<span>Let the ends of the horizontal stick be AB
Let the ends of the vertical stick be CD
Let the two sticks intersect at O.
The perimeter of the kite will be BD+DA+AC+CB
Given: The sticks intersect at right angle at 60cm from the bottom.
The vertical stick bisects the horizontal stick.
Therefore, in triangle BOC, angle BOC = 90 degrees. OC = 60 cm. OB=40 cm.
BC is the hypotenuse of the triangle.
Therefore, according to Pythagorean theorem,
BC^2 = OB^2 + OC^2
BC^2 = 60^2 +40^2
=3600+1600
=5200
BC = sqrt(5200)
=72.1 cm
In triangle AOC, AO=OB and OC is common for both the triangles AOC and BOC.
Therefore AC=BC=72.1 cm
In triangle BOD,
OB=40 cm, OD=30 cm.
According to Pythagorean theorem,
BD^2=OB^2+OD^2
=40^2 + 30^2
=1600+900
=2500
BD= sqrt(2500)
=50cm
In triangle AOD the measures of AO=OB and OD is common for both the triangles AOD and BOD. Therefore, the measure of AD=BD
Perimeter of the kite =BD+DA+AC+CB
=50+50+72.1+72.1
=244.2
Perimeter of the kite =244 cm</span>
