Question

reasoning find the perimeter and area of each triangle. round to the nearest hundredth. 30. 5 in. 59° 31. 18° 12 cm 32. 48° 3.5 ft

Answer:

Problem 30: Right Triangle with one leg 5 in and angle \(59^\circ\)

Step 1: Identify the triangle type and knowns

This is a right triangle (implied by the right angle symbol). Let the right angle be \(C\), one leg \(a = 5\) in, angle \(B=59^\circ\). We can find the other leg \(b\) and hypotenuse \(c\) using trigonometric ratios.

Step 2: Find the other leg (\(b\))

\(\tan(59^\circ)=\frac{b}{a}\), so \(b = a\tan(59^\circ)\)
\(b = 5\times\tan(59^\circ)\approx5\times1.6643\approx8.3215\) in

Step 3: Find the hypotenuse (\(c\))

\(\cos(59^\circ)=\frac{a}{c}\), so \(c=\frac{a}{\cos(59^\circ)}\)
\(c=\frac{5}{\cos(59^\circ)}\approx\frac{5}{0.5150}\approx9.7087\) in

Step 4: Calculate Perimeter (\(P\))

\(P=a + b + c=5 + 8.3215+9.7087\approx23.03\) in

Step 5: Calculate Area (\(A\))

\(A=\frac{1}{2}\times a\times b=\frac{1}{2}\times5\times8.3215\approx20.80\) in²

Problem 31: Right Triangle with one leg 12 cm and angle \(18^\circ\)

Step 1: Identify the triangle type and knowns

Right triangle, one leg \(a = 12\) cm, angle \(A = 18^\circ\) (opposite to leg \(a\)? Wait, no, the right angle is between the legs. Let's assume the leg adjacent to \(18^\circ\) is \(12\) cm. Let the right angle be \(C\), leg \(b = 12\) cm, angle \(A=18^\circ\).

Step 2: Find the other leg (\(a\))

\(\tan(18^\circ)=\frac{a}{b}\), so \(a = b\tan(18^\circ)\)
\(a = 12\times\tan(18^\circ)\approx12\times0.3249\approx3.8988\) cm

Step 3: Find the hypotenuse (\(c\))

\(\cos(18^\circ)=\frac{b}{c}\), so \(c=\frac{b}{\cos(18^\circ)}\)
\(c=\frac{12}{\cos(18^\circ)}\approx\frac{12}{0.9511}\approx12.617\) cm

Step 4: Calculate Perimeter (\(P\))

\(P=a + b + c=3.8988+12 + 12.617\approx28.52\) cm

Step 5: Calculate Area (\(A\))

\(A=\frac{1}{2}\times a\times b=\frac{1}{2}\times3.8988\times12\approx23.39\) cm²

Problem 32: Isosceles? Wait, triangle with base 3.5 ft and angle \(48^\circ\) (assuming it's an isosceles triangle with base 3.5 ft and vertex angle? Wait, no, maybe it's a triangle with base 3.5 ft and base angles \(48^\circ\)? Wait, the diagram shows a triangle with base 3.5 ft and one base angle \(48^\circ\), so it's an isosceles triangle with base \(b = 3.5\) ft, base angles \(48^\circ\). Let's denote the equal sides as \(a\), base \(b = 3.5\) ft.

Step 1: Find the equal sides (\(a\))

Using the law of sines: \(\frac{a}{\sin(48^\circ)}=\frac{b}{\sin(180^\circ - 2\times48^\circ)}=\frac{b}{\sin(84^\circ)}\)
\(a=\frac{b\sin(48^\circ)}{\sin(84^\circ)}\)
\(\sin(48^\circ)\approx0.7431\), \(\sin(84^\circ)\approx0.9945\)
\(a=\frac{3.5\times0.7431}{0.9945}\approx\frac{2.60085}{0.9945}\approx2.615\) ft

Step 2: Calculate Perimeter (\(P\))

\(P = 2a + b=2\times2.615+3.5=5.23 + 3.5 = 8.73\) ft (Wait, no, wait: if base angles are \(48^\circ\), then the vertex angle is \(180 - 2\times48 = 84^\circ\), so the two equal sides are opposite the base angles. Wait, maybe I mixed up. Let's re - express: in a triangle, sides are opposite angles. Let the base be \(b = 3.5\) ft, opposite angle \(A = 84^\circ\), and the other two angles \(B = C=48^\circ\), opposite sides \(a = c\). Then by law of sines: \(\frac{a}{\sin(48^\circ)}=\frac{b}{\sin(84^\circ)}\), so \(a=\frac{b\sin(48^\circ)}{\sin(84^\circ)}\approx\frac{3.5\times0.7431}{0.9945}\approx2.615\) ft. Then perimeter \(P=a + a + b=2\times2.615+3.5 = 5.23+3.5 = 8.73\) ft? Wait, no, that seems small. Wait, maybe the given angle is the vertex angle. Let's assume vertex angle is \(48^\circ\), base \(b = 3.5\) ft, then base angles are \(\frac{180 - 48}{2}=66^\circ\). Then law of sines: \(\frac{a}{\sin(66^\circ)}=\frac{b}{\sin(48^\circ)}\)
\(\sin(66^\circ)\approx0.9135\), \(\sin(48^\circ)\approx0.7431\)
\(a=\frac{3.5\times0.9135}{0.7431}\approx\frac{3.19725}{0.7431}\approx4.30\) ft
Then perimeter \(P = 2\times4.30+3.5=8.6 + 3.5 = 12.10\) ft. There is ambiguity in the diagram, but assuming it's an isosceles triangle with base 3.5 ft and base angles \(48^\circ\) (the angle is at the base), let's recast.

Alternatively, maybe it's a right triangle? No, the diagram doesn't show a right angle. Wait, the original problem says "each triangle", so problem 30 is right - angled, 31 is right - angled, 32: let's assume it's a triangle with base 3.5 ft and height \(h\). If we assume it's an isosceles triangle with base 3.5 ft and vertex angle \(48^\circ\), then the height \(h=\frac{3.5}{2}\tan(66^\circ)\) (since base angles are \(66^\circ\)). \(\frac{3.5}{2}=1.75\), \(\tan(66^\circ)\approx2.246\), so \(h = 1.75\times2.246\approx3.9305\) ft. Area \(A=\frac{1}{2}\times3.5\times3.9305\approx6.88\) ft². Perimeter: the equal sides \(a=\frac{1.75}{\cos(66^\circ)}\approx\frac{1.75}{0.4067}\approx4.30\) ft, so perimeter \(P = 2\times4.30+3.5 = 12.10\) ft.

Final Answers

Problem 30:

• Perimeter: \(\approx23.03\) in
• Area: \(\approx20.80\) in²

Problem 31:

• Perimeter: \(\approx28.52\) cm
• Area: \(\approx23.39\) cm²

Problem 32:

• Assuming isosceles with base \(3.5\) ft and base angles \(48^\circ\) (or vertex angle \(84^\circ\)):
• Perimeter: \(\approx8.73\) ft (if base angles \(48^\circ\)) or \(\approx12.10\) ft (if vertex angle \(48^\circ\))
• Area: If vertex angle \(48^\circ\), \(A\approx6.88\) ft² (calculation as above)

(Note: The exact values depend on the correct interpretation of the triangle's angle placement, which is clearer for problems 30 and 31 as right - angled triangles.)