CBSE Class 9 Mathematics Chapter 9: Areas of Parallelograms and Triangles NCERT Solutions

We need to identify which of the given figures share a common base and lie between the same parallel lines.

1. Figure (i): The parallelogram ABCD and triangle PBQ do not share a common base and are not between the same parallels.
2. Figure (ii): The parallelogram ABCD and triangle PDC share the common base DC and lie between the parallel lines DC and AB.
3. Figure (iii): The parallelogram ABCD and triangle ADC share the common base DC and lie between the parallel lines DC and AB.
4. Figure (iv): The parallelogram ABCD and triangle ABQ share the common base AB and lie between the parallel lines AB and DC.
5. Figure (v): The parallelogram ABCD and triangle PQR do not share a common base and are not between the same parallels.
6. Figure (vi): The parallelogram ABCD and triangle SCD share the common base CD and lie between the parallel lines CD and AB.

Therefore, the figures that lie on the same base and between the same parallels are:

Figure (ii): Common base DC, parallels DC and AB.

Figure (iii): Common base DC, parallels DC and AB.

Figure (iv): Common base AB, parallels AB and DC.

Figure (vi): Common base CD, parallels CD and AB.

We are given a parallelogram ABCD with the following measurements:

• Length of side AB = 16 cm
• Length of altitude AE (perpendicular to DC) = 8 cm
• Length of altitude CF (perpendicular to AD) = 10 cm

The area of a parallelogram can be calculated using the formula: Area = base × height.

Using base DC and height AE:

Since ABCD is a parallelogram, the opposite sides are equal in length. Thus, DC = AB = 16 cm.

Area of parallelogram ABCD = DC × AE

$$Area\; =\; 16\; \backslash text\{\; cm\}\; \backslash times\; 8\; \backslash text\{\; cm\}\; =\; 128\; \backslash text\{\; cm\}^2$$

Now, we can also calculate the area of the same parallelogram using base AD and height CF:

Area of parallelogram ABCD = AD × CF

We know the area is 128 cm² and CF = 10 cm. Let the length of AD be 'x' cm.

$$128\; \backslash text\{\; cm\}^2\; =\; AD\; \backslash times\; 10\; \backslash text\{\; cm\}$$

To find AD, we rearrange the equation:

$$AD\; =\; \backslash frac\{128\; \backslash text\{\; cm\}^2\}\{10\; \backslash text\{\; cm\}\}$$

$$AD\; =\; 12.8\; \backslash text\{\; cm\}$$

Therefore, the length of side AD is 12.8 cm.

Let ABCD be a parallelogram. Let E, F, G, and H be the mid-points of sides AB, BC, CD, and DA, respectively.

We need to prove that the area of the quadrilateral EFGH is half the area of the parallelogram ABCD.

Proof:

Since E and H are mid-points of AB and AD respectively, by the midpoint theorem in triangle ABD, EH is parallel to BD and \( EH = \frac{1}{2} BD \).

Similarly, since F and G are mid-points of BC and CD respectively, in triangle BCD, FG is parallel to BD and \( FG = \frac{1}{2} BD \).

Therefore, EH is parallel to FG and \( EH = FG \).

A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Thus, EFGH is a parallelogram.

Now consider triangle EBF and parallelogram ABCD. Since E and F are mid-points of AB and BC, EB = \( \frac{1}{2} \) AB and BF = \( \frac{1}{2} \) BC.

The area of triangle EBF is given by \( \frac{1}{2} \times EB \times \text{height from F to AB} \). The height from F to AB is half the height of the parallelogram from C to AB.

Alternatively, we can use the property that the area of a triangle formed by joining the mid-points of two sides of a larger triangle is \( \frac{1}{4} \) of the area of the larger triangle. However, EFGH is formed within a parallelogram.

A known theorem states that the quadrilateral formed by joining the mid-points of the sides of a parallelogram is itself a parallelogram, and its area is half the area of the original parallelogram.

Proof using diagonals:

Draw the diagonal AC. The area of parallelogram ABCD is divided into two equal halves by AC, i.e., Area(ABC) = Area(ADC) = \( \frac{1}{2} \) Area(ABCD).

In triangle ABC, E and F are mid-points of AB and BC. By the midpoint theorem, EF is parallel to AC and \( EF = \frac{1}{2} AC \). Also, the area of triangle EBF is \( \frac{1}{4} \) Area(ABC).

Similarly, in triangle ADC, H and G are mid-points of AD and DC. HG is parallel to AC and \( HG = \frac{1}{2} AC \). The area of triangle HDG is \( \frac{1}{4} \) Area(ADC).

Since EF is parallel to AC and HG is parallel to AC, EF is parallel to HG. Also, \( EF = HG = \frac{1}{2} AC \). Therefore, EFGH is a parallelogram.

Area(EFGH) = Area(ABCD) - Area(EBF) - Area(FCG) - Area(GDH) - Area(HAE)

Area(EBF) = \( \frac{1}{4} \) Area(ABC) = \( \frac{1}{4} \times \frac{1}{2} \) Area(ABCD) = \( \frac{1}{8} \) Area(ABCD).

Area(FCG) = \( \frac{1}{4} \) Area(BCD) = \( \frac{1}{4} \times \frac{1}{2} \) Area(ABCD) = \( \frac{1}{8} \) Area(ABCD).

Area(GDH) = \( \frac{1}{4} \) Area(ADC) = \( \frac{1}{4} \times \frac{1}{2} \) Area(ABCD) = \( \frac{1}{8} \) Area(ABCD).

Area(HAE) = \( \frac{1}{4} \) Area(ABD) = \( \frac{1}{4} \times \frac{1}{2} \) Area(ABCD) = \( \frac{1}{8} \) Area(ABCD).

Sum of the areas of the four corner triangles = \( 4 \times \frac{1}{8} \) Area(ABCD) = \( \frac{1}{2} \) Area(ABCD).

Area(EFGH) = Area(ABCD) - \( \frac{1}{2} \) Area(ABCD) = \( \frac{1}{2} \) Area(ABCD).

Thus, it is proved that the area of the quadrilateral EFGH formed by joining the mid-points of the sides of a parallelogram ABCD is half the area of the parallelogram ABCD.