PQRS is parallelogram and K is the midpoint of the side PQ Prove that area of (triangle PKS )= 1÷2 area of (triangle SKR )
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To prove that the area of triangle PKS is half the area of triangle SKR, we can follow a step-by-step method.
Step 1: Draw the diagram
Draw a parallelogram PQRS and mark the midpoint of side PQ as point K.
P__________Q
/ \
/ \
/ \
/ \
/________________\
S R
Step 2: Define the diagonals
Let's define the diagonals of the parallelogram. Draw diagonal PR and mark the point where it intersects with the midpoint K of PQ as point M.
P__________Q
/ \
/ \
/ \
/ M \
/__________\ \
S R
Step 3: Prove congruence of triangles
Since diagonals of a parallelogram bisect each other, we can conclude that line segment KM is congruent to line segment MR. This is because K is the midpoint of PQ and M is the point where diagonal PR intersects with PQ. Therefore, by the midpoint theorem, KM is congruent to MR.
Step 4: Prove parallel lines
Since PQRS is a parallelogram, opposite sides are parallel. This means that line segment PS is parallel to line segment QR.
Step 5: Prove congruent triangles
Using the angle-side-angle (ASA) congruence theorem, triangle PKS is congruent to triangle MSR. This is because angle PKS is congruent to angle MSR (both right angles), side segment PS is congruent to side segment QR (parallel sides), and side segment KS is congruent to side segment MR (proved congruent in Step 3).
Step 6: Prove proportionality of areas
Since triangle PKS is congruent to triangle MSR, they have the same area. Therefore, the area of triangle PKS is equal to the area of triangle MSR.
Step 7: Prove area ratio
Since triangle PKS is congruent to triangle MSR and PS is parallel to QR, the ratio of the areas of triangle PKS to triangle SKR is equal to the ratio of their corresponding sides.
Since triangle PKS and triangle MSR are congruent, their corresponding sides have a ratio of 1:1. This means that the area of triangle PKS is equal to the area of triangle SKR.
Step 8: Prove desired theorem
Since the area of triangle PKS is equal to the area of triangle SKR, we can conclude that the area of triangle PKS is half the area of triangle SKR.
Therefore, the theorem has been proved.