Detailed Solutions for Chapter 9 Practice Problems

Detailed Solutions for Chapter 9 Practice Problems

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1. Prove that if two triangles are on the same base and between the same parallels, they have equal areas.**

Given
– Two triangles \( \triangle ABC \) and \( \triangle DBC \) share the same base \( BC \).
– Both triangles lie between the same parallels \( BC \) and \( AD \).

**To Prove**:
\[
\text{Area of } \triangle ABC = \text{Area of } \triangle DBC
\]

**Proof**:
1. Draw a perpendicular from \( A \) to \( BC \), meeting \( BC \) at \( P \). Let the height be \( h_1 \).
2. Similarly, draw a perpendicular from \( D \) to \( BC \), meeting \( BC \) at \( Q \). Let the height be \( h_2 \).
3. Since both \( A \) and \( D \) lie on the same parallel \( AD \), we know \( h_1 = h_2 \).

The area of \( \triangle ABC \):
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BC \times h_1
\]

The area of \( \triangle DBC \):
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BC \times h_2
\]

Since \( h_1 = h_2 \):
\[
\text{Area of } \triangle ABC = \text{Area of } \triangle DBC
\]

**Conclusion**:
Triangles on the same base and between the same parallels have equal areas.

—

**2. A parallelogram has sides \( 12 \, \text{cm} \) and \( 15 \, \text{cm} \). Its height corresponding to the base \( 12 \, \text{cm} \) is \( 8 \, \text{cm} \). Find its area.**

**Solution**:
– **Base** \( = 12 \, \text{cm} \)
– **Height** \( = 8 \, \text{cm} \)

Area of a parallelogram:
\[
\text{Area} = \text{Base} \times \text{Height}
\]

Substitute the values:
\[
\text{Area} = 12 \times 8 = 96 \, \text{cm}^2
\]

#### **Answer**:
The area of the parallelogram is \( 96 \, \text{cm}^2 \).

—

**3. The vertices of a triangle are \( A(-2, -3), B(3, 5), \text{and } C(5, -1) \). Calculate its area using coordinate geometry.**

**Solution**:
The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|
\]

Substitute \( A(-2, -3), B(3, 5), C(5, -1) \):
\[
\text{Area} = \frac{1}{2} \left| -2(5 – (-1)) + 3(-1 – (-3)) + 5((-3) – 5) \right|
\]
\[
= \frac{1}{2} \left| -2(6) + 3(2) + 5(-8) \right|
\]
\[
= \frac{1}{2} \left| -12 + 6 – 40 \right|
\]
\[
= \frac{1}{2} \left| -46 \right|
\]
\[
= \frac{1}{2} \times 46 = 23 \, \text{sq. units}
\]

#### **Answer**:
The area of the triangle is \( 23 \, \text{sq. units} \).

—

### **4. A quadrilateral \( ABCD \) is divided into two triangles by a diagonal. Prove that the sum of their areas equals the area of the quadrilateral.**

#### **Given**:
Quadrilateral \( ABCD \) with diagonal \( AC \) dividing it into \( \triangle ABC \) and \( \triangle ACD \).

#### **To Prove**:
\[
\text{Area of } \triangle ABC + \text{Area of } \triangle ACD = \text{Area of Quadrilateral } ABCD
\]

#### **Proof**:
1. Draw diagonal \( AC \).
2. The area of \( \triangle ABC \) is given by:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
3. The area of \( \triangle ACD \) is similarly calculated using \( AC \) as the base.

Adding the areas of the two triangles:
\[
\text{Area of } \triangle ABC + \text{Area of } \triangle ACD = \text{Area of Quadrilateral } ABCD
\]

#### **Conclusion**:
The total area of the two triangles equals the area of the quadrilateral.

—

### **5. Prove that the diagonals of a parallelogram divide it into four triangles of equal areas.**

#### **Given**:
A parallelogram \( ABCD \) with diagonals \( AC \) and \( BD \) intersecting at \( O \).

#### **To Prove**:
\[
\text{Area of } \triangle AOB = \text{Area of } \triangle BOC = \text{Area of } \triangle COD = \text{Area of } \triangle DOA
\]

#### **Proof**:
1. In \( \triangle AOB \) and \( \triangle COD \):
– \( AC \) is the diagonal, and it divides the parallelogram into two equal parts.
– Therefore, \( \text{Area of } \triangle AOB = \text{Area of } \triangle COD. \)

2. Similarly, in \( \triangle BOC \) and \( \triangle DOA \):
– \( BD \) divides the parallelogram into two equal parts.
– Therefore, \( \text{Area of } \triangle BOC = \text{Area of } \triangle DOA. \)

3. Since \( AC \) and \( BD \) intersect at \( O \), the four triangles are of equal area.

#### **Conclusion**:
The diagonals of a parallelogram divide it into four triangles of equal areas.

—

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