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*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?

Respuesta :

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

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 = [tex] \frac{1}{2} * base * height [/tex]

A = [tex] \frac{1}{2} * 12 * 8 [/tex] = 48 cm²

2. The area of trapezoid ABCD formula is

A = [tex] \frac{base_1 + base_2}{2}* height [/tex]

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

A = [tex] \frac{17 + 5}{2}* 8 [/tex]

A = 88 cm²

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