
Prove the theorem: parallelogram on the same base and between the same parallel are equal in area
Answer
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Hint: In this question use the Angle-Angle-Side property to show that the two triangles in the given parallelograms are congruent to each other, using this information will help you to approach towards the solution of the question.
Complete step-by-step answer:
According to the given information we have 2 parallelogram ABCD and PQCD having same base CD and lie between the same parallel AP and CD
So we have to prove that the area of the given parallelogram are equal to each other when they have same base and lie between the same parallel
We know that in parallelogram ABCD since AC is parallel to BD where AB is the transversal which is intersecting them
So we can say that by the property of corresponding angles
Similarly in parallelogram PQDC since PC is parallel to QD here PQ is the traversal which is intersecting them
So we can say that by the property of corresponding angles
Since we know that opposite sides of a parallelogram are equal
Therefore AD = BD
Now In triangle APD and triangle BQC
We know that by the above statement
And AD = BC
So by the property of angle-angle-side property i.e. [AAS]
Therefore
So, since the triangle APD and BQC are congruent
Therefore area of triangle APD is equal to the area of triangle BQC
According to the diagram mentioned above we can say that
Area of parallelogram ABCD = Area of triangle APD + Area of PBCD
And since we know that Area of triangle APD = Area of triangle BQC
Therefore Area of parallelogram ABCD = Area of triangle BQC + Area of PBCD
Since we know that Area of parallelogram PQCD = Area of triangle BQC + Area of PBCD
Therefore Area of parallelogram ABCD = Area of parallelogram PQCD
Hence proved.
Note: In the above solution we used a term “congruent “which can be explained as a state of two triangles when the dimension of all three sides and all three angles of two triangles are equal to each other; these triangles are said to be congruent triangles. There are some conditions for congruence of triangles such as Side-Side-Side (SSS) property, Side-Angle-Side (SAS) property, etc. the triangles which satisfy this condition are said to be congruent triangles.
Complete step-by-step answer:
According to the given information we have 2 parallelogram ABCD and PQCD having same base CD and lie between the same parallel AP and CD
So we have to prove that the area of the given parallelogram are equal to each other when they have same base and lie between the same parallel

We know that in parallelogram ABCD since AC is parallel to BD where AB is the transversal which is intersecting them
So we can say that
Similarly in parallelogram PQDC since PC is parallel to QD here PQ is the traversal which is intersecting them
So we can say that
Since we know that opposite sides of a parallelogram are equal
Therefore AD = BD
Now In triangle APD and triangle BQC
We know that by the above statement
And AD = BC
So by the property of angle-angle-side property i.e. [AAS]
Therefore
So, since the triangle APD and BQC are congruent
Therefore area of triangle APD is equal to the area of triangle BQC
According to the diagram mentioned above we can say that
Area of parallelogram ABCD = Area of triangle APD + Area of PBCD
And since we know that Area of triangle APD = Area of triangle BQC
Therefore Area of parallelogram ABCD = Area of triangle BQC + Area of PBCD
Since we know that Area of parallelogram PQCD = Area of triangle BQC + Area of PBCD
Therefore Area of parallelogram ABCD = Area of parallelogram PQCD
Hence proved.
Note: In the above solution we used a term “congruent “which can be explained as a state of two triangles when the dimension of all three sides and all three angles of two triangles are equal to each other; these triangles are said to be congruent triangles. There are some conditions for congruence of triangles such as Side-Side-Side (SSS) property, Side-Angle-Side (SAS) property, etc. the triangles which satisfy this condition are said to be congruent triangles.
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