प्रश्नावली 9.2 से प्रश्न

Steps to Solve

Here's a breakdown of how to approach each question in प्रश्नावली 9.2:

1. Finding AD of the parallelogram ABCD

   • The area of a parallelogram can be calculated in two ways using different bases and corresponding heights.
   • Area = base $\times$ height. In parallelogram ABCD, we can use DC as the base with AE as the height, or AD as the base with CF as the height. Since ABCD is a parallelogram, $AB = DC = 16 \text{ cm}$.
   • Therefore, the area can be expressed as $AE \times DC = CF \times AD$. Substituting given values: $8 \text{ cm} \times 16 \text{ cm} = 10 \text{ cm} \times AD$. Therefore, $AD = \frac{8 \times 16}{10} \text{ cm} = \frac{128}{10} \text{ cm} = 12.8 \text{ cm}$.

2. Showing ar(EFGH) = $\frac{1}{2}$ar(ABCD)

   This questions asks us to proof that the parallelogram formed by joining the midpoints of the sides of another parallelogram is half the area of the original one.

   • E, F, G, and H are midpoints of sides AB, BC, CD, and DA respectively. Joining these points forms a parallelogram EFGH inside ABCD.
   • To prove this, you typically divide the parallelogram into smaller triangles and parallelograms. Then show that the area of EFGH is exactly half of the area of ABCD. This usually involves showing that the sum of the areas of the corner triangles is equal to the area of the inner parallelogram EFGH.

3. Proving a Point Relationship in Parallelogram ABCD

   • Given: P and Q are points on DC and AD respectively such that $ar(APB) = ar(BQC)$.
   • Objective: Unknown, the problem statement stops mid sentence, therefore cannot continue with this question

4. Point P inside Parallelogram ABCD

   (i) Prove $ar(APB) + ar(PCD) = \frac{1}{2} ar(ABCD)$

   • Draw a line through P parallel to AB (and DC). This divides the parallelogram into two smaller parallelograms.
   • Show that $ar(APB)$ is half the area of the bottom parallelogram, and $ar(PCD)$ is half the area of the top parallelogram.
   • Adding these two areas gives half the area of the entire parallelogram ABCD. (ii) Prove $ar(APD) + ar(PBC) = ar(APB) + ar(PCD)$
   • Since the sum of all four smaller areas within the parallelogram must equal the total area, and we know $ar(APB) + ar(PCD) = \frac{1}{2} ar(ABCD)$, then $ar(APD) + ar(PBC)$ must also equal $\frac{1}{2} ar(ABCD)$. Therefore, $ar(APD) + ar(PBC) = ar(APB) + ar(PCD)$

5. Parallelograms PQRS and ABRS

   (i) Prove $ar(PQRS) = ar(ABRS)$

   • Both parallelograms PQRS and ABRS share the same base RS and lie between the same parallel lines (RS and PB).
   • Parallelograms on the same base and between the same parallels are equal in area. (ii) Prove $ar(AXS) = \frac{1}{2} ar(PQRS)$
   • Triangle AXS and parallelogram ABRS are on the same base AS and between the same parallels (AS and BR).
   • The area of the triangle is half the area of the parallelogram. Therefore, $ar(AXS) = \frac{1}{2} ar(ABRS)$. Since $ar(PQRS) = ar(ABRS)$, then $ar(AXS) = \frac{1}{2} ar(PQRS)$.

6. Farmer's Field Problem

   • The farmer's field is in the shape of parallelogram PQRS. Point A is taken on RS and joined to P and Q.
   • The field is divided into three parts: triangle APQ, triangle APS, and triangle AQR.
   • APQ is a triangle and APS, AQR are also triangles
   • To sow wheat and pulses equally, the farmer can sow wheat in triangle APQ and pulses in the combined areas of triangles APS and AQR. However, since $ar(APQ) = \frac{1}{2} ar(PQRS)$ (Triangles on the same base and between the same parallels are half the area of the parallelogram) and $ar(APS) + ar(AQR) = \frac{1}{2} ar(PQRS)$, the farmer could also plant wheat in $APQ$ triangle, and pulses in the combination of the two triangles, $APS$ and $AQR$.