What is the area of the lot in square meters?
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The flat triangular lot depicted above is available for development. If X + X/2 = 150 meters, what is the area of the lot in square meters?
$$A.\ 125$$
$$B.\ 150\sqrt{3}$$
$$C.\ 1,500$$
$$D.\ 1,250\sqrt{3}$$
$$E.\ 2,500\sqrt{3}$$
The OA is D.
If X + X/2 = 150 then, X = 100. Now, we need to calculate the base of triangular lot,
$$100^2=50^2+b^2,\ then,\ b=\sqrt{100^2-50^2}=\sqrt{7500}=50\sqrt{3}$$
Now we can determine the area of the lot of the following way,
$$A_{\triangle}=\frac{1}{2}b\cdot h=\frac{1}{2}\left(50\sqrt{3}\right)\left(50\right)=1,250\sqrt{3}square\ meters$$
Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
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Hi AAPL,AAPL wrote:
The flat triangular lot depicted above is available for development. If X + X/2 = 150 meters, what is the area of the lot in square meters?
$$A.\ 125$$
$$B.\ 150\sqrt{3}$$
$$C.\ 1,500$$
$$D.\ 1,250\sqrt{3}$$
$$E.\ 2,500\sqrt{3}$$
The OA is D.
If X + X/2 = 150 then, X = 100. Now, we need to calculate the base of triangular lot,
$$100^2=50^2+b^2,\ then,\ b=\sqrt{100^2-50^2}=\sqrt{7500}=50\sqrt{3}$$
Now we can determine the area of the lot of the following way,
$$A_{\triangle}=\frac{1}{2}b\cdot h=\frac{1}{2}\left(50\sqrt{3}\right)\left(50\right)=1,250\sqrt{3}square\ meters$$
Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
The approach you followed is perfectly correct.
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We see that a leg of this right triangle is ½ of its hypotenuse. Thus, it must be a 30-60-90 triangle. Recall the ratio of the sides is: s : s√3: 2s. Since here we have s = x/2 and 2s = x, so we have s√3 = x√3/2. Furthermore, we can solve for x since we are given that x/2 + x = 150. Multiplying the equation by 2, we have:
x+ 2x = 300
3x = 300
x = 100
Thus, the base of the triangle = x√3/2 = 100√3/2 = 50√3, and the height = x/2 = 100/2 = 50, and the area is:
1/2(50√3)(50) = 1250√3
Answer: D
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