An isosceles trapezoid ABCD has bases AD = 17 cm, BC = 5cm, and leg AB = 10cm. A line is drawn through vertex B so that it bisects diagonal AC and intersects AD at point M. Find the area of ΔBDM.
48cm2
48cm^2
To find the area of triangle BDM, we need to know the length of its base and height. Let's start by finding the length of base BM.
Since diagonal AC is bisected by the line through vertex B, point M divides the diagonal into two equal parts. Therefore, we have:
AM = MC
Now, let's find the length of diagonal AC. Since we know the lengths of the bases AD and BC, we can use the formula for the length of the diagonals in an isosceles trapezoid:
AC = √[(AD - BC)^2 + AB^2]
Substituting the given values, we have:
AC = √[(17 - 5)^2 + 10^2]
= √[12^2 + 10^2]
= √[144 + 100]
= √244
= 2√61 cm
Since, AM = MC, each part of diagonal AC is half its total length:
AM = MC = AC/2 = (2√61)/2 = √61 cm
Now, let's find the length of base BM. We can use the Pythagorean theorem on triangle ABM:
BM^2 = AB^2 - AM^2
BM^2 = 10^2 - (√61)^2
BM^2 = 100 - 61
BM^2 = 39
BM = √39 cm
Now that we have the length of base BM and the height of triangle BDM, we can calculate its area using the formula for the area of a triangle:
Area of ΔBDM = (1/2) * base * height
Area of ΔBDM = (1/2) * √39 * √61
Area of ΔBDM = (1/2) * √(39 * 61)
Area of ΔBDM = (1/2) * √(2379)
Area of ΔBDM ≈ (1/2) * 48.77
Area of ΔBDM ≈ 24.385 cm^2
Therefore, the area of triangle BDM is approximately 24.385 cm^2.