The question in the image involves the application of quadratic equations to solve a geometry problem. Here's a breakdown of the problem:

Problem Description

• Given:
  • A quadrilateral $ABCD$ where $AB = 10$, $BC = 13$, $CD = 14$, and $DA = 7$.
  • Diagonals $AC$ and $BD$ intersect at point $P$.
  • The area of triangle $APD = 21$ and the area of triangle $BPC = 63$.

(a) Find the area of triangle $BPD$ if $BD^2 = 178 + 84\sqrt{2}$.

(b) Show that the area of quadrilateral $ABCD = 84 + 42\sqrt{2}$.

Solution Approach

Let's start by solving part (a).

1. Find the Area of Triangle $BPD$:

   Given that the diagonals intersect at point $P$, we know that the sum of the areas of the triangles formed by these diagonals is equal to the area of the quadrilateral.

   $\text{Area of Quadrilateral } ABCD = \text{Area of } APD + \text{Area of } BPD + \text{Area of } BPC.$

2. Solve for Part (a):

   • Use the properties of diagonals and the information given to find the area of triangle $BPD$.
   • Then, apply the relationship to calculate the area of quadrilateral $ABCD$ in part (b).

Would you like me to proceed with a detailed solution for part (a) and (b)?

Related Questions

1. What are the properties of a quadrilateral that allow diagonals to intersect at a point?
2. How do you calculate the area of a quadrilateral given its diagonal lengths and intersecting angles?
3. Can you derive the expression for the area of triangle $BPD$ using only the lengths given?
4. How does the Pythagorean theorem apply to the diagonals intersecting at a point in a quadrilateral?
5. What other geometric properties can be derived if we know the areas of individual triangles within a quadrilateral?

Tip:

Remember, the properties of triangles within a quadrilateral often give clues to finding areas and relationships between side lengths.