For those looking for help on Areas of Parallelograms and Triangles Class 9 Math Concepts can find all of them here provided in a comprehensive manner. To make it easy for you we have jotted the Class 9 Areas of Parallelograms and Triangles Maths Formulae List all at one place. You can find Formulas for all the topics lying within the Areas of Parallelograms and Triangles Class 9 Areas of Parallelograms and Triangles in detail and get a good grip on them. Revise the entire concepts in a smart way taking help of the Maths Formulas for Class 9 Areas of Parallelograms and Triangles.
The List of Important Formulas for Class 9 Areas of Parallelograms and Triangles is provided on this page. We have everything covered right from basic to advanced concepts in Areas of Parallelograms and Triangles. Make the most out of the Maths Formulas for Class 9 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Areas of Parallelograms and Triangles Class 9 covering numerous concepts and use them to solve your Problems effortlessly.
1. Area of a parallelogram = base x height
= DC x AE
2. Area of a triangle = \(\frac{1}{2}\)base x height
\(\frac{1}{2}\) BC x AD
3. Area of a trapezium = x (Sum of parallel sides) x Distance between them
\(\frac{1}{2}\) (AB + DC) x AE
4. Area of a rhombus =\(\frac{1}{2}\) \(\frac{1}{2}\) x product of diagonals A B
\(\frac{1}{2}\) x AC x BD
5. 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 (or the vertex) opposite to the common base of each figure He on a line parallel to the base.
Theorem 9.1: Parallelograms on the same base and between the same parallels are equal in area.
ar(ABCD) = ar(EFCD)
Theorem 9.2: Triangles on the same base and between the same parallels are equal in area.
ar(ΔABC) = ar(ΔPBC)
Theorem 9.3: Two triangles having the same base and equal areas lie between the same parallels.
If a triangle and a parallelogram are on the same base and between the same parallels, then
(i) Area of triangle = \(\frac{1}{2}\) x area of the parallelogram
ar(ΔPDC) = \(\frac{1}{2}\) ar(||gmABCD)
(ii) A diagonal of parallelogram divides it into two triangles of equal areas.
ar(ΔABD) = ar(ΔBCD)
(iii) If each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.
(iv) A median AD of a ΔABC divides it into two triangles of equal areas.
ar(ΔABD) = ar(ΔACD)