Answer:
Question 16 = 22
Question 17 = 20 cm²
Step-by-step explanation:
Concepts:
Area of Square = s²
Area of Triangle = bh/2
Diagonals of the square are congruent and bisect each other, which forms a right angle with 90°
Segment addition postulate states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.
Solve:
Question # 16
Step One: Find the total area of two squares
Large square: 5 × 5 = 25
Small square: 2 × 2 = 4
25 + 4 = 29
Step Two: Find the area of the blank triangle
b = 5 + 2 = 7
h = 2
A = bh / 2
A = (7) (2) / 2
A = 14 / 2
A = 7
Step Three: Subtract the area of the blank triangle from the total area
Total area = 29
Area of Square = 7
29 - 7 = 22
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Question # 17
Step One: Find the length of PT
Given:
PT = PR + RT [Segment addition postulate]
PT = (4) + (6)
PT = 10 cm
Step Two: Find the length of S to PT perpendicularly
According to the diagonal are perpendicular to each other and congruent. Therefore, the length of S to PT perpendicularly is half of the diagonal
Length of Diagonal = 4 cm
4 ÷ 2 = 2 cm
Step Three: Find the area of ΔPST
b = PT = 10 cm
h = S to PT = 2 cm
A = bh / 2
A = (10)(2) / 2
A = 20 / 2
A = 10 cm²
Step Four: Find the length of Q to PT perpendicularly
Similar to step two, Q is the endpoint of one diagonal, and by definition, diagonals are perpendicular and congruent with each other. Therefore, the length of Q to PT perpendicularly is half of the diagonal.
Length of Diagonal = 4 cm
4 ÷ 2 = 2 cm
Step Five: Find the area of ΔPQT
b = PT = 10 cm
h = Q to PT = 2 cm
A = bh / 2
A = (10)(2) / 2
A = 20 / 2
A = 10 cm²
Step Six: Combine area of two triangles to find the total area
Area of ΔPST = 10 cm²
Area of ΔPQT = 10 cm²
10 + 10 = 20 cm²
Hope this helps!! :)
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