A pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
Lets concentrate on the quadrilateral ADEG.
let the intersection of AE and DB be K
GE || AD (given)
So triangles ADE and ADG have the same base AD and the same height , since parallel lines are equal distances apart.
So triangle ADE = triangle ADG
but ADK is common to both, so subtracting ...
triangle AKG = triangle DEK
similarly by letting FD and BC intersect at L we can show that triangle FLB = triangle CDL
pentagon ABCDE
= triangle BLD + triangle LCD + triangle ABD + triangle DEK + triangle ADK
= triangle BLD + triangle FLB + triangle ABD + triangle GAK + triangle ADk
= triangle FDG
tri FDG
In given figure X and Y are the mid point of AC and AB respectively QP parallel BC and CYQ and BXP are straight line. Prove that ar Triangle ABP =ar triangle ACQ
Since Y and X are the nid points of AB and AC ; by mid point theorem XY || BC. Now consider the ∆ BCQ and ∆ CBP their areas are equal cz they are on same base BC and between the same parallels
: AR(BCQ) = AR(BCP)
Let the intersection point of QC and BP be O
: add AR(AQOP) both sides.....
Therefore AR (ABP) =AR(ACQ)
Jabbba jabba kappale mayathu
This question is wrong
To prove that the area of pentagon ABCDE is equal to the area of triangle GDF, we can use the concept of similar triangles and the properties of parallel lines.
Let's start by labeling the known points and lines:
- Points:
* A, B, C, D, E: vertices of the pentagon ABCDE
* F: the intersection of line CF and AB produced
* G: the intersection of line EG and BA produced
- Lines:
* EG: a line parallel to DA
* CF: a line parallel to DB
* AB: the side of the pentagon
Now, let's examine the relationship between the triangles GDF and ABC:
1. Similar Triangles:
Since EG is parallel to DA, and CF is parallel to DB, we can apply the similarity property of triangles.
In triangle ABC and triangle GDF, we have:
- Angle GDF = Angle CBA (corresponding angles, due to parallel lines)
- Angle FGD = Angle BAC (corresponding angles, due to parallel lines)
Therefore, triangle GDF is similar to triangle ABC.
2. Ratio of Corresponding Sides:
Since triangle GDF is similar to triangle ABC, the ratio of their corresponding sides will be equal.
By comparing the corresponding sides, we have:
- Side GD corresponds to side BC
- Side DF corresponds to side CA
- Side GF corresponds to side AB
Therefore, the ratio of the area of triangle GDF to the area of triangle ABC is equal to the square of the ratio of their corresponding sides.
3. Equality of Areas:
Since the ratio of the corresponding sides is equal, the square of this ratio will also be equal. Hence, we can conclude that the area of triangle GDF is equal to the area of triangle ABC.
Therefore, the area of pentagon ABCDE is equal to the area of triangle GDF.
Note: In the above proof, we made use of the similarity property of triangles and the fact that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.