An arch for a bridge over a highway is in the form of a semiellipse. the top of the arch is 35 feet above ground (the major axis). what should the span of the bridge be (the length of its minor axis) if the height 25 feet from the center is to be 10 feet above ground?

For given problem:

Put midpoint of ellipse, (0,0) at epicenter of bridge at ground level.

Specified length of vertical major axis = 70=2a

a=35

a^2=1225

Equation of ellipse:

x^2/b^2+y^2=1

plug in coordinates of given point on ellipse(25, 10)

25^2/b^2 + 10^2/a^2 = 1

625/b^2 + 100/1225=1

625/b^2 = 1 - 100/ 1225 = .918

b^2 = 625/.918 ≈ 681

b ≈ 26.09

length of minor axis = 2b = 2(26.09) ≈ 52.16 ft

Span of bridge ≈ 52.16 ft

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.