Name: Problem Based on Midpoint Theorem
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: The line segment in a triangle joining the midpoint of two sides of the triangle is parallel to its third side and is also half of the length of the third side.

Question

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral. [Hint: Join $B D$ and draw perpendicular from $A$ on $B D$ .]

Accepted Answer

Given: A quadrilateral $A B C D$ in which the mid-point of the sides of it are joined in order of form parallelogram $P Q R S .$

To prove: $a r (∥ g m P Q R S) = \frac{1}{2} a r (□ A B C D)$

Construction: Join $B D$ and draw perpendicular from $A$ to $B D$ which intersect $S R$ and $B D$ at $X$ and $Y$ respectively.

Proof: In $∆ A B D$ , $S$ and $R$ are the mid-points of sides $A B$ and $A D$ respectively.

$∴ SR | | BD$

$\Rightarrow SX | | BY$

$\Rightarrow X$ is the mid-point of $A Y$ (Converse of mid-point theorem)

$\Rightarrow A X = X Y . . . (1)$

( $B D$ is the mid-point of $A B$ and $SX | | BY$ )

$S R = \frac{1}{2} B D \dots (2)$ ( $∵$ Mid-point theorem)

Now, $a r (Δ A B D) = \frac{1}{2} \times B D \times A Y$

And $ar (Δ A S R) = \frac{1}{2} \times S R \times A X$

$ar (Δ A S R) = \frac{1}{2} \times (\frac{1}{2} B D) \times (\frac{1}{2} A Y)$ [Using $(1)$ and $(2)$ ]

$\Rightarrow ar (Δ A S R) = \frac{1}{4} \times (\frac{1}{2} B D \times A Y)$

$\Rightarrow ar (Δ A S R) = \frac{1}{4} \times (Δ A B D) \dots (3)$

Similarly,

$ar (Δ C P Q) = \frac{1}{4} ar (Δ C B D) \dots (4)$

$ar (Δ B P S) = \frac{1}{4} ar (Δ B A C) \dots (5)$

$ar (Δ D R Q) = \frac{1}{4} ar (Δ D A C) . .. (6)$

Adding $(3), (4), (5)$ and $(6)$ , we get

$ar (Δ A S R) + ar (Δ C P Q) + cor (Δ B P S) + \cot (Δ D R Q)$

$= \frac{1}{4} ar (Δ A B D) + \frac{1}{4} ar (Δ C B D) + \frac{1}{4} ar (Δ A B C) + \frac{1}{4} ar (Δ D A C)$

$= \frac{1}{4} [a r (Δ A B D) + a r (Δ C B D) + a r (Δ B A C) + a r (Δ D A C)]$

$= \frac{1}{4} [ar (□ A B C D) + ar (□ ABCD)]$

$= \frac{1}{4} \times 2 a r (□ A B C D)$

$= \frac{1}{2} \times ar (□ A B C D)$

$∴ a r (Δ A S R) + a r (Δ C P Q) + a r (Δ B P S) + a r (Δ D R Q)$

$= \frac{1}{2} ar (□ A B C D)$

$\Rightarrow a r (□ A B C D) - a r (| | g m P Q R S) = \frac{1}{2} a r (□ A B C D)$

$\Rightarrow a r (| | g m P Q R S) = a r (□ A B C D) - \frac{1}{2} a r (□ A B C D)$

$\Rightarrow a r (| | g m P Q R S) = \frac{1}{2} a r (□ A B C D)$

Hence, proved.

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral. [Hint: Join $B D$ and draw perpendicular from $A$ on $B D$ .]

Important Questions on Areas of Parallelograms and Triangles

Learn and Prepare for any exam you want !

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral. [Hint: Join BD and draw perpendicular from A on BD.]

Important Questions on Areas of Parallelograms and Triangles

Learn and Prepare for any exam you want !

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral. [Hint: Join $B D$ and draw perpendicular from $A$ on $B D$ .]