Question

In the given figure, ABCD is a rectangle. Its diagonals meet at O. If OC = 3x – 4, OB = x + 4, then find OB, OC, OA, AC, BD and OD​

Answer 1

Answer:

OB = 8

OC = 8

OA = 8

AC = 16

BD = 16

OD = 8

Step-by-step explanation:

Given, rectangle ABCD, with diagonals AC and BD, and OC = 3x – 4, OB = x + 4,

thus, since diagonals of a rectangle are equal, therefore, AC = BD.

Invariably, 2*OC = 2*OB

Thus, $2(3x - 4) = 2(x + 4)$

Solve for x

$6x - 8 = 2x + 8$

Add 8 to both sides

$6x - 8 + 8 = 2x + 8 + 8$

$6x = 2x + 16$

Subtract 2x from both sides

$6x - 2x = 2x + 16 - 2x$

$4x = 16$

Divide both sides by 4

$\frac{4x}{4} = \frac{16}{4}$

$x = 4$

OB = x + 4 = 4 + 4 = 8

OC = 3x - 4 = 3(4) - 4 = 12 - 4 = 8

OA = OC = 8

AC = 2*OC = 2*8 = 16

BD = 2*OB = 2*8 = 16

OD = OB = 8