An isosceles trapezoid ABCD has bases AD=17 cm, BC=5 cm, and side AB=10 cm. Through the vertex B there is a line, which bisects diagonal AC and intersects AD at point M. What is the area of BDM?

Question

Please write explanation, no videos.

Answer 1

draw a diagram

draw height BY and height CZ

let point X be the midpoint of AC

AY=6 and ZD=6 because YZ=5 so AY+ZD must equal 12 and 12/2=6

BAY is a right triangle

10^2=6^2+BY^2

100=36+BY^2

64=BY^2

BY=8, the altitude of triangle BDM

triangles BXC and MXA are congruent by ASA

angle BXC≅angle MXA because they are vertical angles

angle XAM ≅angle XCB because they are alternate interior angles because BC is ll to AD and AC is a transversal

BC≅AM because corresponding parts of congruent triangles are congruent

therefore AM=5

therefore MD=17-5=12

MD is the base of triangle BDM and BY is the height

A=(1/2)*b*h

A=(1/2)*12*8

A=6*8

A=48 sq cm is the area of triangle BDM