In the given figure, which triangle has a greater area: XDC or YBC, if both the rectangles are identical?

Question

Homi · Accepted Answer

The area of a triangle is half the product of its base and its corresponding height.

Step 1 — Calculate the area of $\Delta$ XDC

Let us consider the first rectangle, ABCD. Let the length of the rectangle (DC or AB) be $L$ . Let the width of the rectangle (AD or BC) be $W$ . For $\Delta$ XDC, the base is DC. So, the base of $\Delta$ XDC is $L$ . The vertex X lies on AB, which is parallel to DC. The height of $\Delta$ XDC is the perpendicular distance from X to DC. This height is equal to the width of the rectangle, $W$ . The area of $\Delta$ XDC is given by the formula: (1/2) * base * height.

$\text{Area}(\Delta\text{XDC}) = \frac{1}{2} \times \text{DC} \times \text{AD}$

$= \frac{1}{2} \times L \times W$

$\boxed{\text{Area}(\Delta\text{XDC}) = \frac{1}{2}LW}$

Diagram 1

Step 2 — Calculate the area of $\Delta$ YBC

Let us consider the second rectangle, also ABCD, which is identical to the first. So, its length is $L$ and its width is $W$ . For $\Delta$ YBC, the base is BC. So, the base of $\Delta$ YBC is $W$ . The vertex Y lies on AD, which is parallel to BC. The height of $\Delta$ YBC is the perpendicular distance from Y to BC. This height is equal to the length of the rectangle, $L$ . The area of $\Delta$ YBC is given by the formula: (1/2) * base * height.

$\text{Area}(\Delta\text{YBC}) = \frac{1}{2} \times \text{BC} \times \text{AB}$

$= \frac{1}{2} \times W \times L$

$\boxed{\text{Area}(\Delta\text{YBC}) = \frac{1}{2}LW}$

Diagram 2

Step 3 — Compare the areas

We found that the area of $\Delta$ XDC is $\frac{1}{2}LW$ . We also found that the area of $\Delta$ YBC is $\frac{1}{2}LW$ . Since both areas are equal to $\frac{1}{2}LW$ , they are the same.

Answer

Both triangles have the same area.

Question 6

Answer

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