Pages

MCQ Questions of Quadrilaterals Class IX Maths


1.) Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is
(A) 90º
(B) 95º
(C) 105º
(D) 120º

2. A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is
(A) 55º
(B) 50º
(C) 40º
(D) 25º

3. ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is
(A) 40º
(B) 45º
(C) 50º
(D) 60º

Hint : Diagonal of Rhombus bisect each other at 90º.

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.

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.

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

Hint : Two pair of adjacent agnles are supplementary. 

7. 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) rhombus
(C) parallelogram
(D) quadrilateral whose opposite angles are supplementary

8. If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(A) a square
(B) a rhombus
(C) a rectangle
(D) any other parallelogram

OR

If two parallel lines are intersected intersect by a transversal, then the bisectors of the two pairs of interior angles enclose
(a) a square 
(b) a parallelogram 
(c) a rectangle 
(d) a trapezium

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

Hint : Check Question No. 4.

10.  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) a square
(B) a rectangle
(C) a rhombus
(D) a parallelogram

Hint : If one pair of opposite sides of a quadrilateral in equal and parallel then it is a _________. 

11. The figure formed by joining the mid-points 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 equal and perpendicular
(D) diagonals of ABCD are perpendicular.

Hint : Check Question No. 5.

12. The diagonals AC and BD of a parallelogram ABCD intersect each other at the
point O. If ∠DAC = 32º and ∠AOB = 70º, then ∠DBC is equal to
(A) 24º
(B) 86º
(C) 38º
(D) 32º

Hint : Use the concept of alternate interior angles, linear pair and angle sum property.

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.


Answer Key 

1. (D) 120º
2. (B) 50º
3. (C) 50º
4. (C) diagonals of PQRS are perpendicular
5. (D) diagonals of PQRS are equal.
6. (C) trapezium
7. (D) quadrilateral whose opposite angles are supplementary
8. (C) a rectangle or (c) a rectangle 
9. (B) a rectangle
10. 
(D) a parallelogram
11. (C) diagonals of ABCD are equal and perpendicular
12. (C) 38º
13. (C) opposite angles are bisected by the diagonals

Bisectors of angles B and C of a triangle ABC intersect . . .

Question : Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC=90°+∠A/2

OR

In a ΔABC, if the bisectors of ∠B and ∠C meet at point O, then ∠BOC is equal to 90°+∠A/2

Solution : 


In ΔABC

∠A+∠B+∠C=180°(Angle Sum Property)
Multiply both sides by ½
½×∠A+½×∠B+½×∠C=½×180°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠1+∠2 =90°
∠1+∠2 =90°-∠A/2— (1) 

In ΔOBC
∠BOC+∠1+∠2=180° (ASP)
∠BOC+90°-∠A/2=180° [From eq. (1)]
∠BOC=180°-90°+∠A/2
∠BOC=90°+∠A/2


Similar Questions 

1. If one angles of a triangle 130°, then the angle between the bisectors of the other two angles can be 
(a) 50°
(b) 65°
(c) 145°
(d) 155°

2. Assertion (A) : In a ΔABC, if the bisectors of angles of ∠B and ∠C meet at a point O, then ∠BOC is always an obtuse angle. 
Reason (R) : In a ΔABC, if the bisectors of ∠B and ∠C meet at a point O, then ∠BOC=90°+∠A/2

(A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
(B) Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.




For a particular year, following in the distribution of ages . . .

Case Study Based Question on Statistics
For a particular year, following in the distribution of ages (in years) of primary school teachers in a district :

Age (in years) 15-20 20-25 25-30 30-35 35-40 40-45 45-50
No. of Teachers 10 30 50 50 30 6 4

(a) Determine the class limits of the fourth class interval. 

(b) Find the class marks of the class 45-50. Also determine the class size. 

(c) Write the sum of lower limit of first class interval and upper limit of fifth class interval.

(d) How many primary school teachers are of age more than or equal to 35 years. 


Doubt by Veer

Solution : 

(a) Fourth Class Interval is 30-35
Lower class limit is 30
Upper Class limit is 35


(b) Class Marks & Class size  of the class 45-50

Class Marks (xi)
= [LL+UL]/ 2
= [45+50]/2
= 95/2
= 47.5

Class Size (h)
= UL-LL
= 50-45
= 5 


(c) Lower limit of first class interval (15–20)
= 15

Upper limit of fifth class interval (35–40)
= 40

Sum = 15+40
 = 55 

(d) No. of primary school teachers which are of age more than or equal to 35 years
= 30+6+4
= 40

An angle is 24° less than its complementary angle. . .

Question : An angle is 24° less than its complementary angle. The angles measure is :
(a) 36°
(b) 33°
(c) 66°
(d) 57°

Doubt by Veer

Solution : 

Let the angle be x°
Compelent of x
° = (90°-x°)

ATQ
(90°-x°)-x°=24°
90
°-x°-x°=24°
90
°-2x°=24°
90
°-24°=2x°
66
°=2x°
66
°/2=x°
33
°=x°
x
°=33°

Hence, the correct option is (b) 33°.

If the drops of water are spherical and their diameter is 0.1 cm . . .

Question : If the drops of water are spherical and their diameter is 0.1 cm, 32000 such drops fills a conical glass up to its brim, with diameter equal to height, find the height of the glass.

Doubt by Veer

Solution : 
Diameter of spherical drop (d) = 0.1 cm 
Radius of spherical drop (r) = 0.1/2
=1/20 cm 

For Conical Glass
D=H
2R=H
R=H/2

Volume of 32000 raindrops = volume of conical glass

32000×(4/3)πr³ = (1/3)
πR²H 32000×4r³ = R²H 32000×4(1/20)³ = (H/2)²H 32000×4(1/8000) = (H²/4)H 4×4 = H³/4 4×4×4=H³ H³=(4)³ H=4 cm

Hence, the height of the glass is 4 cm

191125

Question : In the given figure, O is the centre of a circle prove that ∠x+∠ y=∠ z.


Doubt by Kashvi

Solution : Coming Soon






MCQs of Circles

MCQ Questions of Circles

1. AD is a diameter of a circle and AB is a chord. If AD=34 cm, AB=30 cm, the distance of AB from the centre of the circle is :
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

2. In given figure, if OA=5 cm, AB=8 cm and OD is perpendicular to AB, then CD is equal to:
(A) 2 cm
(B) 3 cm
(C) 4 cm
(D) 5 cm

3. If AB=12 cm, BC=16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm

4. In given figure, if ∠ABC=20º, then ∠AOC is equal to:
(A) 20º
(B) 40º
(C) 60º
(D) 10º

5. In given figure, if AOB is a diameter of the circle and AC=BC, then ∠CAB is equal to:
(A) 30º
(B) 60º
(C) 90º
(D) 45º

6. In given figure, if ∠OAB=40º, then ∠ACB is equal to :
(A) 50º
(B) 40º
(C) 60º
(D) 70°

7. In given figure, if ∠DAB=60º, ∠ABD=50º, then ∠ACB is equal to:
(A) 60º
(B) 50º
(C) 70º
(D) 80º

8. ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and
∠ADC=140º, then ∠BAC is equal to:
(A) 80º
(B) 50º
(C) 40º
(D) 30º

9. In given figure, BC is a diameter of the circle and ∠BAO=60º. Then ∠ADC is equal to :
(A) 30º
(B) 45º
(C) 60º
(D) 120º

10. In given figure, ∠AOB=90º and ∠ABC=30º, then ∠CAO is equal to:
(A) 30º
(B) 45º
(C) 90º
(D) 60º

11. In given figure, two congruent circles have centres O and O′. Arc AXB subtends an angle of 75º at the centre O and arc A′ Y B′ subtends an angle of 25º at the centre O′. Then the ratio of arcs A X B and A′ Y B′ is:
(A) 2 : 1
(B) 1 : 2
(C) 3 : 1
(D) 1 : 3

12. In given figure, AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD, respectively. If ∠POQ=150º, then ∠APQ is equal to
(A) 30º
(B) 75º
(C) 15º
(D) 60º

ANSWER KEY
1. (D) 
2. (A) 
3. (C) 
4. (B) 
5. (D)
6. (A) 
7. (C) 
8. (B) 
9. (C) 
10. (D)
11. (C)
12. (B)



Case Study Based Question on Coordinate Geometry

The children of SMPS and DPS are very excited for the football match of inter-school competition, for which the school management has selected the Talkatora football ground and made the booking, as shown in picture below.


The length of the stadium is 120 metres and breadth Is 90 metres. If centre spot is considered as origin, then answer the following questions:
(i) What are the coordinates at which the centre circle meets the halfway line? [1 Mark]

(ii) What will be the coordinates of a player standing 22m away from Centre spot on the straight line joining centre spot to the penalty spot towards the right side goal? [1 Mark]

(iii) What will be the coordinates of the four corners of the football ground? [2 Marks]

OR

What will be the coordinates of the penalty spots of both the goals? [2 Marks]


Solution : 

(i) (0, 9.15), (0,-9.15)
(ii) (22,0)
(iii) (60,45), (60, -45), (-60,45) and (-60,-45)
OR
(49,0) and (-49,0)

If a=√[6-√11] and b=√[6+√11] then . . .

Question : If a=√[6-√11] and b=√[6+√11] then the value of (a+b) is : 
(a) √22
(b) 2√11
(c) √6
(d) √12

Doubt by Veer

Solution : Coming Soon

Answer : 
(a) √22

Factories x4+x2+1 . . .

Questin : Factories x⁴+x²+1

Doubt by Veer

Solution : 
x⁴+x²+1
Adding x² and subtracting x² 
=x⁴+x²+1+x²-x²
=
x⁴+x²+x²+1-x²
=
x⁴+2x²+1-x²
=[
x⁴+2x²+1]-x²
=[(x²)²+2(x²)(1)+(1)²]-x²
=[(x²+1)²-x²] [∵a²+2ab+b²=(a+b)²]
=(x²+1-x)(x²+1+x)
=(x²-x+1)(x²+x+1)