Name: Construction of Rhombus With Special Properties
Uploaded: 2019-07-19
Duration: 7 min 55 s
Description: We usually use five measurements to construct a quadrilateral. Did you know that there are ‘special’ quadrilaterals that can be constructed using a lesser nu...

Question

Prove that the quadrilateral formed by joining the midpoints of an isosceles trapezium is a rhombus.

Accepted Answer

Given, an isosceles trapezium.

Let $ABCD$ be the trapezium, with $AD = BC$ .

$P, Q, R$ and $S$ are the midpoints of $AB, BC, CD$ and $DA$ respectively.

To prove: $PQRS$ is a rhombus.

Draw the diagonals $AC$ and $BD$ .

Consider $∆ ABC$ ,

$PQ$ is the line segment joining the midpoints of $AB$ and $BC$ .

So, by midpoint theorem, we get

$SR = \frac{1}{2} AC, PQ ∥ AC$ $. . . . . . . . (i)$

Consider $∆ ADC$ .

$SR$ is the line segment joining the midpoints of $AD$ and $DC$ .

So, by midpoint theorem, we get

$SR = \frac{1}{2} AC, SR ∥ AC$ $. . . . . . . . (i i)$

From $(i)$ and $(i i)$ , we get

$PQ ∥ SR$ and $PQ = SR = \frac{1}{2} AC$ $. . . . . . . (i i i)$

Consider $∆ ABD$ .

$PS$ is the line segment joining the midpoints of $AB$ and $AD$ .

So, by midpoint theorem, we get

$PS = \frac{1}{2} BD, PS ∥ BD$ $. . . . . . . . . (i v)$

Consider $∆ BCD$ .

$QR$ is the line segment joining the midpoints of the sides $BC$ and $CD$ .

So, by midpoint theorem, we get

$QR = \frac{1}{2} BD, QR ∥ BD$ $. . . . . . . (v)$

From $(i v)$ and $(v)$ , we get

$QR ∥ PS$ and $QR = PS = \frac{1}{2} BD$ $. . . . . . . . (v i)$

Since in an isosceles trapezium, diagonals are equal, from $(i i i)$ and $(v i)$ , we get

$PQ = SR = \frac{1}{2} AC = \frac{1}{2} BD = PS = QR$ .

i.e., $PQ = QR = RS = SP$

Hence, $PQRS$ is a rhombus.

Prove that the quadrilateral formed by joining the midpoints of an isosceles trapezium is a rhombus.

Important Questions on Rectilinear Figures

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