![Example 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2](https://d1avenlh0i1xmr.cloudfront.net/684f721c-7325-40ee-9767-59bda860862aslide4.jpg)
![Example 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3](https://d1avenlh0i1xmr.cloudfront.net/90f422fb-6b43-44b6-a326-599ddbfc93ddslide5.jpg)
![Example 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5](https://d1avenlh0i1xmr.cloudfront.net/eed39b4e-c57b-4302-a864-c9c335ec16a7slide7.jpg)
![Example 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 6](https://d1avenlh0i1xmr.cloudfront.net/44ead589-edc0-4c05-8540-410dc98fb0b1slide8.jpg)
Last updated at April 16, 2024 by Teachoo
Example 2 (Method 1) If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is equal to half the area of parallelogram. Given: A parallelogram ABCD and ABP on the same base AB and between the same parallels To prove: Area of triangle is equal to half the area of parallelogram. ar ( ABP ) = 1/2 ar (ABCD) Construction: Join DP Let DM AB & PN AB Proof: Example 2 (Method 2) If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is equal to half the area of parallelogram. Given: A parallelogram ABCD and ABP on the same base AB and between the same parallels PC & AB To prove: Area of triangle is equal to half the area of parallelogram. ar ( ABP ) = 1/2 ar (ABCD) Proof: In parallelogram ABCD, AB CD So, PC AB We draw a line BQ parallel to AP , i.e. , BQ AP Since BQ AP & PQ AB, Both pairs of opposite sides are parallel ABQP is a parallelogram Parallelograms ABQP & ABCD are on the same base AB and between the same parallel lines AB & PC Area(ABQP) = Area(ABCD) In parallelogram ABQP, BP is the diagonal So, ABP QBP Area( ABP) = Area( QBP) Now, Area( ABP) = Area( QBP) = 1/2 Area(ABQP) Area( ABP) = 1/2 Area(ABQP) Area( ABP) = 1/2 Area(ABCD) Hence proved