Question

a model of an isosceles triangle is shown below.
what is the area, in square yards, of the actual triangle, rounded to the nearest whole number?
448 yd²
199 yd²
168 yd²
none of the above

Explanation:

Step1: Find actual base length

The model base is \( 14 \) in. The scale is \( 3 \) in : \( 4 \) yd. Let actual base be \( b \).
Set up proportion: \( \frac{3}{4}=\frac{14}{b} \) → \( b = \frac{14\times4}{3}=\frac{56}{3}\approx18.67 \) yd? Wait, no—wait, the base of the isosceles triangle: the model has base \( 14\times2 = 28 \) in? Wait, no, the diagram: the height splits the base into two \( 14 \) in? Wait, no, the model's base (AC) is \( 14\times2 = 28 \) in? Wait, the height is \( 16 \) in, and the base segments are \( 14 \) in each? Wait, no, the model: the base (AC) is \( 14 \times 2 = 28 \) in? Wait, no, the diagram shows the height (red dashed) is \( 16 \) in, and the horizontal segment from A to the foot of the height is \( 14 \) in, so total base \( AC = 14 \times 2 = 28 \) in.

So model base \( b_{model} = 28 \) in, model height \( h_{model} = 16 \) in.

Scale: \( 3 \) in : \( 4 \) yd.

For base: \( \frac{3}{4}=\frac{28}{b_{actual}} \) → \( b_{actual} = \frac{28\times4}{3}=\frac{112}{3}\approx37.33 \) yd? Wait, no, wait: maybe I misread the base. Wait, the diagram: the foot of the height is at the midpoint, so the base (AC) is \( 14 \) in? Wait, no, the label "14 in" is under the foot, so the base is \( 14 \) in? Wait, no, the height is \( 16 \) in, and the side AB is \( 18 \) in. Let's recalculate:

Wait, maybe the model's base is \( 28 \) in (since the height splits it into two \( 14 \) in). Let's confirm: in the model, the base (AC) is \( 14 \times 2 = 28 \) in. The height (h_model) is \( 16 \) in.

Now, scale for base: \( 3 \) in : \( 4 \) yd. So actual base \( b_{actual} = \frac{28\times4}{3}=\frac{112}{3}\approx37.33 \) yd? Wait, no, wait—maybe the height is \( 16 \) in, so actual height:

Step2: Find actual height

Model height \( h_{model} = 16 \) in. Let actual height be \( h \).
Proportion: \( \frac{3}{4}=\frac{16}{h} \) → \( h = \frac{16\times4}{3}=\frac{64}{3}\approx21.33 \) yd.

Wait, no—wait, the scale is \( 3 \) in represents \( 4 \) yd. So for any length \( L_{model} \) (in inches), actual length \( L_{actual} = \frac{4}{3} \times L_{model} \) (in yards).

So model base (AC) is \( 28 \) in (since two \( 14 \) in segments). So \( b_{actual} = \frac{4}{3} \times 28 = \frac{112}{3}\approx37.33 \) yd.
Model height \( h_{model} = 16 \) in, so \( h_{actual} = \frac{4}{3} \times 16 = \frac{64}{3}\approx21.33 \) yd.

Step3: Calculate area of triangle

Area of triangle: \( A = \frac{1}{2} \times base \times height \).
\( A = \frac{1}{2} \times \frac{112}{3} \times \frac{64}{3} \)? Wait, no—wait, no, I messed up the base. Wait, the model's base: looking at the diagram, the horizontal segment from A to the foot is \( 14 \) in, so total base (AC) is \( 14 \times 2 = 28 \) in. Correct.

Wait, but let's recalculate with correct scale application:

For base: \( 28 \) in. \( \frac{3 \text{ in}}{4 \text{ yd}} = \frac{28 \text{ in}}{b} \) → \( b = \frac{28 \times 4}{3} = \frac{112}{3} \approx 37.33 \) yd.

For height: \( 16 \) in. \( \frac{3}{4} = \frac{16}{h} \) → \( h = \frac{16 \times 4}{3} = \frac{64}{3} \approx 21.33 \) yd.

Then area \( A = \frac{1}{2} \times 37.33 \times 21.33 \approx \frac{1}{2} \times 796.67 \approx 398.33 \)? No, that's not matching. Wait, maybe I misread the base. Wait, the model's base is \( 14 \) in (not \( 28 \))? Wait, the diagram: the foot of the height is at the midpoint, and the segment from A to foot is \( 14 \) in, so total base is \( 14 \times 2 = 28 \) in. But maybe the scale is applied to each dimension: base \( 28 \) in, height \( 16 \) in.

Wait, let's do it again:

Scale: \( 3 \) in = \( 4 \) yd → \( 1 \) in = \( \frac{4}{3} \) yd.

So actual base: \( 28 \) in \( \times \frac{4}{3} \) yd/in = \( \frac{112}{3} \) yd.

Actual height: \( 16 \) in \( \times \frac{4}{3} \) yd/in = \( \frac{64}{3} \) yd.

Area = \( \frac{1}{2} \times \frac{112}{3} \times \frac{64}{3} = \frac{1}{2} \times \frac{7168}{9} = \frac{3584}{9} \approx 398.22 \). No, that's not one of the options. Wait, maybe the base is \( 14 \) in (not \( 28 \))? Wait, maybe the diagram's base is \( 14 \) in (the entire base, not split). Wait, the height is \( 16 \) in, and the base is \( 14 \) in? But then the triangle would have sides \( 18 \) in, \( 18 \) in, and base \( 14 \) in? Wait, using Pythagoras: \( 14/2 = 7 \), \( 7^2 + 16^2 = 49 + 256 = 305 \), \( 18^2 = 324 \). Not equal. So my initial assumption is wrong.

Wait, maybe the model's base is \( 14 \) in (the entire base), and the height is \( 16 \) in. Then:

Actual base: \( b = \frac{4}{3} \times 14 = \frac{56}{3} \approx 18.67 \) yd.

Actual height: \( h = \frac{4}{3} \times 16 = \frac{64}{3} \approx 21.33 \) yd.

Area = \( \frac{1}{2} \times \frac{56}{3} \times \frac{64}{3} = \frac{1}{2} \times \frac{3584}{9} = \frac{1792}{9} \approx 199.11 \) yd², which rounds to \( 199 \) yd². Ah! So I messed up the base: the model's base is \( 14 \) in (not \( 28 \)). The height splits the base into two \( 7 \) in? Wait, no, the diagram: the label "14 in" is under the foot, so the base is \( 14 \) in (the entire base), not split. So the base is \( 14 \) in, height \( 16 \) in. Then:

Model base \( b_{model} = 14 \) in, model height \( h_{model} = 16 \) in.

Actual base \( b = \frac{4}{3} \times 14 = \frac{56}{3} \) yd.

Actual height \( h = \frac{4}{3} \times 16 = \frac{64}{3} \) yd.

Area = \( \frac{1}{2} \times \frac{56}{3} \times \frac{64}{3} = \frac{56 \times 32}{9} = \frac{1792}{9} \approx 199.11 \approx 199 \) yd².

Step2: Find actual height (corrected)

Model height: \( 16 \) in. Scale: \( 3 \) in : \( 4 \) yd. So \( h_{actual} = \frac{16 \times 4}{3} = \frac{64}{3} \) yd.

Step3: Calculate area

Area = \( \frac{1}{2} \times b_{actual} \times h_{actual} = \frac{1}{2} \times \frac{56}{3} \times \frac{64}{3} = \frac{1792}{9} \approx 199 \) yd².

Answer:

199 yd² (option: 199 yd²)