Question

in the figure, BC = CD = DE and is the mid-point of cd. Which of the following is the area of APC?
(I)1/3 ar(ABC)
(ii)1/2ar(ABD)
(III)1/6ar(ABC)
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Answer 1

Answer:

it's 1/3

Step-by-step explanation:

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Answer 2

Answer:  the area of triangle APC is equal to 1/4 of the area of triangle ABC and 1/2 of the area of triangle ABD. Therefore, the answer is (ii) 1/2 ar(ABD).

Step-by-step explanation: The area of a triangle can be calculated using the formula:

Area = (base * height) / 2

We can use this formula to find the areas of triangles ABC, ABD, and APC.

Since BC = CD = DE, we can conclude that triangles ABC, ABD, and ACD are isosceles triangles. This means that their bases are equal in length and their heights are equal in length.

Since D is the midpoint of CD, we can conclude that CD is half the length of BC.

So, the height of triangle ABC is equal to the base of triangle ABD, and the height of triangle ABD is equal to half the height of triangle ABC.

Using the formula, we can calculate the area of triangle ABC:

Area(ABC) = (BC * BC) / 2 = (2 * BC * BC) / 2 = BC^2

Using the formula, we can calculate the area of triangle ABD:

Area(ABD) = (BD * BD) / 2 = (BC / 2 * BC / 2) / 2 = BC^2 / 4

Using the formula, we can calculate the area of triangle APC:

Area(APC) = (PC * PC) / 2 = (BC / 2 * BC / 2) / 2 = BC^2 / 4

So, the area of triangle APC is equal to 1/4 of the area of triangle ABC and 1/2 of the area of triangle ABD. Therefore, the answer is (ii) 1/2 ar(ABD).