Explanation:
6.
the figure can be considered as a combination of a 16×8 rectangle and a 10×8 right-angled triangle on the left side.
for both, perimeter and area, we need to find the length of the missing third side of the right-angled triangle, as this is a part of the long baseline of the overall figure.
as it is a right-angled triangle, we can use Pythagoras :
c² = a² + b²
"c" being the Hypotenuse (the side opposite of the 90° angle, in our case 10 ft). "a" and "b" being the legs (in our case 8 ft and unknown).
10² = 8² + (leg2)²
100 = 64 + (leg2)²
36 = (leg2)²
leg2 = 6 ft
that means the bottom baseline is 16 + 6 = 22 ft long.
a.
Perimeter = 10 + 16 + 8 + 22 = 56 ft
b.
Area is the sum of the area of the rectangle and the area of the triangle.
area rectangle = 16×8 = 128 ft²
area triangle (in a right-angled triangle the legs can be considered baseline and height) = 8×6/2 = 24 ft²
total Area = 128 + 24 = 152 ft²
7.
it was important that the bottom left angle of the quadrilateral was indicated as right angle (90°). otherwise this would not be solvable.
but so we know, it is actually a rectangle.
that means all corner angles are 90°.
therefore, the angle AMT = 90 - 20 = 70°.
the diagonals split these 90° angles into 2 parts that are equal in both corners of the diagonal, they are just up-down mirrored.
a.
so, the angle HTM = AMT = 70°.
b.
MEA is an isoceles triangle.
so, the angles AME and EAM are equal.
the angle AME = AMT = EAM = 70°.
the sum of all angles in a triangle is always 180°.
therefore,
the angle MEA = 180 - 70 - 70 = 40°
c.
both diagonals are equally long, and they intersect each other at their corresponding midpoints.
so, when AE = 15 cm, then AH = 2×15 = 30 cm.
TM = AH = 30 cm.