Tag: mid-point and its converse

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Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

Mid-point theorem states that:

  1. The line segment joining the mid-points of two sides of a triangle is not parallel to the third side and equal to half the length of the third side.

  2. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to one-third the length of the third side.

  3. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to the length of the third side.

  4. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mid-point theorem states that

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Hence, option D is correct.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In $\Delta ABC$, AB$ =5$cm, $BC=8$cm and $CA=7$cm. If D and E are respectively, the mid-points of AB and BC, then determine the length of DE.

  1. $3.5$ cm
  2. $2.5$ cm
  3. $2.8$ cm
  4. $2.0$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mid-Point Theorem : 
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

On applying the midpoint theorem, we get
 $DE = \cfrac {AC}{2} = \cfrac 72 = 3.5$cm

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

Suppose $ABCD$ is a rhombus. A straight line passing through $C$ meet $AD$ which is produced at $P$ and meet $AB$ produced at $Q$. Therefore if $DP=\dfrac {1}{2}AB$, then find the ratio between $BQ$ and $AB$?

  1. $2:1$
  2. $1:1$
  3. $1:3$
  4. $3:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ ABCD $ is a rhombus. $ AB =BC = CD = DA $


$ \displaystyle \frac{DP}{AB} = \frac{1}{2} \Rightarrow DP =1 AB =2 $

In rhombus  $ \angle \theta _1 = \angle \theta _2 $

$ \angle Q $ is common for the $ \triangle BCQ \, and \, \triangle APQ $

$ \because \angle APQ = \angle BCQ $

$ \because \triangle BCQ $ is similar to $ \triangle APQ $ by $ AAA $ property. 

$ \displaystyle \frac{AP}{BC} = \frac{AQ}{BQ} = \frac{AD + DP}{BC} = \frac{3}{2} $

$ \displaystyle \frac{AQ}{BQ} = \frac{3}{2} \Rightarrow \frac{AB + BQ}{BQ} = \frac{3}{2} $

$ \displaystyle \frac{AB}{BQ} = \frac{3}{2} - 1 = \frac{1}{2} $

$ \displaystyle \frac{BQ}{AB} = \frac{2}{1} $

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$.

  1. $(\dfrac{3}{4}, \dfrac{3}{4}, 0)$
  2. $(1, \dfrac{3}{4}, 0)$
  3. $(0, 1, 1)$
  4. $(1, 1, 1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

using formula incenter $x= \dfrac{ax _{1}+bx _{2}+cx _{3}}{a+b+c}$ 


    $y=\dfrac{ay _{1}+by _{2}+cy _{3}}{a+b+c}$ 
    $z=\dfrac{az _{1}+bz _{2}+cx _{3}}{a+b+c}$
    $a=5$    $x=\dfrac{5(0)+4(3)+3(0)}{5+4+3}$       ;     $x=1$


    $b=4$      $y=\dfrac{5(0)+4(0)+3(3)}{5+4+3}$       ; $y=\dfrac{3}{4}$


    $c=3$      $z=\dfrac{5(0)+4(0)+3(0)}{5+4+3}$        ; $z=0$

 incentre =$(1,\dfrac{3}{4},0)$ 

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The mid-points of the sides of a triangle are $D(6,1),E(3,5)$ and $F(-1,-2)$ then vertex opposite to D is 

  1. $(-4,2)$
  2. $(-4,5)$
  3. $(2,2)$
  4. $(10,8)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the vertices be A, B, C. Midpoints are D(6,1), E(3,5), F(-1,-2). The vertex opposite to D is A. Using the property that the quadrilateral formed by midpoints is a parallelogram, A = E + F - D = (3-1-6, 5-2-1) = (-4, 2).

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

ABC is an isosceles triangle with AB=AC. D,E, F are mid point of sides BC,AB and AC respectively then line segment $A D \perp E F$ and is bisected by it.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an isosceles triangle with AB=AC, the median AD to the base BC is also the altitude. EF, connecting midpoints of AB and AC, is parallel to BC. Thus, AD is perpendicular to EF and bisected by it.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

Each side of $\triangle ABC$ is 12 units. D is the foot of the perpendicular dropped from A on BC and E is the mid point of AD. The length of BE in the same units is: 

  1. $\sqrt{18}$
  2. $\sqrt{28}$

  3. 6

  4. 7.93
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an equilateral triangle with side 12, the altitude AD = 12 * sqrt(3)/2 = 6 * sqrt(3) approx 10.39. E is the midpoint of AD, so AE = 3 * sqrt(3). In right triangle ABE, BE^2 = AE^2 + AB^2 is not correct; rather, use triangle BDE where BD=6 and DE=3*sqrt(3). BE^2 = 6^2 + (3*sqrt(3))^2 = 36 + 27 = 63. BE = sqrt(63) approx 7.937.