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Answer 1

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Answer 2

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Answer:Given: ACB = 90° and CD | AB

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = ap

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have 1/2 x BC x AC = 1/2 x AB x CD

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have 1/2 x BC x AC = 1/2 x AB x CD => 1/2 x a x b = 1/2 x c x p

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have 1/2 x BC x AC = 1/2 x AB x CD => 1/2 x a x b = 1/2 x c x p => ab = cp

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have 1/2 x BC x AC = 1/2 x AB x CD => 1/2 x a x b = 1/2 x c x p => ab = cp Hence , proved.

Answer:Given: ACB = 90° and CD | ABTo Prove: cp = apProof: Area of triangle = 1/2 x base x height So, Ar( ∆ ABC) = 1/2 x BC x AC -------------1 and ar(∆ ABC ) = 1/2 x AB x CD -----------2 Now, from equ. 1 and 2 , we have 1/2 x BC x AC = 1/2 x AB x CD => 1/2 x a x b = 1/2 x c x p => ab = cp Hence , proved.