Grades 9-10 Video Solutions 2022-2022_9-10

Video Transcription

Question 27. The lengths of the diagonals of squares ABCD and EFGB are 7 cm and 10 cm, respectively.  Point P is the intersection of the diagonals of square ABCD. What is the area of triangle FPD?  Take a look at these squares and the triangle. We can draw out the diagonals, like so. This is 10 and this is 7.  We notice something interesting. The diagonals make a triangle that holds the triangle FPD.  So we can draw this triangle. We can easily find the area of this triangle by doing 1 half of base times height. So 1 half times 7 times 10, which gives us 35.  Next, we can conveniently notice that point P is the intersection of the diagonals of square ABCD.  So A is the midpoint between D and B. This means that the shaded region FPD is exactly half the area of our drawn triangle.  So all we have to do is take 35, divide by 2, and this gives us our answer, which is E, 17.5 cm2.

Video Summary

The problem involves calculating the area of triangle FPD formed by the diagonals of squares ABCD and EFGB, with lengths of 7 cm and 10 cm, respectively. Since P is the intersection of the diagonals of square ABCD, A is the midpoint of D and B, splitting the triangle into two equal parts. This makes the area of triangle FPD half of the area of the larger triangle formed by the diagonals. The area of this larger triangle is \( \frac{1}{2} \times 7 \times 10 = 35 \). Thus, the area of triangle FPD is \( \frac{35}{2} = 17.5 \, \text{cm}^2 \).

Keywords

triangle area

diagonals intersection

squares ABCD EFGB

geometry problem

triangle FPD