Chapter 9 – Areas of Parallelogram and Triangles
Sahay Sir > Question Answers > NCERT Class 9 > Maths > Chapter 9 - Areas of Parallelogram and Triangles
2. In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels?
01
Jul
2. In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels? 2. In which of the following figures (Fig. 9.3) you find two polygons on the same base and between the same parallels? July 1, 2021 Category: Chapter 9 - Areas of Parallelogram and [...]
1. The median of a triangle divides it into two
01
Jul
1. The median of a triangle divides it into two 1. The median of a triangle divides it into two July 1, 2021 Category: Chapter 9 - Areas of Parallelogram and Triangles , Maths , NCERT Class 9 ,
Example 4 : In Fig. 9.22, ABCD is a quadrilateral and BE ∣∣ AC and also BE meets DC produced at E. Show that area of ΔADEis equal to the area of the quadrilateral ABCD.
15
Oct
Example 4 : In Fig. 9.22, ABCD is a quadrilateral and BE ∣∣ AC and also BE meets DC produced at E. Show that area of ΔADEis equal to the area of the quadrilateral ABCD. ABCD is a quadrilateral and BE ∣∣ AC and also BE meets DC produced at E. Show that area of [...]
Example 3 : Show that median of a triangle divides it into two triangles of equal area.
15
Oct
Example 3 : Show that median of a triangle divides it into two triangles of equal area. Example 3 : Show that median of a triangle divides it into two triangles of equal area. October 15, 2020 Category: Chapter 9 - Areas of Parallelogram and Triangles , Maths , NCERT Class 9 ,
Example 2 : If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that area of the triangle is equal to half of the area of parallelogram.
15
Oct
Example 2 : If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that area of the triangle is equal to half of the area of parallelogram. Example 2 : If a triangle and a parallelogram lie on the same base and between the same parallels then [...]
Example 1 : In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle. Also, AL⊥DC. Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = DC × AL
15
Oct
Example 1 : In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle. Also, AL⊥DC. Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = DC × AL ABCD is a parallelogram and EFCD is a rectangle. Also AL⊥DC. Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = [...]
16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
15
Oct
16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. 16. In Fig.9.29 ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. October 15, 2020 Category: Chapter 9 - Areas of Parallelogram and Triangles , Maths , [...]
15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Prove that ABCD is a trapezium.
15
Oct
15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Prove that ABCD is a trapezium. 15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Prove that ABCD is a trapezium. October 15, 2020 [...]
14. In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).
15
Oct
14. In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR). 14. In Fig.9.28 AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR). October 15, 2020 Category: Chapter 9 - Areas of Parallelogram and Triangles , Maths , NCERT Class 9 ,
13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) = ar (△ACY).
15
Oct
13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) = ar (△ACY). 13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) [...]