The bases of isosceles trapezoid ABCD are 17 cm and 25 cm and its base angles (angles BAD and CDA) are 45 degrees. Determine the height of the trapezoid.
I need this by tommorrow morning. Thanks!
2006-12-17 12:11:10 · 4 answers · asked by florida2georgiaguy 2 in Science & Mathematics ➔ Mathematics
Since the trapezoid is isosceles, the 17 cm side is centered above the 25 cm side. If you drop two perpendiculars from the 17 cm side to the 25 cm side, you have an extra
(25-17)/2 = 4 cm
on each side. This gives you two right triangles with a 45° angle. So the leg = the height of the triangle (and hence the height of the trapezoid).
The height of the trapezoid is 4 cm.
2006-12-17 12:47:30 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Are the slanted sides equal in length, i.e. is it an isosceles trapezoid? If so, then you can do this: Draw a line perpendicular to the bases, through the upper corner of the trapezoid that is at the end of the side of length 40. This creates a right triangle with hypotenuse 40, and base equal to (28-24) / 2. Now use the Pythagorean theorem on that right triangle to find the height.
2016-05-23 03:10:20 · answer #2 · answered by Anonymous · 0⤊ 0⤋
the difference between the bases=25-17=8cm
since it is an isosceles trapezium
distributing the difference the base of the triangle=4cm
height/base=tan45*
h=4tan45*=4cm
2006-12-17 12:16:30 · answer #3 · answered by raj 7 · 1⤊ 0⤋
If the base angles are 45 degs, wouldn't it be a rectangle?
2006-12-17 12:14:50 · answer #4 · answered by Rama 2 · 0⤊ 2⤋