triangles with common vertex have same height or same base how explain
Answers
Answer:
Consider the figure presented above. Can you see that ΔBCE and ΔADF will be congruent? This is easy to show.
We have: BC = AD (opposite sides of a parallelogram are equal)
∠BCE = ∠ADF (corresponding angles)
∠BEC = ∠AFD (corresponding angles)
By the ASA criterion, the two triangles are congruent, which means that their areas are equal. Now,
area(ABCD) = area(ABED) + area(ΔBCE)
Similarly,
area(ABEF) = area(ABED) + area(ΔADF)
Clearly,
area(ABCD) = area(ABEF)
Next, consider a parallelogram ABCD and a rectangle ABEF on the same base and between the same parallels:
Parallelogram and Rectangle - Same base same parallel
Clearly, their areas will be equal. Now, the length and height (width) of the rectangle have been marked as l and w respectively. Therefore,
area(ABCD) = area(ABEF) = l × w
This means that the area of any parallelogram is equal to the product of its base and its height (the height of a parallelogram can be defined as the distance between its base and the opposite parallel).
Now, consider the following figure, which shows a parallelogram ABCD and a triangle ABE on the same base AB and between the same parallels:
Parallelogram and Triangle - Same base same parallel
What will be the relation between the areas of these two figures? Let us complete the parallelogram ABEF, as shown below:
Parallelogram and Triangles - Same base same parallel
Now, we have:
area(ΔABE) = ½ area(ABFE)
= ½ area(ABCD)
Thus, the area of the triangle is exactly half of the area of the parallelogram. Let us define the height of a triangle as the distance between the base and the parallel through the opposite vertex. We can therefore say that the area of the triangle will be:
Area = ½ × base × height
Note that any of the three sides of the triangle can be taken as the base, but then the height will change accordingly.
Step-by-step explanation:
इश्क टू एंगल से उस सेम बेसन सेम हाइट इन व्हिच ऑफ द फॉलोइंग इज करेक्ट