12 multiple-choice questions, progressively harder.
A triangular skylight has sides 13 dm, 18 dm, and 25 dm. What is its area to the nearest tenth of a square decimeter?
Solution
Correct answer: B
The sides are positive and the shorter pair totals 31 dm, more than 25 dm. The semiperimeter is 28 dm. The differences are 15 dm, 10 dm, and 3 dm.
A triangle's two shorter sides measure 21 mm and 27 mm. The third side is 47.6 mm. What is the smallest of the three differences from the semiperimeter?
Correct answer: D
The positive shorter lengths total 48 mm, more than 47.6 mm, so the triangle is valid. Subtracting the largest side gives the smallest difference.
A triangle has semiperimeter 27 units. Two of its sides measure 16 and 20 units. Which is the remaining side length?
Correct answer: C
The full perimeter is 54 units, so subtract the two known lengths.
The lengths are positive and 16 plus 18 exceeds 20.
The positive side lengths of a triangle are a,b,ca,b,ca,b,c. Which expression for 2(s−a)2(s-a)2(s−a) uses just the side lengths?
Correct answer: A
Start with two copies of the semiperimeter and subtract two copies of aaa.
Two valid triangular panels each have perimeter 56 cm.
Panel A has sides 18 cm, 18 cm, and 20 cm.
Panel B has sides 15 cm, 19 cm, and 22 cm.
Which comparison of their areas is correct?
The positive shorter sides total 36 cm in A and 34 cm in B, exceeding the longest side in each case. Both semiperimeters are 28 cm. In square centimeters, compare the squares of the areas.
The areas are positive, so A has the greater area.
The triangle shown has sides 13 m, 14 m, and 23 m. What is the exact perpendicular height to the 23 m side?
The positive lengths pass the strict check because 27 exceeds 23. The semiperimeter is 25 m. Its differences from the sides are 12 m, 11 m, and 2 m.
Double the area and divide by the chosen base.
A triangular mat has sides 60 cm, 0.8 m, and 1.1 m. What is the area to the nearest hundredth of a square meter?
Convert 60 cm to 0.6 m. The positive shorter lengths total 1.4 m, more than 1.1 m. The semiperimeter is 1.25 m, giving differences of 0.65 m, 0.45 m, and 0.15 m.
A valid triangle has side lengths 7, 9, and 14 units. Angle CCC is opposite the 14-unit side.
The Law of Cosines gives 2abcosC=−662ab\cos C=-662abcosC=−66 when lengths are expressed in these units. Which expression equals 16K216K^216K2?
The sides are positive and 16 exceeds 14. Squaring the sine area expression and multiplying by 16 gives the following result.
Replace sine squared with one minus cosine squared. The sign of the cosine disappears when it is squared.
Two identical triangular pennants each have sides 16 cm, 19 cm, and 27 cm. Find the sum of their areas to the nearest square centimeter.
Each triangle has positive sides, and 35 exceeds 27. Each has semiperimeter 31 cm and differences of 15 cm, 12 cm, and 4 cm.
Double the exact area and round the total.
A valid triangle has K2=2646K^2=2646K2=2646 when its area is measured in square centimeters. Which exact area results?
The area is positive, so take the positive square root. Factor 2646 as 441 times 6.
A valid triangle has side lengths 18 cm, 26 cm, and 31 cm. What is the difference between the largest and smallest of the three side differences from the semiperimeter?
The positive shorter pair totals 44 cm, exceeding 31 cm. The smallest side gives the largest difference from sss, and the largest side gives the smallest difference from sss. The two copies of sss cancel when these differences are subtracted.
Two students name the same three sides of a valid triangle. The second student interchanges aaa and ccc. How does this change the area from Heron's formula?
Interchanging the two labels leaves sss unchanged. The factors s−as-as−a and s−cs-cs−c trade places without changing the squared area.
The positive square root therefore gives the same area.
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