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In trapezium ABCD a line EF cuts the diagonal AC in O such that AO/OC = 2/3 and EF is parallel to BC.In what ratio does EF cut AB and CD?
- a)1:2
- b)3:4
- c)2:3
- d)1:4
Correct answer is option 'C'. Can you explain this answer?
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In trapezium ABCD a line EF cuts the diagonal AC in O such that AO/OC ...
Solution:
Given that,
AO/OC = 2/3 and EF is parallel to BC
We have to find the ratio in which EF cuts AB and CD.
Let's join E and C and extend it to meet AB at G.
Now, we have two similar triangles AOG and COE.
By the property of similar triangles,
AO/OC = AG/CE = 2/3
If we take CG as x, then AG will be (2/3)x and CE will be (1/3)x.
Now, let's join E and F and extend it to meet DC at H.
By the property of parallel lines,
∠DBC = ∠FEH (Alternate angles)
Also, ∠BDC = ∠HEF (Alternate angles)
Therefore, ΔBDC ~ ΔFEH (By AA similarity)
By the property of similar triangles,
DC/FE = BC/EF
DC/(DC + EF) = BC/EF
DC/BC = EF/(DC + EF)
DC/BC = EF/BD
DC/BC = EF/(BC + CD)
BC/CD = EF/DC
CD/BC = DC/EF
Let's assume EF cuts AB at G1 and CD at H1.
Now, let's take BC/CD as x. Then, CD = x and BC = 1.
Therefore, DC = x + 1.
From the above equation, we get
CD/BC = DC/EF
x/(1) = (x + 1)/EF
EF = (x + 1)/x
Now, let's consider the triangle AEF.
By the property of similar triangles,
AG1/GE = AO/OE (From ΔAOG and ΔEOC)
AG1/(EF - x) = 2/3 (As AO/OC = 2/3)
AG1 = (2/3)(EF - x)
Similarly, let's consider the triangle DEH.
By the property of similar triangles,
DH1/HE = DC/CE (From ΔDCB and ΔEFC)
DH1/(EF - x) = x/3
DH1 = (x/3)(EF - x)
Now, let's find the ratio in which EF cuts AB and CD.
Ratio = AG1/DH1
= [(2/3)(EF - x)]/[(x/3)(EF - x)]
= 2x/3x
= 2:3
Therefore, the ratio in which EF cuts AB and CD is 2:3.
Hence, the correct option is (C).
Given that,
AO/OC = 2/3 and EF is parallel to BC
We have to find the ratio in which EF cuts AB and CD.
Let's join E and C and extend it to meet AB at G.
Now, we have two similar triangles AOG and COE.
By the property of similar triangles,
AO/OC = AG/CE = 2/3
If we take CG as x, then AG will be (2/3)x and CE will be (1/3)x.
Now, let's join E and F and extend it to meet DC at H.
By the property of parallel lines,
∠DBC = ∠FEH (Alternate angles)
Also, ∠BDC = ∠HEF (Alternate angles)
Therefore, ΔBDC ~ ΔFEH (By AA similarity)
By the property of similar triangles,
DC/FE = BC/EF
DC/(DC + EF) = BC/EF
DC/BC = EF/(DC + EF)
DC/BC = EF/BD
DC/BC = EF/(BC + CD)
BC/CD = EF/DC
CD/BC = DC/EF
Let's assume EF cuts AB at G1 and CD at H1.
Now, let's take BC/CD as x. Then, CD = x and BC = 1.
Therefore, DC = x + 1.
From the above equation, we get
CD/BC = DC/EF
x/(1) = (x + 1)/EF
EF = (x + 1)/x
Now, let's consider the triangle AEF.
By the property of similar triangles,
AG1/GE = AO/OE (From ΔAOG and ΔEOC)
AG1/(EF - x) = 2/3 (As AO/OC = 2/3)
AG1 = (2/3)(EF - x)
Similarly, let's consider the triangle DEH.
By the property of similar triangles,
DH1/HE = DC/CE (From ΔDCB and ΔEFC)
DH1/(EF - x) = x/3
DH1 = (x/3)(EF - x)
Now, let's find the ratio in which EF cuts AB and CD.
Ratio = AG1/DH1
= [(2/3)(EF - x)]/[(x/3)(EF - x)]
= 2x/3x
= 2:3
Therefore, the ratio in which EF cuts AB and CD is 2:3.
Hence, the correct option is (C).
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In trapezium ABCD a line EF cuts the diagonal AC in O such that AO/OC ...
By bpt theorem ration of both are same
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In trapezium ABCD a line EF cuts the diagonal AC in O such that AO/OC = 2/3and EF is parallel to BC.In what ratio does EF cut AB and CD?a)1:2b)3:4c)2:3d)1:4Correct answer is option 'C'. Can you explain this answer?
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