pls explain this to me
(no spamming, no silly answers)
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hi dude,
just go through the theorems.
First the similarity between the TRIANGLE ABC and XBY is proved using AA similarity criterion
then by using the theorem.... they have come to a statement... with respect to the areas.
then they have just substituted the values that they got.
just take ur note and solve it by looking at the solution... then automatically u will be able to do it.
Anyway,
All the best for tmrw's exam!
just go through the theorems.
First the similarity between the TRIANGLE ABC and XBY is proved using AA similarity criterion
then by using the theorem.... they have come to a statement... with respect to the areas.
then they have just substituted the values that they got.
just take ur note and solve it by looking at the solution... then automatically u will be able to do it.
Anyway,
All the best for tmrw's exam!
RagaviM203:
hi dude
hope the answer was helpful
Answered by
0
I got it as they gave for we write that XYparallel to AC
Next angle BXY =angle A and angle BYX =angleC
Because they are corresponding angles
So by A.A similarity ∆ABC similar to ∆XBY
Now by ratio theorem ar∆ABC/ar∆XBY =(AB/XB) square
Here comes the confusion in the question they have given that the triangle is divide into 2 equal parts
So now arAXCY =ar∆BXY_____1
So now arABC =arAXCY +artriangle BXY
Now by 1 ar∆ABC=2 at XBY
So ar∆ABC/arXBY=2/1_____2
Now from one and two( AB/XB) square= 2/1as root goes to the other side √1 is always 1 and √ 2
So =√2/1
Now the reciprocal comes that is XB/AB=1/√2
Now to get AX we are subtracting one on both sides
Now 1- XB/AB =1-1/√2
Now by LCM
U will get the answer that's all bro
Hope u mark this as brainliest
Bcoz I texted this with so much of my hard work and precious time
Next angle BXY =angle A and angle BYX =angleC
Because they are corresponding angles
So by A.A similarity ∆ABC similar to ∆XBY
Now by ratio theorem ar∆ABC/ar∆XBY =(AB/XB) square
Here comes the confusion in the question they have given that the triangle is divide into 2 equal parts
So now arAXCY =ar∆BXY_____1
So now arABC =arAXCY +artriangle BXY
Now by 1 ar∆ABC=2 at XBY
So ar∆ABC/arXBY=2/1_____2
Now from one and two( AB/XB) square= 2/1as root goes to the other side √1 is always 1 and √ 2
So =√2/1
Now the reciprocal comes that is XB/AB=1/√2
Now to get AX we are subtracting one on both sides
Now 1- XB/AB =1-1/√2
Now by LCM
U will get the answer that's all bro
Hope u mark this as brainliest
Bcoz I texted this with so much of my hard work and precious time
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