Question

The top surface of a pouch is in the shape of a
rectangle LMNO with sides 9 cm and 6 cm.
(1) A zip is to be sewn along OH such that H is a
point on LM and HM = 2 cm. Find the length of
the zip
(ii) A second zip is to be sewn along NK such that
NK is the perpendicular from N to OH. Calculate
the length of the second zip.​

Answer 1

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 = $\sqrt 85cm$

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

$6/\sqrt85 = NK /9\\NK = 9(6/\sqrt85)\\NK = 5.86 cm$