Maths

7. ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD at F. If ar (DFB) = 3 cm^2, find the area of the parallelogram ABCD. 7. ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. [...]

6. O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO). 6. O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO). July 1, 2021 Category: Chapter 9 - Areas of Parallelogram and [...]

5. ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig. 9.15), prove that ar (AER) = ar (AFR) 5. ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF [...]

4. In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), then prove that ar (BPQ) = 1/2 ar (ABC). 4. In ∆ ABC D is the mid-point of AB and P is any point on BC. If CQ [...]

3. The area of the parallelogram ABCD is 90 cm^2 (see Fig.9.13). Find 3. The area of the parallelogram ABCD is 90 cm^2 (see Fig.9.13). Find July 1, 2021 Category: Chapter 9 - Areas of Parallelogram and Triangles , Maths , NCERT Class 9 ,

2. X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 9.12). Prove that ar (LZY) = ar (MZYX) 2. X and Y are points on the side LN [...]

1. In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD). 1. In Fig.9.11 PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR [...]

5. In Fig. 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar ( EFGD). 5. In Fig. 9.8 ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar ( EFGD). July 1, 2021 Category: Chapter [...]

4. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 1/4 ar ( ABC). 4. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 1/4 ar ( ABC). July 1, 2021 Category: Chapter 9 - [...]

3. PQRS is a parallelogram whose area is 180 cm^2 and A is any point on the diagonal QS. The area of ∆ ASR = 90 cm^2. 3. PQRS is a parallelogram whose area is 180 cm^2 and A is any point on the diagonal QS. The area of ∆ ASR = 90 cm^2. July [...]