the base BC of a triangle ABC is divided at D so that BD=1/3 BC .prove that ar(triangle ABD)=1/2 ar (triangle ADC)
Not informative concept vise.Worst answer everrrrr.Shame of the person who gave this solution.
To prove that the area of triangle ABD is half the area of triangle ADC, we can use the fact that the areas of two triangles with a common base are directly proportional to their corresponding altitudes.
Let's denote the altitude of triangle ABD from vertex A as h_1 and the altitude of triangle ADC from vertex A as h_2.
Since triangle ABD and triangle ADC share the same base (BC), their altitudes can be represented as h_1 and h_2 respectively.
By the given condition, it is stated that BD = (1/3)BC. From this, we can conclude that the ratio of the lengths of BD to BC is 1:3.
Since triangle ABD and triangle ADC share the same height (h_1 = h_2), the ratio of their corresponding altitudes can also be expressed as 1:3.
Now, let's calculate the area of triangle ABD and triangle ADC using their corresponding altitudes and the common base BC.
Area of triangle ABD (denoted as ar(ABD)) = (1/2) * BC * h_1
Area of triangle ADC (denoted as ar(ADC)) = (1/2) * BC * h_2
Since h_1 = h_2, we can rewrite the area of triangle ABD and triangle ADC as follows:
ar(ABD) = (1/2) * BC * h_1
ar(ADC) = (1/2) * BC * h_1
Since the altitudes h_1 and h_2 have a ratio of 1:3, we can substitute h_1 with 3h_1 in the area of triangle ADC:
ar(ADC) = (1/2) * BC * 3h_1
= (3/2) * BC * h_1
Comparing the area of triangle ABD (ar(ABD)) and the modified area of triangle ADC [(3/2) * BC * h_1], we can see that ar(ABD) is exactly half of (3/2) * BC * h_1.
ar(ABD) = (1/2) * (3/2) * BC * h_1
= (3/4) * BC * h_1
Since (3/4) * BC * h_1 is equivalent to (1/2) * BC * h_1, we have proved that:
ar(ABD) = (1/2) * ar(ADC)
Hence, the area of triangle ABD is half the area of triangle ADC, as required.
To prove that the area of triangle ABD is half the area of triangle ADC, we can use the concept of similar triangles. Here's how you can prove it step by step:
Step 1: Draw triangle ABC.
Step 2: Draw a line segment from vertex B to a point D on the base AC, such that BD = 1/3 BC.
Step 3: Draw a line segment from vertex A to point D.
Step 4: Label the intersection point of AD and BC as E.
Step 5: We need to prove that the area of triangle ABD is half the area of triangle ADC.
Step 6: To prove this, we'll show that triangles ABD and ADC are similar.
Step 7: Notice that triangle ABD and triangle AEC share the same height, as the height is measured perpendicular to the base BC.
Step 8: Since BD = 1/3 BC, we can also say that DE = 2/3 BC.
Step 9: Now, consider triangles ABD and AEC. They share the same height and have a common angle at vertex A.
Step 10: By the SAS (Side-Angle-Side) similarity criterion, triangles ABD and AEC are similar.
Step 11: By the property of similar triangles, corresponding sides are in proportion.
Step 12: In triangle ABD, the corresponding side to BC is BD, and in triangle AEC, the corresponding side to BC is EC.
Step 13: Since BD = 1/3 BC, we can conclude that EC = 2/3 BC.
Step 14: Now, in triangle ADC, the corresponding side to BC is DC.
Step 15: Since BC = BE + EC, we substitute BC with BE + EC.
Step 16: We know that EC = 2/3 BC, so we can rewrite BC as BE + 2/3 BC.
Step 17: Simplifying this equation, we get 1/3 BC = BE.
Step 18: This tells us that BD = BE.
Step 19: Triangle ABD and triangle BDE share the same base BD.
Step 20: Since they share the same base and have equal heights, their areas are equal.
Step 21: Therefore, the area of triangle ABD = the area of triangle BDE.
Step 22: Since BE = 1/3 BC and EC = 2/3 BC, the ratio of the areas of triangle BDE and triangle ADC is 1/2.
Step 23: Therefore, the area of triangle ABD = 1/2 the area of triangle ADC.
Thus, we have proved that the area of triangle ABD is half the area of triangle ADC.
isn't the height of the triangle the same, so the only thing different about the two new triangles is the base?
Area ABD=1/2 Base*height= 1/2 1/3BC*h
area ADC= 1/2 Base*height= 1/2 2/3BC*h
dividefirst area by second
ratio of areas= 1/2