Question 41417: I have been given the following problem and can't seem to figure out the answer:
You are given triangle abc, and and abc in that triangle is a right angle.
You are also given pqrs is a rectangle inside the triangle, so that ps is parallel to ab. This problem is trying to prove that (PQ) squared is equal to AQ x BR
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I have been given the following problem and can't seem to figure out the answer:
You are given triangle abc, and and abc
NO NOT CORRECT ACB IS THE RIGHT ANGLE.
in that triangle is a right angle.
You are also given pqrs is a rectangle inside the triangle, so that ps is parallel to ab. This problem is trying to prove that (PQ) squared is equal to AQ x BR
HOPE YOU GOT THE FIGURE WITH YOU...P IS AC,S IS ON BC AND Q & R ARE ON AB. ANGLE ACB=90,OTHER ANGLES ARE A AND B
PS IS || TO AB...GIVEN
BC IS TRANVERSAL
HENCE ANGLE CSP=ANGLE CBA=B.........CORRESPONDING ANGLES
PCS IS RIGHT ANGLED TRIANGLE AT C .HENCE
ANGLE CPS =90-ANGLE CSP =90-B
ANGLE APQ + ANGLE QPS + ANGLE CPS =180
ANGLE APQ =180-90-(90-B)=180-90-90+B=B
HENCE IN TRIANGLES APQ AND SBR,WE HAVE
ANGLE APQ =B = ANGLE SBR...PROVED ABOVE
ANGLE AQP=90=ANGLE SRB...PQRS IS A RECTANGLE
HENCE THE 2 TRIANGLES ARE SIMILAR
SO
AQ/SR=PQ/BR
BUT PQ=SR .....PQRS IS RECTANGLE
SO
AQ/PQ=PQ/BR
AQ*BR=PQ^2
|
|
|
