Tag: mid point theorem

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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

State true or false:

In triangle $ ABC $, $ P $ is the mid-point of side $ BC $. A line through $ P $ and Parallel to $ CA $ meets $ AB $ at  point  $ Q $; and a line through $ Q $ and parallel to $ BC $ meets median $ AP $ at point $ R $. Can it be concluded that, $ BC= 4QP $

  1. True

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

Given: $\triangle ABC$, P is mid point of BC, $QR \parallel BC$ and $PQ \parallel AC$

Since, $ PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In $\triangle ABP$
Since, $QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, $QR = \frac{1}{2} BP$ (Mid point theorem)
$QR = \frac{1}{2} (\frac{1}{2} BC)$ (P is midpoint of BC)
$BC = 4 QR$

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

In triangle $ ABC $; $ D $ and $ E $ are mid-points of the sides $ AB $ and $ AC $ respectively. Through $ E $, a straight line is drawn parallel to $ AB $ to meet $ BC $ at $ F $. Quadrilateral $ BDEF $ is a parallelogram.If $ AB= 16 $ cm, $ AC= 12 $ cm and $ BC= 18 $ cm, find the perimeter of the parallelogram $ BDEF $.

  1. 36 cm

  2. 44 cm

  3. 34 cm

  4. 54 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: D and F are mid points of AB and AC respectively.
Hence, by mid point theorem, $DF \parallel BC$

Also, given $BD \parallel EF$
Since, opposite sides are parallel to each other. Hence, $BDEF$ is a parallelogram

Perimeter of BDEF = $2 (BD + BF)$ (Opposite sides of parallelogram are equal)
Perimeter of BDEF = $AB + BC$ (D and F are mid points of AB and BC respectively)
Perimeter of BDEF = $16 + 18$
Perimeter of BDEF = $34$ cm

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

In triangle $ ABC $; $ M $ is mid-point of $ AB $, $ N $ is mid-point of $ AC $ and $ D $ is any point in base $ BC $. Then:

  1. MN bisects AD

  2. MN divides AD in the ratio 1:3

  3. MN divides AD in the ratio 1:2

  4. MN divides AD in the ratio 1:4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

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

$P, Q, R$ and $S$ are the mid-points of sides $AB. BC, CD$ and $DA$ respectively of rhombus $ABCD$. Show that $PQRS$ is a rectangle.
Under what condition will $PQRS$ be a square ?

  1. When $ABCD $ is a square.
  2. When $ABCD$ is a parallelogram
  3. When $ABCD$ is a rectangle
  4. When $ABCD$ is a square or a rectangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid points of $AB, BC, CD, DA$ respectively.
Join: $AC$ and $BD$

In $\triangle ABC$
$P$ is mid point of $AB$ and $Q$ is mid point of $AC$.
Thus, by mid point theorem, $PQ \parallel AC$ and $PQ = \dfrac{1}{2} AC$
Similarly, In $\triangle ACD$,
$S$ is mid point of $AD$ and $R$ is mid point of $CD$.
Thus, by mid point theorem, $SR \parallel AC$ and $SR = \dfrac{1}{2} AC$
Hence, $PQ \parallel SR$ and $PQ = SR$

Similarly, $PS = QR$ and $PS \parallel QR$
Thus, the opposite sides of $PQRS$ are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $AC \perp BD$ thus, $PS \perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of $PQRS$ are equal and parallel and the sides meet each other at right angles. Hence, $PQRS$ is a rectangle.

If $PQRS$ had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if $ABCD$ is a square.

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

Tangents $PA$ and $PB$ drawn to ${ x }^{ 2 }+{ y }^{ 2 }=9$ from any arbitrary point $'P'$ on the line ${ x }+{ y }=25$. Locus of midpoint of chord $AB$ is

  1. $25({ x }^{ 2 }+{ y }^{ 2 })=9(x+y)$
  2. $25({ x }^{ 2 }+{ y }^{ 2 })=3(x+y)$
  3. $5({ x }^{ 2 }+{ y }^{ 2 })=3(x+y)$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the point on the line $x+y=25$ be $P(a,b)$
Thus equation of chord of contact $AB$ from point $P$ to the circle is given by,
$T  =0 \Rightarrow ax+by = 9$  (i)
Let mid point of $AB$ be $R(h,k)$.
Now equation of chord $AB$ with mid point $R$ is given by,
$T = S _1 \Rightarrow hx+ky = h^2+k^2$ (ii)
Both line (i) and (ii) represents the same line $AB$
$\therefore \displaystyle \frac{a}{h}=\frac{b}{k} = \frac{9}{h^2+k^2}$
$\Rightarrow  a=\cfrac{9h}{h^2+k^2}, b = \cfrac{9k}{h^2+k^2}$
Also point $(a,b)$ lie on the line $x+y = 25$
$\Rightarrow a+b = 25\Rightarrow \cfrac{9h}{h^2+k^2}+ \cfrac{9k}{h^2+k^2}=25$
$ \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by $25(x^2+y^2) = 9(x+y)$

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

In $\Delta ABC$, point P,Q and R are the mid points of the sides AB, BC and CA respectively. If area of $\Delta ABC$ is 32 sq units, then area of $\Delta PQR$ is

  1. $8$ sq cm
  2. $16$ sq cm
  3. $64$ sq cm
  4. $24$ sq cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.

$ΔABC\sim ΔPQR$

In $ΔABC$, $P$ and $R$ are mid points of $AB$ and $AC$ respectively.

$∴ PR || BC$ (midpoint theorem)

In $ΔABC$ and $ΔAPR$:

$∠A$ is common and $∠APR = ∠ABC$ (corresponding angles)

Therefore, $ΔABC\sim ΔAPR$ (AA similarity)

In $ΔABC$ and $ΔPQR$, since $P, Q, R$ are the midpoints of $AB, BC$ and $AC$ respectively,

$PR =\dfrac {1}{2}BC$; (midpoint theorem)

$∴ ΔABC\sim ΔPQR$ (SSS similarity)

$\dfrac { Ar(ΔPQR) }{ Ar(ΔABC) } =\dfrac { PR^{ 2 } }{ BC^{ 2 } } =\left( \dfrac { 1 }{ 2 }  \right) ^{ 2 }=\dfrac { 1 }{ 4 }$ 

Now, it is given that area of $ΔABC$ is $32$ sq. units. Therefore, we have:

$\dfrac { Ar(ΔPQR) }{ 32 } =\dfrac { 1 }{ 4 } \\ Ar(ΔPQR)=\dfrac { 32 }{ 4 } \\ Ar(ΔPQR)=8$

Hence, the area of $ΔPQR$ is $8$ sq units.
Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In $\triangle ABC, D$ is a point on AB and E is a point on BC such that DE || AC and $ar (DBE) = \dfrac {1}{2} ar (ABC)$. Find $\dfrac{AD}{AB}$

  1. $\dfrac{1 - \sqrt 2}{2}$
  2. $\dfrac{\sqrt 2 - 1}{\sqrt 2}$
  3. $\dfrac{\sqrt 2 - 1}{2}$
  4. $\dfrac{\sqrt 2 + 1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$

$DE \parallel BC$
$ \cfrac{AD}{BD} = \cfrac {AE}{EC} $ Basic proportionality theorm
$ AD = \cfrac{BD}{2}$
$ \cfrac {AD}{BD} = \cfrac{1}{3}$
$ \cfrac {area(DBE)}{area(ABC)} = \left (\cfrac {AD}{AB} \right)^{2} $

$ \sqrt {\cfrac {area(DBE)}{area(ABC)}} = \left (\cfrac {AD}{AB} \right) $
$ 1 - \cfrac{AD}{AB}$
$ = 1- \cfrac{1}{\sqrt {2}}$
$ = \cfrac {\sqrt{2} - 1}{\sqrt{2}}$

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

In any triangle ABC state whether following statements are true or false:
(1) the bisectors of the angles A, B, and C meet in a point,
(2) the medians, i.e. the lines joining each vertex to the middle point of the opposite side, meet in a point, and
(3) the straight lines through the middle points of the sides perpendicular to the sides meet in a point.

  1. True

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

The angle bisectors meet at the incenter, the medians meet at the centroid, and the perpendicular bisectors of the sides meet at the circumcenter. All three statements are standard geometric properties.