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The top surface of a pouch is in the shape of arectangle LMNO with sides 9 cm and 6 cm.(1) A zip is to be sewn along OH such that H is apoint on LM and HM = 2 cm. Find the length ofthe zip(ii) A second zip is to be sewn along NK such thatNK is the perpendicular from N to OH. Calculatethe length of the second zip. |
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Answer» Answer: Part 1. OH. = 9.22 cm Part 2. NK = 5.86 cm Step-by-step explanation: we KNOW that A rectangle has OPPOSITE sides parallel and congruent and the measure of the internal angles is equal to 90 degrees each Part 1) Find the length of the zip OH In the right triangle OLH find out the length side of the hypotenuse OH Applying the Pythagoras Theorem OH sq = LH sq + LO sq We Have, LH = LM - HM LH = 9 - 2 = 7 cm LO = MN = 6 cm substitute the values OH sq = 7 sq + 6 sq OH sq = 85 OH = OH = 9.22 cm Part 2) Find the length of the zip NK we know that The measure of ANGLE LOH is equal to the measure of angle ONK, because triangle KON is a right triangle In the right triangle LOH find the cosine of angle LOH cos(∠LOH)=LO/OH -----> adjacent side to angle LOH divided by the hypotenuse substitute the values cos(∠LOH)=6/√85 -----> equation A In the right triangle KON find the cosine of angle ONK cos(∠ONK)=NK/ON -----> adjacent side to angle ONK divided by the hypotenuse substitute the values cos(∠ONK)=NK/9 -----> equation B Remember that cos(∠LOH)=cos(∠ONK) equate equation A and equation B |
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