Question

) ABCD is a
quadrilateral,
up of two
made up
isosceles triangles
ABD and CBD (see
fig.). The diagonals
meet at O. BD = 12
cm; AO = 8 cm; AC =
12.5 cm. If AB = AD
and CB=CD, find the
lengths of the sides
of the quadrilateral
and its area.
It’s answer is =

Answer 1

Answer:

Pardon the blurriness.

In this diagram, I've extended side AB to point E, where it intersects the perpendicular that passes through point C. I've also added the perpendicular BF. I've labelled the lengths of the perpendiculars as h , and the lengths of the segments EB and CF as a .

Note that each of these makes a useful right triangle. In particular, triangle ACE is a right triangle whose hypotenuse is AC, one of the diagonals we were given: AC = 20. We also know that AB = 12, so AE = 12+a . Finally, EC = h . The pythagorean theorem gives us:

(12+a)2+h2=202

144+24a+a2+h2=400

a2+h2+24a=256

The other interesting triangle is BDF. This is also a right triangle, and its hypotenuse is BD, the other diagonal we were given: BD = 16. We also know that CD = 12, so DF = 12−a , and of course BF = h . Again, the pythagorean theorem gives us:

(12−a)2+h2=162

144−24a+a2+h2=256

a2+h2−24a=112

Subtracting these two equations gives us:

48a=144

a=3

You can quickly find h as well:

h2+(12−3)2=162

h2+81=256

h2=175

h=57–√

So the area of the parallelogram is just AB times h, or

12∗57–√=607–√

(You can see that the area is the base times the height, because if you cut off triangle BCF and moved it over so that BC met with AD, you'd produce a rectangle.)

Answer 2

Answer:

please show your figure with Question