ABCD is a trapezium whose BC parallel AD and AD = 4 cm. Diagonals AC and BD are intersected at O in such way that #[AO]/[OC] = [DO]/[OB] = 1/2#, then what is the length of BC?

Question

Mason Adams · Accepted Answer

#BC=8#

Given #BC# parallel to #AD, and AD=4#
Let #BC=m#
Let #angleAOD=x, => angleCOB=x#
Let #angleOAD=y, => angleOCB=y#
Let #CO=2a, => OA=1a#

#=> DeltaOAD, and DeltaOCB# are similar,
#=> (OA)/(AD)=(OC)/(CB)#
#=> (1a)/4=(2a)/m#
#=> m=8#

Hence, #BC=m=8#

Caroline Aucoin · Answer

undefined Let $ x $ be the length of BC. Since BC is parallel to AD and ABCD is a trapezium, we know that AD and BC are the non-parallel sides. Let E be the point of intersection of AC and BD.

Given that [AO]/[OC] = 1/2, we can say that AO is 1/3 of AC and OC is 2/3 of AC. Similarly, since [DO]/[OB] = 1/2, we can say that DO is 1/3 of BD and OB is 2/3 of BD.

Since AC and BD are the diagonals of the trapezium, they bisect each other. Therefore, AE = EC and DO = OB.

Let's denote the length of AD as 4 cm. Using the above information, we can express the lengths of AE, EO, DO, and OB in terms of x:

1. AE = EC = 4 cm (since AD = 4 cm)
2. EO = OC = 2/3 * AC = 2/3 * (AD + DC) = 2/3 * (4 + x) = 8/3 + 2x/3 cm
3. DO = OB = 1/3 * BD = 1/3 * (BC + CD) = 1/3 * (x + 4) = x/3 + 4/3 cm

Since EO and DO are equal, we have:

$$ 8/3 + 2x/3 = x/3 + 4/3 $$

$$ 6 + 2x = x + 4 $$

$$ 2x - x = 4 - 6 $$

$$ x = 2 $$

Therefore, the length of BC is 2 cm.

Claire Akin · Answer

undefined Given that $ \frac{{AO}}{{OC}} = \frac{1}{2} $ and $ \frac{{DO}}{{OB}} = \frac{1}{2} $, and $ AD = 4 $ cm, we can use the properties of similar triangles to find the lengths of the sides of trapezium $ ABCD $.

Since $ \frac{{AO}}{{OC}} = \frac{1}{2} $, we can infer that triangles $ ABO $ and $ COO $ are similar. Similarly, since $ \frac{{DO}}{{OB}} = \frac{1}{2} $, triangles $ DOB $ and $ BOA $ are similar.

Let $ OC = x $. Then, $ AO = 2x $ and $ BO = 4x $ (since $ OB = 2AO $). Since $ AD = 4 $ cm, and $ AD = OC + AO $, we have $ 4 = x + 2x = 3x $, which implies $ x = \frac{4}{3} $ cm.

Now, we can find $ BO = 4x = \frac{16}{3} $ cm. And since $ BO = BC + CO $, we have $ BC = BO - OC = \frac{16}{3} - \frac{4}{3} = \frac{12}{3} = 4 $ cm.

Therefore, the length of $ BC $ is $ 4 $ cm.