Help anyone can help me do 16 and 17 question, I will mark brainlest. The no 16 question is find the area of the shaded region​

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

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!! :)

Please let me know if you have any questions