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

NO videos please, only written explanations please.

Answer 1

This really requires a diagram. Normally I would use the video to take a picture of a diagram, but since you don't want that, I suggest that you follow the directions here to draw your own so you can follow along

To draw to scale, turn a piece of paper sideways.

(1) Near the lower edger of the paper, draw a line segment 17 cm long and label it AD.

(2) Go 8 cm up from that line and draw another line parallel to it. (We'll prove why 8 cmn is correct shortly.) Mark the center of AD and mark a center point on the line above it. Mark 2.5 cm to the left of that center on the upper line and label it B and 2.5 cm to the right of the center point on the upper line and label it C. Now BC is 5 cm long and is centered above AD, parallel to AD.

We did this because we are told ABCD is a *trapezoid* and that means the two bases are parallel, and the trapezoid is *isosceles* so we know it is left-right symmetrical,

(3) Join AB and CD to complete the trapezoid. If your drawing is to scale, they will each be 10 cm long or at least close.

(4) Join the diagonal AC. Measure it and mark its midpoint. Label the midpoint of AC as E

(5) Draw a line segment through BE and continue it down until it meets the line AD. When BE meets AD, call that point M. Connect BD.

(6) Finally we are going to need the height between the two parallel bases. Draw (if possible with a different lighter colour, two perpendiculars down from B and C to meet AD. Call the points where the perpendiculars meet AD F and G.

Mark on the diagram the lengths we know, AD = 17, BC = 5, AB = CD = 10.

BCGF which we just drew iis a rectangle. BC is parallel to FG by original definition of the trapezoid, BF and CG are parallel because the are both at right angles to FG, and angle BFG is a right angle.

So length or FG = length of BC = 5

The remaining parts of AD, AF + GD, must add up to 17- 5 = 12.

The whole diagram must be left-right symmetrical because ABCD is isosceles. We could prove that if we needed to: Triangle ABF congruent to triangle DCG by hypotenuse leg.

So measures AF = GD = 12/2 = 6.

Now for the height BF = CG, look at right triangle DCG. Hypotenuse DC = 10, Base DG = 6. You could work out Pythagoras, x^2 + 6^2 = 10^2, OR you could realize we have the double of a 3-4-5 triangle. So CG = sqrt(100-36) = sqrt(64) OR CG = 2*4 and either way CG = 8 (proved that assumption in drawing is correct.)

Now we have the height of BDM, BF = CG = 8

We still need the base DM

Look at triangles AEM and CEB,

Angle AEM = angle CEB because they are vertically opposite.

Angle BCE = Angle MAE because MA and BC are parallel (MA is part of AD and BC is parallel to AD as given in out trapezoid) and these are alternate interior angles. Angle CBE = angle MDE similarly, or by subtracting from 180 degrees.

Measure CE = AE by definition because we are given that BM *bisects* AC

Therefore triangle AEM is *congruent* to triangle CEB by SAS

This gives AM = BC = 5

Therefore DM = 17 - 5 = 12

Area triangle BDM = 1/2 base * height = 1/2 * 12 * 8 = 6* 8 = 48 cm^2

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