogram is a square.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 61:

​(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, assertion (A) is true.
We can easily prove reason (R). So, (R) is also true.

But, reason (R) does not give assertion (A).

Hence, the correct answer is (b)

Question 62:

Match the following columns:

Column I	Column II
(a) Angle bisectors of a parallelogram form a	(p) parallelogram
(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a	(q) rectangle
(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a	(r) square
(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a	(s) rhombus

(a) .....,
(b) .....,
(c) .....,
(d) .....,

Answer 62:

(a) - (q), (b) - (r), (c) - (s), (d) - (p)

Question 63:

Match the following columns:

Column I	Column II
(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7 cm. If P and Q are the mid-points of AD and BC respectively, then PQ =	(p) equal
(b) In the given figure, PQRS is a || gm whose diagonals intersect at O. If PR = 13 cm, then OR =	(q) at right angles
(c) The diagonals of a square are	(r) 8.5 cm
(d) The diagonals of a rhombus bisect each other	(s) 6.5 cm

(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer 63:

​(a) - (r), (b) - (s), (c) - (p), (d) - (q)

Explanation:

(a) PQ = $\frac{1}{2}$(AB+ CD) = $\frac{1}{2}$(17) = 8.5 cm
 
(b) OR = $\frac{1}{2}$(PR) = ​$\frac{1}{2}$(13) = 6.5 cm