Question

ABCD is a rectangle formed by the points A( -1, -1 ), B( -1, 4), C(5,4), D(5, -1).P,Q,R and S are the mid points of AB,BC,CD and DA respectively.Prove that PQRS is a rhombus.

Answer 1

Answer:

Proved

Step-by-step explanation:

Given

$A = (-1,-1)$

$B = (-1,4)$

$C = (5,4)$

$D = (5,-1)$

Required

Prove that PQRS is a rhombus

P = midpoint of AB

So:

$P = \frac{1}{2}(-1-1,-1+4)$

$P = \frac{1}{2}(-2,3)$

$P = (-1,\frac{3}{2})$

Q = midpoint of BC

$Q = \frac{1}{2}(-1+5,4+4)$

$Q = \frac{1}{2}(4,8)$

$Q = (2,4)$

R = midpoint of CD

$R =\frac{1}{2}(5+5,4-1)$

$R =\frac{1}{2}(10,3)$

$R = (5,\frac{3}{2})$

S = midpoint of DA

$S = \frac{1}{2}(5-1,-1-1)$

$S = \frac{1}{2}(4,-2)$

$S = (2,-1)$

So, we have:

$P = (-1,\frac{3}{2})$

$Q = (2,4)$

$R = (5,\frac{3}{2})$

$S = (2,-1)$

To show that PQRS is a rhombus, the distance between PQ, QR, RS and SW must be equal.

Distance is calculated as:

$D = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$

For PQ

$D_1 = \sqrt{(-1-2)^2 + (\frac{3}{2}-4)^2}$

$D_1 = \sqrt{(-3)^2 + (-\frac{5}{2})^2}$

$D_1 = \sqrt{9 + \frac{25}{4}}$

$D_1 = \sqrt{\frac{61}{4}}$

For QR

$D_2 = \sqrt{(2-5)^2 + (4-\frac{3}{2})^2}$

$D_2 = \sqrt{(-3)^2 + (\frac{5}{2})^2}$

$D_2 = \sqrt{9 + \frac{25}{4}}$

$D_2 = \sqrt{\frac{61}{4}}$

For RS

$D_3 = \sqrt{(5-2)^2 + (\frac{3}{2}-(-1))^2}$

$D_3 = \sqrt{(5-2)^2 + (\frac{3}{2}+1)^2}$

$D_3 = \sqrt{(3)^2 + (\frac{5}{2})^2}$

$D_3 = \sqrt{9 + \frac{25}{4}}$

$D_3 = \sqrt{\frac{61}{4}}$

For SP

$D_4 = \sqrt{(2-(-1))^2+(\frac{3}{2}-(-1))^2}$

$D_4 = \sqrt{(2+1))^2+(\frac{3}{2}+1)^2}$

$D_4 = \sqrt{(3)^2 + (\frac{5}{2})^2}$

$D_4 = \sqrt{9 + \frac{25}{4}}$

$D_4 = \sqrt{\frac{61}{4}}$

From the calculations above:

$D_1 =D_2 =D_3 =D_4 = \sqrt{\frac{61}{4}}$

Hence: PQRS is a rhombus