The Mid-point Theorem Questions in English

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta ABC$,$X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.
Therefore,$XY \parallel BC$ and $XY = \frac{1}{2} BC$.
In quadrilateral $XBCY$,we have one pair of opposite sides $XY$ and $BC$ that are parallel to each other $(XY \parallel BC)$.
$A$ quadrilateral with at least one pair of parallel opposite sides is defined as a trapezium.
Thus,$XBCY$ is a trapezium.

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of its length.
Let the sides of $\Delta ABC$ be $AB$,$BC$,and $AC$.
Since $P$,$Q$,and $R$ are midpoints of sides $AB$,$BC$,and $AC$ respectively,the sides of $\Delta PQR$ are $PQ = \frac{1}{2} AC$,$QR = \frac{1}{2} AB$,and $PR = \frac{1}{2} BC$.
The perimeter of $\Delta PQR = PQ + QR + PR = \frac{1}{2} AC + \frac{1}{2} AB + \frac{1}{2} BC = \frac{1}{2} (AB + BC + AC)$.
Given that the perimeter of $\Delta ABC = AB + BC + AC = 23 \, cm$.
Therefore,the perimeter of $\Delta PQR = \frac{1}{2} \times 23 = 11.5 \, cm$.

(D) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta XYZ$,$A$ and $B$ are the midpoints of $XY$ and $XZ$ respectively.
Therefore,$AB = \frac{1}{2} \times YZ$.
Given $AB = 7.5 \, cm$.
Substituting the value: $7.5 = \frac{1}{2} \times YZ$.
$YZ = 7.5 \times 2 = 15 \, cm$.

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta ABC$,since $P$ and $Q$ are midpoints of $AB$ and $AC$ respectively,we have $PQ = \frac{1}{2} BC$.
Given that $BC + PQ = 21 \text{ cm}$.
Substituting $PQ = \frac{1}{2} BC$ into the equation,we get $BC + \frac{1}{2} BC = 21$.
This simplifies to $\frac{3}{2} BC = 21$.
Multiplying both sides by $\frac{2}{3}$,we get $BC = 21 \times \frac{2}{3} = 14 \text{ cm}$.

(N/A) $1$. Since $M$ and $N$ are midpoints of $PQ$ and $PR$ respectively,by the Midpoint Theorem,$MN \parallel QR$ and $MN = \frac{1}{2} QR$.
$2$. Consider $\Delta MXN$ and $\Delta M N R$. Both triangles lie between the same parallel lines $MN$ and $QR$.
$3$. The base of $\Delta MXN$ is $MN$ and the base of $\Delta MNR$ is $MN$. Since they share the same base and are between the same parallels,$ar(MXN) = ar(MNR)$.
$4$. In $\Delta PQR$,$MN$ is parallel to $QR$. The height of $\Delta MNR$ with respect to base $MN$ is half the height of $\Delta PQR$ with respect to base $QR$ because $M$ and $N$ are midpoints.
$5$. $ar(MNR) = \frac{1}{2} \times MN \times h_{MNR} = \frac{1}{2} \times (\frac{1}{2} QR) \times (\frac{1}{2} h_{PQR}) = \frac{1}{4} \times (\frac{1}{2} \times QR \times h_{PQR}) = \frac{1}{4} ar(PQR)$.
$6$. Since $ar(MXN) = ar(MNR)$,it follows that $ar(MXN) = \frac{1}{4} ar(PQR)$.

(B) Let the base of the parallelogram $ABCD$ be $b = CD$ and its height be $h$.
The area of parallelogram $ABCD$ is given by $\operatorname{ar}(ABCD) = \text{base} \times \text{height} = b \times h$.
In $\triangle PBC$,the base is $PC$ and the height is $h$ (the same as the parallelogram).
Since $P$ is the midpoint of $CD$,$PC = \frac{1}{2} CD = \frac{1}{2} b$.
The area of $\triangle PBC$ is $\operatorname{ar}(PBC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (\frac{1}{2} b) \times h = \frac{1}{4} bh$.
Now,the ratio $\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = (bh) : (\frac{1}{4} bh) = 1 : \frac{1}{4} = 4 : 1$.

(B) $1$. In $\Delta ABD$,$AM$ is a median because $M$ is the midpoint of $BD$. Therefore,$\operatorname{ar}(ADM) = \operatorname{ar}(ABM) = \frac{1}{2} \operatorname{ar}(ABD)$.
$2$. In $\Delta ADM$,$AN$ is a median because $N$ is the midpoint of $MD$. Therefore,$\operatorname{ar}(AND) = \operatorname{ar}(ANM) = \frac{1}{2} \operatorname{ar}(ADM)$.
$3$. Given $\operatorname{ar}(AND) = 20\, cm^2$,we have $\operatorname{ar}(ADM) = 2 \times 20 = 40\, cm^2$.
$4$. Since $\operatorname{ar}(ADM) = \frac{1}{2} \operatorname{ar}(ABD)$,then $\operatorname{ar}(ABD) = 2 \times 40 = 80\, cm^2$.
$5$. In $\Delta ABC$,$AD$ is a median,so $\operatorname{ar}(ABC) = 2 \times \operatorname{ar}(ABD) = 2 \times 80 = 160\, cm^2$.