Areas of Parallelograms and Triangles
To help you visualize better and prepare comprehensively, here are more practice problems along with explanations to strengthen the concepts.
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### **Problem 1: Prove a Property of a Triangle within a Parallelogram**
#### **Problem Statement**:
Prove that a triangle formed by joining any vertex of a parallelogram to the midpoints of the opposite side has one-fourth the area of the parallelogram.
#### **Solution with Diagram**:
1. **Steps**:
– Draw a parallelogram \( ABCD \).
– Mark the midpoints \( P \) and \( Q \) of sides \( BC \) and \( CD \), respectively.
– Join \( A \) to \( P \) and \( Q \) to form \( \triangle APQ \).
2. **Proof**:
– Since \( P \) and \( Q \) are midpoints, the line segment \( PQ \) is parallel to \( AB \) and half its length.
– The area of \( \triangle APQ \) is proportional to the base \( PQ \) and the height from \( A \) to \( PQ \).
– The height is half the height of the parallelogram because \( PQ \) divides \( ABCD \) into equal halves.
– Hence, \( \text{Area of } \triangle APQ = \frac{1}{4} \times \text{Area of Parallelogram ABCD.} \)
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### **Problem 2: Calculate the Area of a Quadrilateral Using Triangles**
#### **Problem Statement**:
Find the area of a quadrilateral \( ABCD \) with vertices \( A(2, 3), B(6, 7), C(10, 3), D(6, -1) \).
#### **Solution with Diagram**:
1. **Steps**:
– Divide the quadrilateral \( ABCD \) into two triangles: \( \triangle ABC \) and \( \triangle ACD \).
– Use the formula for the area of a triangle in coordinate geometry:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|
\]
2. **Area of \( \triangle ABC \)**:
\[
\text{Vertices: } A(2, 3), B(6, 7), C(10, 3)
\]
Substitute into the formula:
\[
\text{Area} = \frac{1}{2} \left| 2(7-3) + 6(3-3) + 10(3-7) \right|
\]
\[
= \frac{1}{2} \left| 2(4) + 6(0) + 10(-4) \right|
\]
\[
= \frac{1}{2} \left| 8 – 40 \right| = \frac{1}{2} \times 32 = 16 \, \text{sq. units.}
\]
3. **Area of \( \triangle ACD \)**:
\[
\text{Vertices: } A(2, 3), C(10, 3), D(6, -1)
\]
Substitute into the formula:
\[
\text{Area} = \frac{1}{2} \left| 2(3 – (-1)) + 10(-1 – 3) + 6(3 – 3) \right|
\]
\[
= \frac{1}{2} \left| 2(4) + 10(-4) + 6(0) \right|
\]
\[
= \frac{1}{2} \left| 8 – 40 \right| = \frac{1}{2} \times 32 = 16 \, \text{sq. units.}
\]
4. **Total Area of Quadrilateral**:
\[
\text{Area of } ABCD = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD = 16 + 16 = 32 \, \text{sq. units.}
\]
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### **Problem 3: Relationship Between Parallelogram and Triangle Areas**
#### **Problem Statement**:
In a parallelogram \( ABCD \), diagonal \( AC \) divides it into two triangles. Prove that the area of \( \triangle ABC \) is equal to the area of \( \triangle ADC \).
#### **Solution**:
1. **Steps**:
– The diagonal \( AC \) divides the parallelogram \( ABCD \) into \( \triangle ABC \) and \( \triangle ADC \).
– Both triangles share the same diagonal \( AC \) as the base.
– Since \( AC \) is a diagonal, it divides the parallelogram into two equal parts.
– The height of both triangles is the perpendicular distance between \( AC \) and the opposite vertices.
2. **Conclusion**:
The areas of \( \triangle ABC \) and \( \triangle ADC \) are equal because they share the same base and height.
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### **Problem 4: Mixed Concept Problem with Coordinates**
#### **Problem Statement**:
Prove that the midpoints of the sides of a triangle form a parallelogram, and find its area using coordinate geometry if the vertices of the triangle are \( A(0, 0), B(4, 6), C(8, 0) \).
#### **Solution**:
1. **Steps**:
– Find the midpoints of sides \( AB \), \( BC \), and \( AC \):
\( M_1 \): Midpoint of \( AB = \left( \frac{0+4}{2}, \frac{0+6}{2} \right) = (2, 3) \).
\( M_2 \): Midpoint of \( BC = \left( \frac{4+8}{2}, \frac{6+0}{2} \right) = (6, 3) \).
\( M_3 \): Midpoint of \( AC = \left( \frac{0+8}{2}, \frac{0+0}{2} \right) = (4, 0) \).
– Prove that \( M_1M_2M_3M_4 \) is a parallelogram by verifying that opposite sides are parallel and equal.
– Use the area formula for coordinate geometry to find the area of the parallelogram.
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