CBSE Class 9 Maths NCERT Solutions: Areas of Parallelograms and Triangles

To determine if figures lie on the same base and between the same parallels, we need to examine each figure individually.

1. Figure (i): This figure shows two triangles and a parallelogram. The triangle and parallelogram share the base DC and lie between the parallel lines DC and AB. Thus, they lie on the same base and between the same parallels. Common base: DC, Parallels: DC and AB.
2. Figure (ii): This figure shows two triangles. They share the base QR and lie between the parallel lines QR and PS. Thus, they lie on the same base and between the same parallels. Common base: QR, Parallels: QR and PS.
3. Figure (iii): This figure shows a triangle and a parallelogram. They share the base AD and lie between the parallel lines AD and BQ. Thus, they lie on the same base and between the same parallels. Common base: AD, Parallels: AD and BQ.
4. Figure (iv): This figure shows two triangles. They share the base AB and lie between the parallel lines AB and PQ. Thus, they lie on the same base and between the same parallels. Common base: AB, Parallels: AB and PQ.
5. Figure (v): This figure shows two triangles. They share the base AB and lie between the parallel lines AB and PQ. Thus, they lie on the same base and between the same parallels. Common base: AB, Parallels: AB and PQ.
6. Figure (vi): This figure shows two triangles. They share the base CD and lie between the parallel lines CD and RS. Thus, they lie on the same base and between the same parallels. Common base: CD, Parallels: CD and RS.

We are given a parallelogram ABCD with the following information:

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

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

Using base DC and height AE, the area of parallelogram ABCD is:

$$Area\; =\; DC\; \backslash times\; AE$$

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

Substituting the values:

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

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

$$Area\; =\; AD\; \backslash times\; CF$$

We know the area is 128 cm² and CF = 10 cm. We need to find AD.

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

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 the side AD is 12.8 cm.

Answer: AD = 12.8 cm

Given: A parallelogram ABCD, and E, F, G, H are the mid-points of sides AB, BC, CD, and DA respectively.

To Prove: Area (EFGH) = \frac{1}{2} Area (ABCD)

Proof:

Let the vertices of the parallelogram be A, B, C, and D. Since E, F, G, and H are mid-points of the sides AB, BC, CD, and DA respectively, we can join these mid-points to form the quadrilateral EFGH.

Consider the diagonal AC of the parallelogram ABCD. This diagonal divides the parallelogram into two congruent triangles, ΔABC and ΔADC.

In ΔABC, E is the mid-point of AB and F is the mid-point of BC. By the Mid-point Theorem, EF is parallel to AC and EF = \frac{1}{2} AC.

Similarly, in ΔADC, H is the mid-point of AD and G is the mid-point of CD. By the Mid-point Theorem, HG is parallel to AC and HG = \frac{1}{2} AC.

Since EF is parallel to AC and HG is parallel to AC, EF is parallel to HG.

Also, since EF = \frac{1}{2} AC and HG = \frac{1}{2} AC, we have EF = HG.

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

Now consider ΔAEH and ΔCFG. Since ABCD is a parallelogram, AD || BC and AB || DC. Also, AD = BC and AB = DC.

Since E and H are mid-points, AE = \frac{1}{2} AB and AH = \frac{1}{2} AD.

Since F and G are mid-points, CF = \frac{1}{2} BC and CG = \frac{1}{2} DC.

Because AB = DC, AE = \frac{1}{2} AB = \frac{1}{2} DC = DG. Similarly, AH = \frac{1}{2} AD = \frac{1}{2} BC = BF.

Consider ΔAEH and ΔCGF. AE = CG (as shown above), AH = CF (as shown above), and ∠A = ∠C (opposite angles of a parallelogram).

By SAS congruence, ΔAEH ≅ ΔCGF. Therefore, Area(ΔAEH) = Area(ΔCGF).

Similarly, consider ΔEBF and ΔHDG. EB = \frac{1}{2} AB = \frac{1}{2} DC = DG. BF = \frac{1}{2} BC = \frac{1}{2} AD = DH. ∠B = ∠D (opposite angles of a parallelogram).

By SAS congruence, ΔEBF ≅ ΔHDG. Therefore, Area(ΔEBF) = Area(ΔHDG).

The area of parallelogram ABCD can be expressed as the sum of the areas of the four smaller triangles and the central quadrilateral EFGH:

$$Area(ABCD)\; =\; Area(\Delta AEH)\; +\; Area(\Delta EBF)\; +\; Area(\Delta CGF)\; +\; Area(\Delta HDG)\; +\; Area(EFGH)$$

Since Area(ΔAEH) = Area(ΔCGF) and Area(ΔEBF) = Area(ΔHDG), we can write:

$$Area(ABCD)\; =\; 2\; \backslash times\; Area(\Delta AEH)\; +\; 2\; \backslash times\; Area(\Delta EBF)\; +\; Area(EFGH)$$

Also, we know that EFGH is a parallelogram. The area of parallelogram ABCD can be seen as the sum of the areas of the four triangles formed by joining the mid-points:

$$Area(ABCD)\; =\; Area(\Delta AEH)\; +\; Area(\Delta EBF)\; +\; Area(\Delta CGF)\; +\; Area(\Delta HDG)\; +\; Area(EFGH)$$

A key property is that the area of the parallelogram formed by joining the mid-points of a larger parallelogram is half the area of the larger parallelogram. This can be proven by considering the diagonals and the properties of triangles formed.

Alternatively, consider the diagonal BD. It divides the parallelogram into two congruent triangles, ΔABD and ΔCDB.

In ΔABD, E is the mid-point of AB and H is the mid-point of AD. By the Mid-point Theorem, EH is parallel to BD and EH = \frac{1}{2} BD.

In ΔCDB, F is the mid-point of BC and G is the mid-point of CD. By the Mid-point Theorem, FG is parallel to BD and FG = \frac{1}{2} BD.

Thus, EH || FG and EH = FG. This confirms EFGH is a parallelogram.

The area of ΔAEH is \frac{1}{4} Area(ΔABD) because E and H are mid-points. Similarly, Area(ΔEBF) = \frac{1}{4} Area(ΔABC), Area(ΔCGF) = \frac{1}{4} Area(ΔADC), and Area(ΔHDG) = \frac{1}{4} Area(ΔADC).

Since Area(ΔABD) = Area(ΔCDB) = \frac{1}{2} Area(ABCD), and Area(ΔABC) = Area(ΔADC) = \frac{1}{2} Area(ABCD).

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

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

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

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

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

The area of the central parallelogram EFGH is the remaining area:

$$Area(EFGH)\; =\; Area(ABCD)\; -\; (Area(\Delta AEH)\; +\; Area(\Delta EBF)\; +\; Area(\Delta CGF)\; +\; Area(\Delta HDG))$$

$$Area(EFGH)\; =\; Area(ABCD)\; -\; \backslash frac\{1\}\{2\}\; Area(ABCD)$$

$$Area(EFGH)\; =\; \backslash frac\{1\}\{2\}\; Area(ABCD)$$

Thus, the area of quadrilateral EFGH is half the area of parallelogram ABCD.