When the catenary ye=a cosh (x/a) is revolved around the x-axis, it sweeps out a surface of revolution called a catenoid. find the area of the surface generated when y= cosh x on [−ln15,ln15] is rotated around the x-axis.

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.