Areas of Parallelograms and Triangles Notes: Class 9 Maths Chapter 9
Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles covers figures on the same base and between the same parallels, and the key theorems relating their areas.
Overview of Topics & Subtopics (Click to Jump)
1. Figures on the Same Base and Between the Same Parallels
Short Note:
• Two figures are said to be on the same base and between the same parallels if they have a common side (base) and the vertices opposite to the base lie on a line parallel to the base.
2. Parallelograms on the Same Base and Between the Same Parallels
Short Note:
• Parallelograms on the same base and between the same parallels are equal in area.
• A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
3. Triangles on the Same Base and Between the Same Parallels
Short Note:
• Two triangles on the same base and between the same parallels are equal in area.
• A triangle and a parallelogram on the same base and between the same parallels have area of the triangle equal to half the area of the parallelogram.
• The median of a triangle divides it into two triangles of equal area.
NCERT Textbook Exercise Questions
In the given figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Answer: Area of parallelogram = base × height. Using DC as base: Area = DC × AE = 16 × 8 = 128 cm² (since DC = AB = 16 cm). Using AD as base: Area = AD × CF ⇒ 128 = AD × 10 ⇒ AD = 12.8 cm.
Show that a diagonal of a parallelogram divides it into two triangles of equal area.
Answer: Diagonal AC divides parallelogram ABCD into ΔABC and ΔACD. Since AB = DC and BC = AD (opposite sides) and AC is common, ΔABC ≅ ΔACD by SSS, so they have equal area, each half the area of the parallelogram.
D, E and F are respectively the mid-points of the sides BC, CA and AB of ΔABC. Show that (i) BDEF is a parallelogram, (ii) ar(DEF) = ¼ ar(ABC).
Answer: By the Mid-point Theorem, FE || BD and FE = BD (= ½BC), so BDEF is a parallelogram. Since ΔABC is divided into four congruent triangles by joining the mid-points, each has area equal to one-fourth of ar(ABC), so ar(DEF) = ¼ ar(ABC).
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(AOD) = ar(BOC).
Answer: ΔABD and ΔABC lie on the same base AB and between the same parallels AB and DC, so ar(ABD) = ar(ABC). Subtracting the common area ar(AOB) from both gives ar(AOD) = ar(BOC).
