Question

In rectangle ABCE, AC is the diagonal. BD⟂AC at O. AB = 13, OA=5. Find the length of OD and OB

plz answer with explanation

Answer 1

ABCD is a parallelogram. The diagonals AC and BD intersect at O. If AB = 6 cm, = AC = 11 cm and BO =3.5 cm what is the length of AD?

The diagonals AC and BD bisect each other, therefore AO = OC = 5.5 cm.

The area of triangle AOB = COD = [s(s-a)(s-b)(s-c)]^0.5

Here a = 6, b = 5.5, c = 3.5, so 2s = 15 or s = 7.5. So

[s(s-a)(s-b)(s-c)]^0.5 = [7.5(7.5–6)(7.5–5.5)(7.5–3.5)]^0.5

= [7.5*1.5*2*4]^0.5

= 90^0.5 sq cm

Area of triangle ADC = 2 x area of triangle AOB = 2*90^0.5 sq cm, or

2*90^0.5 = [(s1(s1-p)(s1-q)(s1-r)]^2, squaring both sides we get

4*90 = [s1(s1-p)(s1-q)(s1-r)]

360 = s1(s1–11)(s1–6)(s1-AD) … (1)

2s1 = AD + 11 + 6 = AD + 17, or

s1 = 8.5 + 0.5AD … (2). Put the value of s1 from (2) in (1) to get

360 = (8.5 + 0.5AD)(8.5 + 0.5AD-11)(8.5 + 0.5AD-6)(8.5 + 0.5AD-AD)

360 = (8.5 + 0.5AD)(0.5AD-2.5)(0.5AD+2.5)(8.5 - 0.5AD)

360 = (8.5^2 - 0.5AD^2)(0.25AD^2–6.25)

= - (8.5^2 - 0.5AD^2)(6.25 - 0.25AD^2)

= - (72.25 -0.5AD^2)(6.25 - 0.25AD^2)

360 = - [451.5625 -18.0625AD^2 - 3.125AD^2 + 0.125AD^3], or

-360 = 451.5625 -18.0625AD^2 - 3.125AD^2 + 0.125AD^3, or

0.125AD^3 - 21.1875 AD^2 + 451.5625 + 360 = 0, or

0.125AD^3 - 21.1875 AD^2 + 811.5625 = 0, or

AD^3 - 168.5 AD^2 + 6492.5 = 0, or

By trial and error, AD = 6.3272

Check: AD^3 - 168.5 AD^2 + 6492.5

6.3272^3 - 168.5*6.3272^2 + 6492.5

253.3- 6745.6 + 6492.5 = 0.2 as against 0

Hence AD = 6.3272 cm.

Answer 2

Assuming you really mean the rectangle is ABCD, not ABDE:

Since DB is perpendicular to AC, it must be square.

If OA = 5, then OB = OC = OD = 5

This isn't possible because the triangle made by ABO would be a right triangle and AO^2 + BO^2 would equal AB^2. It does not.

5^2 + 5^2 =AB^2 = 7.07, not 13