Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Trigonometric Identities

Mathematics Aptitude Test for Class 10 on “Trigonometric Identities – 2”.

1. If sec θ – tan θ = M then sec θ + tan θ = (frac {1}{M}).
a) False
b) True
Answer: b
Clarification: The appropriate trigonometric identity used here is sec2 θ – tan2 θ = 1.
(sec θ – tan θ) (sec θ + tan θ) = 1
M (sec θ + tan θ) = 1
(sec θ + tan θ) = (frac {1}{M})

2. Find the correct trigonometric identity.
a) cos2 θ = 1 – sin2 θ
b) cos2 θ = 1 + sin2 θ
c) tan2 θ + sec2 θ = 1
d) tan2 θ = sec2 θ + 1
Answer: a
Clarification: The appropriate trigonometric identity used here is sin2 θ + cos2 θ = 1.
cos2 θ = 1 – sin2 θ

3. Evaluate (sec θ – tan θ) (sec θ + tan θ).
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: = (sec θ – tan θ) (sec θ + tan θ) = sec2 θ – tan2 θ
= 1
The identity used here is sec2 θ – tan2 θ = 1

4. Evaluate (sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin , ⁡⁡A}}).
a) cos A + tan A
b) cos A – tan A
c) tan A – cot A
d) sec A + tan A
Answer: d
Clarification: (sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin⁡ , ⁡A}} = sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin , ⁡⁡A}} . sqrt {frac {1 – sin , ⁡A}{1 – sin , ⁡⁡A}})
= (frac {sqrt {(1 + sin , ⁡A)^2}}{sqrt {1 – sin^2} , ⁡A} )
= (frac {1 + sin⁡ , ⁡A}{sqrt {cos^2} , ⁡A})     (∵ sin2 A + cos2 A = 1)
= (frac {1 + sin⁡ , ⁡A}{cos⁡ , ⁡A})
= sec A + tan A

5. Evaluate (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2.
a) 1
b) 0
c) (cosec2 θ – cot2 θ)2
d) (cosec θ + cot θ)2
Answer: d
Clarification: (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2 = 1 (cosec θ + cot θ)2
= (cosec θ + cot θ)2

6. (1 – sin2 A) (1 + tan2 A) equals to _____
a) – Sec2 θ Tan2 θ
b) – Sec2 θ Tan2 θ
c) 1
d) 0
Answer: c
Clarification: (1 – sin2 A) (1 + tan2 A) = cos2 A . sec2 A
= cos2 A . (frac {1}{cos^2 A})
= 1

7. Evaluate (cosec A – 1) (cosec A + 1) (sec2 A – 1).
a) 0
b) 1
c) (frac {4}{3})
d) (frac {3}{4})
Answer: b
Clarification: (cosec A – 1) (cosec A + 1) (sec2 A – 1) = (cosec2 A – 1) (sec2 A – 1)
= cot2A . tan2 A
= (frac {1}{tan^2 A}) . tan2 A
= 1

8. (sin A + cos A)2 is equal to _____
a) 1 + 2sin A cos A
b) 1 – 2sin A cos A
c) 2sin A cos A – 1
d) 2sin A cos A + 1
Answer: a
Clarification: (sin A + cos A)2 = sin2 A + cos2 A + 2sin A cos A
= 1 + 2sin A cos A

9. Evaluate tan2 A + (1 + sec A) (sec A – 1).
a) 3 tan2 A
b) 0
c) 2tan2 A
d) 1
Answer: c
Clarification: tan2 A + (1 + sec A) (sec A – 1) = tan2 A + (sec A + 1) (sec A – 1)
= tan2 A + (sec2 A – 1)
= tan2 A + tan2 A
= 2tan2 A

10. (1 + cosec⁡ θ) (1 – cosec⁡ θ) + cot2⁡ θ is _____
a) Cot ⁡θ
b) 0
c) 1
d) Tan ⁡θ
Answer: b
Clarification: (1 + cosec⁡ θ) (1 – cosec θ) + cot2 ⁡θ = (1 – cosec2 θ) + cot2 θ
= -cot2⁡ θ + cot2⁡ θ     (∵ cosec2 A – cot2 A = 1)
= 0

To practice Mathematics Aptitude Test for Class 10,

[CLASS 10] Mathematics MCQs on Geometry – Area of Triangle

Mathematics Multiple Choice Questions & Answers on “Geometry – Area of Triangle”.

1. What will be the area of a triangle whose vertices are (3, 1), (0, 4) and (5, 9)?
a) 5 units
b) -5 units
c) -15 units
d) 15 units
Answer: c
Clarification: We know that, area of triangle = (frac {1}{2}){x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2 )}
The coordinates of vertices of the triangle are (3, 1), (0, 4) and (5, 9).
The area of triangle = (frac {1}{2}){3(4 – 9) + 0(9 – 1) + 5(1 – 4) } = (frac {1}{2}) {-15 + 0 – 15} = (frac {-30}{2}) = -15 units
Since area of triangle cannot be zero. So area of triangle is 15 units.

2. What will be the area of the quadrilateral ABCD whose vertices are A (8, 6), B (9, 0), C (1, 2) and D (3, 4)?
a) -20 units
b) 26 units
c) 23 units
d) 3 units
Answer: b
Clarification: Area of quadrilateral = Area of ∆ABC + Area of ∆ADC
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}
The coordinates of vertices of the triangle are (8, 6), (9, 0) and (1, 2).
The area of triangle = (frac {1}{2}){8(0 – 2) + 9(2 – 6) + 1(6 – 0)} = (frac {1}{2}) { – 16 – 36 + 6} = (frac {-46}{2}) = – 23 units
The area of triangle cannot be zero. So, Area of ∆ABC = 23 units
We know that, area of triangle = (frac {1}{2}){x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)}
The coordinates of vertices of the triangle are (8, 6), (3, 4) and (1, 2).
The area of triangle = (frac {1}{2}) {8(4 – 2) + 3(2 – 6) + 1(6 – 4)} = (frac {1}{2}) {16 – 12 + 2} = (frac {6}{2}) = 3 units
So, Area of ∆ADC = 3 units
Area of quadrilateral = Area of ∆ABC + Area of ∆ADC = 23 + 3 = 26 units

3. What will be the area of triangle PQR formed by joining the midpoints of the sides of the triangle whose vertices are (3, 3), (9, 9) and (4, 6)?
a) (frac {1}{2}) units
b) (frac {3}{2}) units
c) (frac {3}{5}) units
d) (frac {5}{2}) units
Answer: b
Clarification:

ABC is the triangle formed by the coordinates A (3, 3), B (9, 9) and C (4, 6) and DEF is the triangle formed by joining the midpoints of AC, BC, AB.
D is the midpoint of AC. Coordinates of D = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {3+4}{2}, frac {3+6}{2} ) = ( frac {7}{2}, frac {9}{2} ) )
E is the midpoint of AB. Coordinates of E = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {3+9}{2}, frac {3+9}{2} ) = ( frac {12}{2}, frac {12}{2} ) ) = (6, 6)
F is the midpoint of BC. Coordinates of F = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {9+4}{2}, frac {9+6}{2} ) = ( frac {13}{2}, frac {15}{2} ) )
The area of triangle formed by D ( ( frac {7}{2}, frac {9}{2} ) ), E (6, 6) and F( ( frac {13}{2}, frac {15}{2} ) )
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
The coordinates of vertices of the triangle are ( ( frac {7}{2}, frac {9}{2} ) ), (6, 6) and ( ( frac {13}{2}, frac {15}{2} ) ).
The area of triangle = (frac {1}{2} { frac {7}{2}) (6 – (frac {15}{2})) + 6((frac {15}{2} – frac {9}{2})) + (frac {13}{2} ( frac {9}{2} – 6 ) } ) = (frac {1}{2} { – frac {21}{4}) + 18 – (frac {39}{4} } = frac {3}{2} ) units

4. If the points A, B, C is collinear then the area of the triangle will be zero.
a) True
b) False
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are (0, 0), (0, 4) and (0, 9).
The area of triangle = (frac {1}{2}) {0(4 – 9) + 0(9 – 0) + 0(0 – 4) } = (frac {1}{2}){0} = 0 units
Hence, the area of triangle will be zero if the points are collinear.

5. What will be the area of triangle formed by joining the points P (2, 3), Q (4, 6) and R (6, 9)?
a) 0
b) 2
c) 3
d) 4
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are (2, 3), (4, 6) and (6, 9).
The area of triangle = (frac {1}{2}) {2(6 – 9) + 4(9 – 3) + 6(3 – 6) } = (frac {1}{2}){0} = 0 units
Hence, the area of triangle is zero, the points are collinear.

6. What will be the value of p, for which the points (-3, p), (-1, 2) and (1, 1) are collinear?
a) p = -4
b) p = 4
c) p = 3
d) p = -3
Answer: c
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
The vertices of the triangle are (-3, p), (-1, 2) and (1, 1)
The area of triangle = (frac {1}{2}) {-3(2 – 1) + (-1)(1 – p) + 1(p – 2)} = (frac {1}{2}) {-3 – 1 + p + p – 2} = (frac {-6 + 2p}{2}) = – 3 + p
Since, the points are collinear; the area of triangle will be zero.
-3 + p = 0
p = 3

7. The points A (3, 0), B (4, 5) and C (6, 7) are collinear.
a) True
b) False
Answer: b
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are A (3, 0), B (4, 5) and C (6, 7)
The area of triangle = (frac {1}{2}) {3(5 – 7) + 4(7 – 0) + 6(0 – 5)} = (frac {1}{2}) {-8} = – 4 units
Since, the area of triangle is not zero.
Therefore, the points are not collinear.

8. Three points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear only when (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) } = 0.
a) False
b) True
Answer: b
Clarification: Consider three points (-3, 3), (-1, 2) and (1, 1)
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2)}
The area of triangle = (frac {1}{2}) {-3(2 – 1) + (-1)(1 – 3) + 1(3 – 2)} = (frac {1}{2}) {-3 – 1 + 3 + 3 – 2} = (frac {0}{2}) = 0
Hence if the points are collinear the area of triangle is zero.

9. If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC, then its area is given by (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }
a) True
b) False
Answer: a
Clarification:

A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC.
Draw AL, CM, BN perpendicular to x – axis.
Then, ML = (x1 – x2), LN = (x3 – x1) and MN = (x3 – x2).
∴ area of ∆ABC = ar(trap.BMLA) + ar(trap.ALNC) + ar(trap.BMNC)
= {(frac {1}{2}) (AL + BM) × ML} + {(frac {1}{2}) (AL + CN) × LN} – {(frac {1}{2}) (CN + BM) × MN}
= {(frac {1}{2}) (y1 + y2 ) × (x1 – x2)} + {(frac {1}{2}) (y1 + y3 ) × (x3 – x1)} – {(frac {1}{2}) (y2 + y3 ) × (x3 – x2)}
= (frac {1}{2}) {x1 (y1 + y2 – y1 – y3 ) + x2 (y2 + y3 – y1 – y2 ) – x3 (y1 + y3 – y2 – y3 ) }
= (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }

10. If the points A (a, 0), B (0, b) and C (c, 0) are collinear then a = -c.
a) False
b) True
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1(y2 – y3 ) + x2(y3 – y1 ) + x3(y1 – y2 )}
The three points are A (a, 0), B (0, b) and C (c, 0)
Since the points are collinear, the area of triangle will be zero.
The area of triangle = (frac {1}{2}) {a(b – 0) + 0(c – 0) + c(0 – b) } = (frac {1}{2}) {ab + 0 – bc}
(frac {1}{2}) {ab – bc} = 0
{ab – bc} = 0
a = c

[CLASS 10] Mathematics MCQs on Zeros and Coefficients of Polynomial

Mathematics Multiple Choice Questions & Answers on “Zeros and Coefficients of Polynomial – 1”.

1. The zeros of the polynomial 18x2-27x+7 are ___________
a) (frac {7}{6}, frac {1}{3})
b) (frac {-7}{6}, frac {1}{3})
c) (frac {7}{6}, frac {-1}{3})
d) (frac {7}{3}, frac {1}{3})
Answer: a
Clarification: 18x2-27x+7=0
18x2-21x-6x+7=0
3x(6x-7)-1(6x-7)=0
(6x-7)(3x-1)=0
x=(frac {7}{6}, frac {1}{3})
The zeros are (frac {7}{6}) and (frac {1}{3}).

2. What will be the polynomial if its zeros are 3, -3, 9 and -9?
a) x4-80x2+729
b) x4-90x2+729
c) x4-90x2+79
d) x4-100x2+729
Answer: b
Clarification: The zeros of the polynomial are 3, -3, 9 and -9.
Then, (x-3), (x+3), (x-9) and (x+9) are the factors of the polynomial.
Multiplying the factors, we have
(x-3) (x+3) (x-9) (x+9)
(x2-9) (x2-81) (By identity (x-a)(x+a)=x2-a2)
(x4-9x2-81x2+729)
x4-90x2+729

3. The sum and product of zeros of a quadratic polynomial are 10 and (frac {5}{2}) respectively. What will be the quadratic polynomial?
a) 2x2-20x+10
b) 2x2-x+5
c) 2x2-20x+5
d) x2-20x+5
Answer: c
Clarification: The sum of the polynomial is 10, that is, α+β = 10
The product of the polynomial is (frac {5}{2}) i.e. αβ = (frac {5}{2})
∴ f(x)=x2-(α+β)x+αβ
f(x)=x2-10x+(frac {5}{2})
f(x)=2x2-20x+5

4. If α and β are the zeros of x2+20x-80, then the value of α+β is _______
a) -15
b) -5
c) -10
d) -20
Answer: d
Clarification: α and β are the zeros of x2+20x-80.
Sum of zeros or α+β = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {-20}{1}) = -20

5. If α and β are the zeros of 3x2-5x-15, then the value of αβ is _______
a) -5
b) -10
c) -15
d) -20
Answer: a
Clarification: α and β are the zeros of 3x2-5x-15.
Product of zeros or αβ = (frac {constant , term}{coefficient , of , x^2} = frac {-15}{3}) = -5

6. What will be the value of other zero, if one zero of the quadratic polynomial is 5 and the sum of the zeros is 10?
a) 10
b) 5
c) -5
d) -10
Answer: b
Clarification: One zero of the quadratic polynomial is 5. ∴ the factor of the polynomial is (x-5)
Let us assume the other zero to be b. ∴ the other factor of the polynomial is (x-b)
Multiplying the factors, we have (x-5)(x-b)
x2-5x-bx+5b
x2-(5+b)x+5b
The sum of zeros is 10.
∴ (frac {-coefficient , of , x}{coefficient , of , x^2})=10
(frac {-(-5-b)}{1})=10
5+b=10
b=5
The equation becomes x2-10x+25.
Therefore, the other zero is 5.

7. The value of a and b, if the zeros of x2+(a+5)x-(b-4) are -5 and 9 will be _________
a) 47, -5
b) -5, 47
c) -9, 49
d) -4, 45
Answer: c
Clarification: The zeros of the polynomial are -5 and 9.
Hence, α=-5, β=9
The polynomial is x2+(a+5)x-(b-4).
Sum of zeros or α+β=-5+9 = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {a+5}{1})
-4=a+5
a = -9
Product of zeros or αβ = -45 = (frac {constant , term}{coefficient , of , x^2} = frac {-(b-4)}{1})
-45=-b+4
b=49

8. What will be the value of k, if one zero of x2+(k-3)x-16=0 is additive inverse of other?
a) 4
b) -4
c) -3
d) 3
Answer: d
Clarification: Since, one zero of the polynomial is the additive inverse of the other.
Hence, the sum of roots will be zero.
The polynomial is x2+(k-3)x-16=0
Sum of zeros or α+β=(frac {-coefficient , of , x}{coefficient , of , x^2} = frac {k-3}{1})=0
k-3=0
k=3

9. If α and β are the zeros of 10x2+20x-80, then the value of (frac {1}{alpha } + frac {1}{beta }) is _______
a) (frac {5}{4})
b) (frac {1}{5})
c) (frac {3}{4})
d) (frac {1}{4})
Answer: d
Clarification: (frac {1}{alpha } + frac {1}{beta } = frac {alpha +beta }{alpha beta })
α+β=(frac {-20}{10})=-2
αβ=(frac {-80}{10})=-8
∴ (frac {alpha +beta }{alpha beta } = frac {-2}{-8} = frac {1}{4})

10. If α and β are the zeros of x2+35x-75, then _______
a) α+β<αβ
b) α+β>αβ
c) α+β=αβ
d) α+β≠αβ
Answer: b
Clarification: The given polynomial is x2+35x-75.
The sum of zeros, α + β = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {-35}{1}) = -35
The product of zeros, αβ = (frac {constant , term}{coefficient , of , x^2}) = -75
Clearly, sum of zeros is greater than product of zeros.

[CLASS 10] Maths MCQs on Trigonometry Application – Height & Distance

Mathematics Multiple Choice Questions & Answers on “Trigonometry Application – Height and Distance”.

1. The distance between two objects can be known using trigonometric ratios.
a) True
b) False

Answer: a
Clarification: Finding the distance between two objects is one of the different applications of trigonometry. Trigonometry is also used to find the height or length of an object too.

2. One of the applications of trigonometry is used for astronomers to find the distance between two planets.
a) False
b) True

Answer: b
Clarification: The length or height or distance between two objects can be known using trigonometry. Trigonometry is applied in astronomy, navigation, architecture, etc.

3. _____ is drawn from the eye of an observer to the targeted object.
a) A parallel line
b) Line of sight
c) Elevation line
d) Depression line

Answer: b
Clarification: The line of sight is a line that is drawn from the eye of an observer to the targeted object viewed by the observer and the line of sight is required to form the angle of elevation.

4. What happens to the angle of elevation if the height of a tower, the distance between the tower and the observer is doubled?
a) Doubled
b) halved
c) Tripled
d) Remains the same

Answer: d
Clarification: The angle of elevation is independent of the height of an object and the distance between the object and the observer. So, the angle of elevation remains the same when the height of a tower, the distance between the tower and the observer is doubled.

5. The angle of depression is always a/an _____ angle.
a) right
b) obtuse
c) complete
d) acute

Answer: d
Clarification: The angle of elevation is an acute angle. The angles that are less than 90° are called as acute angles. The angle of depression is also an acute angle.

6. A pole stands vertically on the floor. From a point, it is 120 meters away from the foot of the pole and the angle of elevation is 45° then find the height of the pole.
a) 100 meters
b) 180 meters
c) 120 meters
d) 10 meters
Answer: c

7. A slope is built against a wall which makes an angle 30° with the ground and the height of the wall is 2 meters. Find the length of the slope in meters.
a) 2
b) 4
c) 1
d) 3
Answer: b

8. A slide should be built in a park that has to be set up at 10 meters height and also 10 meters away from the support. What should be the angle of elevation?
a) 45°
b) 30°
c) 60°
d) 90°
Answer: a

9. A tree trunk of height 12 meters is erected with the support of three metal chains. An angle of 60° is made by each metal chain with the tree trunk. Find the length of the metal chain.
a) 32 meters
b) 10meters
c) 24 meters
d) 18 meters
Answer: c

10. A ship has to cross a sea by making an angle of 30° and with a distance of 200 meters to reach the shore then find the width of the sea.
a) 300 meters
b) 100 meters
c) 110 meters
d) 50 meters
Answer: b

[CLASS 10] Mathematics MCQs on Geometry – Section Formula

Mathematics Multiple Choice Questions & Answers on “Geometry – Section Formula”.

1. What will be the coordinates of the point which divides the line segment joining the points A(-2, 2) and B(-1, 5) in the ratio 2:5?
a) ((frac {-4}{3}, frac {-20}{9}))
b) ((frac {-4}{3}, frac {20}{9}))
c) ((frac {4}{3}, frac {20}{9}))
d) ((frac {4}{3}, frac {-20}{9}))
Answer: b
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-2, 2) and B(-1, 5) and the ratio is 2:5
∴ x = (frac {2(-1)+5(-2)}{2+7} = frac {-2-10}{9} = frac {-12}{9} = frac {-4}{3})
y = (frac {2(5)+5(2)}{2+7} = frac {10+10}{9} = frac {20}{9} = frac {20}{9})
Hence, the point is ((frac {-4}{3}, frac {20}{9})).

2. What will be the coordinates of the midpoint of the line segment joining the points (-5, 10) and(15, 2)?
a) (-5, -6)
b) (-5, 6)
c) (5, 6)
d) (5, -6)
Answer: c
Clarification: Midpoint lies in the center of the line segment
Hence, it divides the line in the ratio 1:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-5, 10) and B(15, 2) and the ratio is 1:1
∴ x = (frac {1(-5)+1(15)}{2} = frac {-5+15}{2} = frac {10}{2}) = 5
y = (frac {1(2)+1(10)}{2} = frac {2+10}{2} = frac {12}{2}) = 6
Hence, the point is (5,6).

3. In what ratio does the point ((frac {-19}{3}, frac {7}{3})) divide the line segment joining A(3, 7) and B(-11, 0)?
a) 1:2 (externally)
b) 1:2 (internally)
c) 2:1 (externally)
d) 2:1 (internally)
Answer: d
Clarification: Let the ratio in which the point ((frac {-19}{3}, frac {7}{3})) divides the line segment joining the points A(3, 7) and B(-11, 0) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(3, 7) and B(-11, 0) and the ratio is k:1
∴ x = (frac {k(-11)+1(3)}{k+1} = frac {-11k+3}{k+1})
y = (frac {k(0)+1(7)}{k+1} = frac {7}{k+1})
Since, the point is ((frac {-19}{3}, frac {7}{3})).
∴ (frac {-19}{3} = frac {-11k+3}{k+1})
-19(k + 1) = 3(-11k + 3)
-19k – 19 = -33k + 9
-19k + 33k = 19 + 9
14k = 28
k = (frac {28}{14}) = 2
The ratio is 2:1.

4. What will be the value of y, if the ratio in which the point ((frac {3}{4}), y) divides the line segment joining the points A(-1, 4) and B(6, 5)is 1:3?
a) y = (frac {9}{2})
b) y = (frac {5}{2})
c) y = (frac {9}{4})
d) y = (frac {5}{2})
Answer: a
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are (-1, 4)and B(6, 5)in the ratio 1:3
∴ x = (frac {1(6)+3(-1)}{1+3} = frac {6-3}{4} = frac {3}{4})
y = (frac {1(6)+3(4)}{1+3} = frac {6+12}{4} = frac {18}{4})
Therefore y = (frac {9}{2})

5. What will be ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4)?
a) 1:2 (internally)
b) 1:2 (externally)
c) 2:1 (externally)
d) 2:1 (internally)
Answer: b
Clarification: Let the ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4) be k:1.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(0, -1) and B(-3, -4) and the ratio is k:1.
∴ x = (frac {k(-3)+1(0)}{k+1} = frac {-3k}{k+1})
y = (frac {k(-4)+1(-1)}{k+1} = frac {-4k-1}{k+1})
Since, the point ((frac {-3k}{k+1}, frac {-4k-1}{k+1} )) lies on the line 3x+y-11 = 0.
3 ((frac {-3k}{k+1} + frac {-4k-1}{k+1} ))-11 = 0
3(-3k) + (-4k – 1) – 11(k + 1) = 0
-9k – 4k – 1 – 11k – 11 = 0
-24k – 12 = 0
-24k = 12
k = (frac {12}{-24} = frac {-1}{2})
The ratio is 1:2 (externally).

6. In what ratio is the line segment joining the points A(-5, 2) and B(3, 9) divided by the x-axis?
a) 2:5 (internally)
b) 2:5 (externally)
c) 2:9 (externally)
d) 2:9 (internally)
Answer: c
Clarification: Let the ratio in which the x-axis divides the line segment joining the points A(-5, 2) and B(3, 9) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-5, 2) and B(3, 9) and the ratio is k:1
∴ x = (frac {k(3)+1(-5)}{k+1} = frac {3k-5}{k+1})
y = (frac {k(9)+1(2)}{k+1} = frac {9k+2}{k+1})
Since, the point is on x-axis.
Hence, the y-coordinate will be zero.
∴ 0 = (frac {9k+2}{k+1})
0 = 9k+2
k = (frac {-2}{9})
The ratio in which the y-axis cuts the line segment joining the points A(-5, 2) and B(3, 9) will be 2:9 (externally).

7. In what ratio is the line segment joining the points A(2, 4) and B(6, 5) divided by the y-axis?
a) 2:1 (internally)
b) 2:1 (externally)
c) 3:1 (internally)
d) 3:1 (externally)
Answer: d
Clarification: Let the ratio in which the y-axis divides the line segment joining the points A(2, 4) and B(6, 5) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(2, 4) and B(6, 5) and the ratio is k:1
∴ x = (frac {k(6)+1(2)}{k+1} = frac {6k+2}{k+1})
y = (frac {k(5)+1(4)}{k+1} = frac {5k+4}{k+1})
Since, the point is on y-axis.
Hence, the x-coordinate will be zero.
∴ 0 = (frac {6k+2}{k+1})
0 = 6k + 2
k = (frac {-6}{2}) = -3
The ratio in which the y-axis cuts the line segment joining the points A(2, 4) and B(6, 5) will be 3:1 (externally).

8. What will be the coordinates of B, if the point C((frac {29}{7}, frac {46}{7} )), divides the line segment joining A (5, 8) and B (a, b) in the ratio 2:5?
a) a = 2, b = 3
b) a = -2, b = 3
c) a = 2, b = -3
d) a = -2, b = -3
Answer: a
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(5, 8)and B(a, b)in the ratio 2:5
∴ x = (frac {2(a)+5(5)}{2+5} = frac {2a+25}{7})
y = (frac {2(b)+5(8)}{2+5} = frac {2b+40}{7})
But the coordinates of C are ((frac {29}{7}, frac {46}{7} ))
Therefore, (frac {2a+25}{7} = frac {29}{7})
a = 2
(frac {2b+40}{7} = frac {46}{7})
b = 3

9. What will be the length of the median through the vertex A, if the coordinates of the vertices of ∆ABC are A(2, 5), B(5, 0), C(-2, 5)?
a) (sqrt {frac {113}{3}}) units
b) (sqrt {frac {13}{2}}) units
c) (sqrt {frac {113}{2}}) units
d) (sqrt {frac {13}{2}}) units
Answer: b
Clarification:

The median through A will bisect the line BC.
Hence, D is the midpoint of BC
Coordinates of D = ((frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {5-2}{2}, frac {0-5}{2} ) =( frac {3}{2}, frac {-5}{2} ))
Distance between A and D = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2-frac {3}{2})^2+ (5+frac {5}{2})^2} )
= ( sqrt {(frac {1}{2})^2+ (frac {15}{2})^2} )
= ( sqrt {frac {1}{4}+ frac {225}{4}} )
= ( sqrt {frac {113}{2}} ) units

10. What will be the coordinates of the fourth vertex S, if P(-1, -1), Q(2, 0), R(2, 3) are the three vertices of a parallelogram?
a) (-5, -12)
b) (5, -12)
c) (5, 12)
d) (-5, 12)
Answer: c
Clarification:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.
∴ O is the mid-point SQ and PR.
Midpoint of PR
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are P(-1, -1) and R(2, 3) and the ratio is 1:1
∴ x = (frac {1(-1)+1(2)}{2} = frac {-1+2}{2} = frac {1}{2})
y = (frac {1(3)+1(-1)}{2} = frac {3-1}{2} = frac {2}{2}) = 1
Hence, the coordinates of O is (5, 6)
Midpoint of QS
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are Q(2, 0) and S(a, b) and the ratio is 1:1
∴ x = (frac {1(a)+1(2)}{2} = frac {a+2}{2})
y = (frac {1(b)+1(0)}{2} = frac {b}{2})
The coordinates of O is (5, 6)
Therefore, (frac {a+2}{2}) = 5
a = 8
(frac {b}{2}) = 6, b = 12
The coordinates of S are (5, 12).

11. What will be the value of a and b, if (-5, a), (-3, -3), (-b, 0) and (-3, 3) are the vertices of the parallelogram?
a) a = 0, b = -1
b) a = -1, b = 1
c) a = 1, b = 1
d) a = 0, b = 1
Answer: d
Clarification:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.
∴ O is the mid-point SQ and PR.
Midpoint of PR
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are P(-5, a) and R(-b, 0) and the ratio is 1:1
∴ x = (frac {1(-b)+1(-5)}{2} = frac {-b-5}{2})
y = (frac {1(0)+1(a)}{2} = frac {a}{2})
Midpoint of QS
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are Q(-3, -3) and S(-3, 3) and the ratio is 1:1
∴ x = (frac {1(-3)+1(-3)}{2} = frac {-6}{2}) = -3
y = (frac {1(3)+1(-3)}{2} = frac {0}{2}) = 0
Therefore, (frac {-b-5}{2}) = -3
b = 1
(frac {a}{2}) = 0
a = 0

12. What will be the centroid of the ∆ABC whose vertices are A(-2, 4), B(0, 0) and C(4, 2)?
a) ((frac {2}{3}), 2)
b) ((frac {2}{3}), 1)
c) ((frac {2}{5}), 2)
d) ((frac {1}{3}), 2)
Answer: a
Clarification: We know, xcentroid = (frac {x_1+x_2+x_3}{3}) and ycentroid = (frac {y_1+y_2+y_3}{3})
xcentroid = (frac {-2+0+4}{3} = frac {2}{3})
ycentroid = (frac {4+0+2}{3}) = 2
The coordinates of the centroid are ((frac {2}{3}), 2).

13. The coordinates of one end of the diameter AB of a circle are A (-2, -3) and the coordinates of diameter are (-2, 0). What will be the coordinates of B?
a) (2, -3)
b) (-2, 3)
c) (2, 3)
d) (-2, -3)
Answer: b
Clarification: We know that the diameter is twice the radius.
Hence, the center is the midpoint of the diameter.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-2, -3) and center is (-2, 0) and the ratio is 1:1
Let the coordinates of other side of the radius be (x, y).
∴ -2 = (frac {1(-2)+1(x)}{2} = frac {-2+x}{2})
-4 = -2 + x
-4 + 2 = x
x = -2
0 = (frac {1(-3)+1(y)}{2} = frac {-3+y}{2})
0 = -3 + y
y = 3
Hence, the point is (-2, 3).

14. The coordinates of the ends of the diameter AB of a circle are A (-4, 7) and B(4, 7). What will be the coordinates of the center of the circle?
a) (0, -8)
b) (0, 8)
c) (0, 7)
d) (0, -7)
Answer: c
Clarification: We know that the diameter is twice the radius.
Hence, the center is the midpoint of the diameter.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-4, 7) and B(4, 7) and the ratio is 1:1
∴ x = (frac {1(-4)+1(4)}{2} = frac {0}{2}) = 0
y = (frac {1(7)+1(7)}{2} = frac {7+7}{2} = frac {14}{2}) = 7
Hence, the point is (0, 7).

15. The two vertices of ∆ABC are given by A(-3, 0) and B(-8, 5) and its centroid is (-2, 1).What will be the coordinates of the third vertex C?
a) (-5, -2)
b) (5, 2)
c) (-5, 2)
d) (5, -2)
Answer: d
Clarification: The two vertices of triangle are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).
We know, xcentroid = (frac {x_1+x_2+x_3}{3}) and ycentroid = (frac {y_1+y_2+y_3}{3})
Now, xcentroid = (frac {-3-8+x_3}{3})
xcentroid = -2
-2 = (frac {-3 – 8 + x_3}{3})
-6 = -3 – 8 + x3
5 = x3
Now, ycentroid = (frac {0+5+y_3}{3})
ycentroid = 1
1 = (frac {0 + 5 + y_3}{3})
3 = 5 + y3
-2 = y3
The third coordinate is (5, -2).

[CLASS 10] Mathematics MCQs on Geometrical Meaning of Zeros of Polynomial

Mathematics Multiple Choice Questions for Schools on “Geometrical Meaning of Zeros of Polynomial”.

1. The graph of the polynomial 4x2-8x+3 cuts the x-axis at ________ and ________ points.
a) ((frac {3}{4}), 0), ((frac {1}{2}), 0)
b) ((frac {3}{2}), 0), ((frac {1}{2}), 0)
c) ((frac {3}{2}), 0), ((frac {1}{6}), 0)
d) ((frac {7}{2}), 0), ((frac {3}{2}), 0)

Answer: b
Clarification: The graph of the polynomial cuts the x-axis. Only the zeros of the polynomial cut the x-axis.
4x2-8x+3=0
4x2-6x-2x+3=0
2x(2x-3)-1(2x-3)=0
(2x-3)(2x-1)=0
x=(frac {3}{2}, frac {1}{2})
Hence, the graph of the polynomial cuts the x-axis at ((frac {3}{2}), 0) and ((frac {1}{2}), 0).

2. The graph of the polynomial 2x2-8x+5 cuts the y-axis at __________
a) (6, 0)
b) (0, 7)
c) (0, 5)
d) (8, 9)

Answer: c
Clarification: The graph of the polynomial 2x2-8x+5 cuts the y-axis.
Hence, the value of x will be 0.
y(0)=2(0)2-8(0)+5
y=5
The graph cuts the y-axis at (0,5)

3. How many points will the graph of x2+2x+1 will cut the x-axis?
a) 3
b) 1
c) 2
d) 0

Answer: d
Clarification: The graph of x2+2x+1 does not cut the x-axis, because it has imaginary roots.
x2+2x+1=0
x2+x+x+1=0
x(x+1)+(x+1)=0
(x+1)(x+1)=0
x=-1, -1

4. The graph of the quadratic polynomial -x2+x+90 will open upwards.
a) False
b) True

Answer: a
Clarification: The graph of the polynomial will have a downward opening since, a
The graph for the same can be observed here,
” alt=”” width=”512″ height=”434″ data-src=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q4.png” data-srcset=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q4.png 512w, 2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q4-300×254.png 300w” data-sizes=”(max-width: 512px) 100vw, 512px” />

5. If the graph of a polynomial cuts the x-axis at 3 points, then the polynomial is ______
a) Linear
b) Quadratic
c) Cubic
d) Biquadratic

Answer: c
Clarification: Since, the graph of the polynomial cuts the x-axis at 3 points, hence, it will be a cubic polynomial. A polynomial is said to be linear, quadratic, cubic or biquadratic according to the degree of the polynomial.

6. What will be the nature of the zeros of a quadratic polynomial if it cuts the x-axis at two different points?
a) Real
b) Distinct
c) Real, Distinct
d) Complex

Answer: c
Clarification: The zeros of the quadratic polynomial cut the x-axis at two different points.
∴ b2 – 4ac ≥ 0
Hence, the nature of the zeros will be real and distinct.

7. The graph of a quadratic polynomial cuts the x-axis at only one point. Hence, the zeros of the quadratic polynomial are equal and real.
a) True
b) False

Answer: a
Clarification: If the graph meets x-axis at one point only, then the quadratic polynomial has coincident zeros. Also, the discriminant of the quadratic polynomial is zero, therefore roots will be real.

8. A real number is called zeros of the polynomial p(x) if _________
a) p(α)=4
b) p(α)=1
c) p(α)≠0
d) p(α)=0

Answer: d
Clarification: A number is called zero of polynomial when it satisfies the equation of the polynomial.

9. If a < 0, then the graph of ax2+bx+c, has a downward opening.
a) True
b) False

Answer: a
Clarification: The leading coefficient of the polynomial is less than zero, hence, it has downward opening. For example, the graph of -x2 is
” alt=”” width=”521″ height=”385″ data-src=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9.png” data-srcset=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9.png 521w, 2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9-300×222.png 300w” data-sizes=”(max-width: 521px) 100vw, 521px” />

10. A polynomial is said to be linear, quadratic, cubic or biquadratic according to the degree of the polynomial.
a) False
b) True

Answer: b
Clarification: The degree of the polynomial is the highest of the degree of the polynomial. Hence, a polynomial with highest degree one is linear, two as quadratic and so on.

11. Which of the following is a polynomial?
a) x2+2x+5
b) √x+2x+4
c) x(frac {2}{3})+10x
d) 5x+(frac {5}{x})

Answer: a
Clarification: An expression in the form of (x)=a0+a1x+a2x2+…+anxn, where an≠0, is called a polynomial where a1, a2 … an are real numbers and each power of x is a non-negative integer.
In case of √x+2x+4 , the power of √x is not an integer. Similarly for x(frac {2}{3})+10x, (frac {2}{3}) is a fraction.
Now, 5x+(frac {5}{x}) in this case the power of x is a negative integer. Hence it is not a polynomial.

12. The biquadratic polynomial from the following is ______
a) (x2+3)(x2-3)
b) x2-7
c) x7+x6+x5
d) 5x-3

Answer: a
Clarification: A biquadratic polynomial has highest power 4.
Hence, the polynomial with the highest power as 4 is x4-9 or (x2+3)(x2-3).

13. Which of the following is not a polynomial?
a) x2+5x+10
b) √x+2x+4
c) x10+10x
d) 5x+4

Answer: b
Clarification: An expression in the form of (x)=a0+a1x+a2x2+…+anxn, where an≠0, is called a polynomial where a1, a2 … an are real numbers and each power of x is a non-negative integer.
In case of √x+2x+4, the power of x is not an integer.
Therefore it is not a polynomial.

14. If the zeros of a polynomial are 3 and -5, then they cut the x-axis at ____ and _____ points.
a) (8, 0) and (-4, 0)
b) (3, -3) and (-5, 5)
c) (-3, 0) and (5, 0)
d) (3, 0) and (-5, 0)

Answer: d
Clarification: Since, the zeros of the polynomial are 3 and -5.
Therefore, x = 3 and x = -5 and they cut the x-axis so the y-coordinate will be zero.
Hence, the points it cuts the x-axis will be (3, 0) and (-5, 0).

15. If the graph of the quadratic polynomial is completely above or below the x-axis, then the nature of roots of the polynomial is _____
a) Real and Distinct
b) Distinct
c) Real
d) Complex

Answer: d
Clarification: Since, the graph is completely above or below the x-axis, hence, it has no real roots. If a polynomial has real roots only then it cuts the x-axis. If it lies above or below, the roots are complex in nature.