Question

*An isosceles trapezoid ABCD has bases AD = 17cm, BC = 5cm, and leg AB = 10 cm. A line is drawn through vertex B so that it bisects diagonal
AC
and intersect
AD
at point M. Find the area of ΔBDM. What is the area of ABCD?

Answer 1

The diagram is attached below

The midpoint of AC is E. The ΔMAE is congruent to ΔBCE, and the sides MA = BC = 5 cm. Also MD = 17 -5 = 12 cm

From the figure we can see that the height of the trapezoid is 8cm

The height of trapezoid is the difference between the x axis and the BC that is 8cm

The triangle BDM has base 12 cm and height 8 cm.

Area of triangle = $\frac{1}{2} * base * height$

A = $\frac{1}{2} * 12 * 8$ = 48 cm²

2. The area of trapezoid ABCD formula is

A = $\frac{base_1 + base_2}{2}* height$

base 1= 17cm , base2= 5cm , height = 8cm

A = $\frac{17 + 5}{2}* 8$

A = 88 cm²

Answer 2

Asked and answered elsewhere.
brainly.com/question/9635743