ML Aggarwal Section Formula MCQ Solutions ICSE Class-10 Maths Ch-11. We Provide Step by Step Answer of Exercise-11 **Section Formula** MCQs Questions for ICSE Class-10 APC Understanding Mathematics. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Section Formula MCQ Solutions ICSE Class-10 Maths Ch-11

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-11 | Section Formula |

Writer / Book | Understanding |

Topics | Solutions of MCQ |

Academic Session | 2024-2025 |

### ML Aggarwal Section Formula MCQ Solutions ICSE Class-10 Maths Ch-11

Choose the correct answer from the given four options (1 to 10) :

**Question -1 **** ****The points A (9, 0), B (9, 6), C ( – 9, 6) and D ( – 9, 0) are the vertices of a**

**(a) rectangle**

**(b) square**

**(c) rhombus**

**(d) trapezium**

**Answer -1**

A (9, 0), B (9, 6), C (-9, 6), D (-9, 0)

AB² = (x_{2} – x_{1})² + (y_{2} – y_{1})²

= (9 – 9)^{2 }+ (6 – 0)^{2}

= 0^{2} + 6^{2}

or 0^{2} + 36

= 36

CD^{2} = (-9 + 9)^{2} + (6 – 0)^{2}

= 0^{2} + 6^{2}

or 0 + 36

= 36

BC^{2} = (9 + 9)^{2} + (6 – 6)^{2}

= 18^{2} + 0^{2}

or 324 + 0

= 324

AD^{2} = (9 + 9)^{2} + (0)^{2}

= 18^{2} + 0^{2}

or 324 + 0

= 324

**AB = CD and BC = AD so it is rectangle**

**Question -2 ****If P(a/3, 4) segment joining the points Q ( – 6, 5) and R ( – 2, 3), then the value of a is**

**(a) – 4**

**(b) – 6**

**(c) 12**

**(d) – 12**

**Answer -2**

**P(a/3,4)** is mid-point of the line segment

joining the points Q (-6, 5) and R (-2, 3)

∴ a/3 = (-6-2)/2

= -8/2 = -4

a = -4 × 3

⇒ a = -12

**Question-3 ****If the end points of a diameter of a circle are A ( – 2, 3) and B (4, – 5), then the coordinates of its centre are**

**(a) (2, – 2)**

**(b) (1, – 1)**

**(c) ( – 1, 1)**

**(d) ( – 2, 2)**

**Answer -3**

End points of a diameter of a circle are (-2, 3) and B (4,-5)

then co-ordinates of the centre of the circle

= {(- 2 + 4)/2, (3 – 5)/2} or (2/2, -2/2)

= (1, -1) (b)

**Question -4 ****If one end of a diameter of a circle is (2, 3) and the centre is ( – 2, 5), then the other end is**

**(a) ( – 6, 7)**

**(b) (6, – 7)**

**(c) (0, 8)**

**(d) (0, 4)**

**Answer -4**

One end of a diameter of a circle is (2, 3) and centre is (-2, 5)

Let (x, y) be the other end of the diameter

(2 + x)/2 = – 2

⇒ 2 + x = – 4

⇒ x = -4-2 = – 6

And (3+y)/2 = 5

⇒ 3 + y = 10

⇒ y = 10 – 3 = 7

∴ Co-ordinates of other end are (-6, 7)

**Question -5 ****If the mid-point of the line segment joining the points P (a, b – 2) and Q ( – 2, 4) is R (2, – 3), then the values of a and b are**

**(a) a = 4, b = – 5**

**(b) a = 6, b = 8**

**(c) a = 6, b = – 8**

**(d) a = – 6, b = 8**

**Answer -5**

the mid-point of the line segment joining the

points P (a, b – 2) and Q (-2, 4) is R (2, -3)

2 = (a – 2)/2

⇒ a – 2 = 4

⇒ a = 4 + 2 = 6

– 3 = (b – 2 + 4)/2 = (b + 2)/2

⇒ b + 2 = – 6

⇒ b = – 6 – 2 = – 8

∴ a = 6,

b = – 8

**Question -6 ****The point which lies on the perpendicular bisector of the line segment joining the points A ( – 2, – 5) and B (2, 5) is**

**(a) (0, 0)**

**(b) (0, 2)**

**(c) (2, 0)**

**(d) ( – 2, 0)**

**Answer -6**

the line segment joining the points A (-2, -5) and B (2, -5), has mid-point

= {(-2 + 2)/2, (-5 + 5)/2} = (0, 0)

(0, 0) lies on the perpendicular bisector of AB. (a)

**Question -7 ****The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are**

**(a) (x, y)**

**(b) (y, x)**

**(c) (x/2, y/2)**

**(d) (y/2, x/2)**

**Answer-7**

In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)

The point which is equidistant from O, A and B is the mid-point of AB.

∴ Coordinates are {(0 + 2x)/2, (2y + 0)/2} or (x, y) (a)

**Question -8 ****The fourth vertex D of a parallelogram ABCD whose three vertices are A ( – 2, 3), B (6, 7) and C (8, 3) is**

**(a) (0, 1)**

**(b) (0, – 1)**

**(c) ( – 1, 0)**

**(d) (1, 0)**

**Answer -8**

ABCD is a ||gm whose vertices A (-2, 3), B (6, 7) and C (8, 3).

The fourth vertex D will be the point on which diagonals AC and BD bisect each other at O.

∴ Co-ordinates of O are (-2+8)/2, (3+3)/2

or (6/2, 6/2)

or (3, 3)

Let co-ordinates of D be (x, y), then

3 = (x + 6) /2 (Finding x of D)

6 = x + 6

⇒ x = 6 – 6 = 0

And 3 = (y + 7) /2 (Finding y of D)

⇒ y + 7 = 6

⇒ y = 6 – 7 = – 1

∴ Co-ordinates of D are (0, – 1)

**Question -9 ****The points which divides the line segment joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally lies in the**

**(a) Ist quadrant**

**(b) IInd quadrant**

**(c) IIIrd quadrant**

**(d) IVth quadrant**

**Answer -9**

A point divides line segment joining the points

A (7, -6) and B (3, 4) in the ratio 1 : 2 internally.

Let (x, y) divides it in the ratio 1 : 2

∴ x = (mx_{2} + nx_{1})/(m + n)

= (1 × 3 + 2 × 7)/(1 + 2)

or (3 + 14)/3

= 17/3

y = (my_{2} + ny_{1})/(m + n)

= (1 × 4 + 2 × (-6)/(1 + 2)

or (4 – 12)/3

= -8/3

We see that x is positive and y is negative.

∴ It lies in the fourth quadrant.

**Question -10 ****The centroid of the triangle whose vertices are (3, – 7), ( – 8, 6) and (5, 10) is**

**(a) (0, 9)**

**(b) (0, 3)**

**(c) (1, 3)**

**(d) (3, 3)**

**Answer -10**

Centroid of the triangle whose Vertices are (3, -7), (-8, 6) and (5, 10) is

{(3–8+5)/3, (-7+6+10)/3} or (0, 9/3)

or (0, 3) (b)

— : End of ML Aggarwal Section Formula MCQ Solutions ICSE Class-10 Maths Ch-11 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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