2) * (0.82)AREA = 118.3 squared units orAREA = 1/2(base x height )AREA = 1/2 (BC * AD)AREA = 1/2 ( 18.2 * 13)AREA = 1/2 (236.6)AREA = 118.3PERIMETER = AC + BC + AB.......
x=15.87
Now that we have the hypotunuse of one triangle (remember the big triangle was split into two triangles) we know that it's also the hypotunuse of the other triangle.
So to find perimeter, we add up the three sides of the triangle, which are, the two hypotenuses, and the other side BC
Perimeter=15.87 + 15.87 + 18.2
Perimeter=49.94 (rounded to the nearest tenth it's 49.9)
So the question says that the two sides AB and AC are of the same length. Since angle BAC is 70 degrees, then the other two angles add up to 110 degrees (180 - 70) as the sum of the angles of a triangle is 180. But since AB and AC are the same, we know the remaining two angles are the same so angles ABC and ACB are 55 degrees (110/2). From here we use trig (angles are all in degrees).
tan55 = 13/BD. So BD = 13/(tan55). DC is exactly the same. So add up BD and DC to get the length of BC = 26/(tan55) = 18.2 units
Area of the triangle is altitude * length of base /2 = 13 * 18.2 / 2= 118.3 square units
Trig again for the hypotenuse so we can find the perimeter of the triangle. sin55 = 13/AB
So AB = 13/sin(55). AC is the same so AC + AB = 26/sin(55) = 31.7
So the perimeter is AC + AB + BC = 31.7 + 18.2 = 49.9 units
