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Distances & Areas

12 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0232 International

Problem 1 — Distances & Areas

The midpoint of segment ABAB where A(2,4)A(2, 4) and B(8,10)B(8, 10) is:

Show answer & worked solution
  1. A. (3,5)(3, 5)
  2. B. (4,6)(4, 6)
  3. C. (5,7)(5, 7)✓ correct
  4. D. (6,14)(6, 14)

M= ⁣(102, 142)=(5,7)M = \!\left(\dfrac{10}{2},\, \dfrac{14}{2}\right) = (5, 7).

Problem #0233 International

Problem 2 — Distances & Areas

The area of the rectangle with vertices A(0,0),B(5,0),C(5,3),D(0,3)A(0, 0), B(5, 0), C(5, 3), D(0, 3) is:

Show answer & worked solution
  1. A. 88
  2. B. 1010
  3. C. 1515✓ correct
  4. D. 3030

Width =5= 5, height =3= 3. Area =15= 15.

Problem #0231 International

Problem 3 — Distances & Areas

The distance between A(1,2)A(1, 2) and B(4,6)B(4, 6) is:

Show answer & worked solution
  1. A. 33
  2. B. 44
  3. C. 55✓ correct
  4. D. 77

d=(4−1)2+(6−2)2=9+16=5d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5.

Problem #0930 US SAT

Problem 4 — Distance Between Two Points

The distance between A=(1,2)A = (1, 2) and B=(4,6)B = (4, 6) is:

Show answer & worked solution
  1. A. 44
  2. B. 55✓ correct
  3. C. 77
  4. D. 25\sqrt{25}

AB=(4−1)2+(6−2)2=9+16=25=5AB = \sqrt{(4-1)^{2} + (6-2)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5.

Problem #0937 US SAT

Problem 5 — Area of a Triangle

The area of a triangle with base 1010 and height 66 is:

Show answer & worked solution
  1. A. 6060
  2. B. 3030✓ correct
  3. C. 1616
  4. D. 1515

A=12⋅10⋅6=30A = \dfrac{1}{2} \cdot 10 \cdot 6 = 30.

Problem #0237 International

Problem 6 — Distances & Areas

The points A(1,2)A(1, 2), B(3,6)B(3, 6), C(5,10)C(5, 10) are:

Show answer & worked solution
  1. A. collinear✓ correct
  2. B. vertices of a right triangle
  3. C. vertices of an isosceles triangle
  4. D. vertices of an equilateral triangle

Area =12 ∣1(6−10)+3(10−2)+5(2−6)∣=12 ∣−4+24−20∣=0= \dfrac{1}{2}\,|1(6 - 10) + 3(10 - 2) + 5(2 - 6)| = \dfrac{1}{2}\,|-4 + 24 - 20| = 0. Hence collinear.

Problem #0236 International

Problem 7 — Distances & Areas

The distance from the origin to the point P(−3,4)P(-3, 4) is:

Show answer & worked solution
  1. A. 11
  2. B. 55✓ correct
  3. C. 77
  4. D. 2525

d=(−3)2+42=25=5d = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5.

Problem #0238 International

Problem 8 — Distances & Areas

The perimeter of the triangle with vertices A(0,0)A(0, 0), B(3,0)B(3, 0), C(0,4)C(0, 4) is:

Show answer & worked solution
  1. A. 77
  2. B. 1010
  3. C. 1212✓ correct
  4. D. 2020

∣AB∣=3|AB| = 3, ∣AC∣=4|AC| = 4, ∣BC∣=9+16=5|BC| = \sqrt{9 + 16} = 5. Perimeter =3+4+5=12= 3 + 4 + 5 = 12.

Problem #0234 International

Problem 9 — Distances & Areas

A triangle has vertices A(0,0)A(0, 0), B(6,0)B(6, 0), C(0,4)C(0, 4). Its area equals:

Show answer & worked solution
  1. A. 66
  2. B. 1010
  3. C. 1212✓ correct
  4. D. 2424

Base =6= 6, height =4= 4. Area =12⋅6⋅4=12= \dfrac{1}{2} \cdot 6 \cdot 4 = 12.

Problem #0235 International

Problem 10 — Distances & Areas

The area of the triangle with vertices A(1,1)A(1, 1), B(4,5)B(4, 5), C(7,2)C(7, 2) is:

Show answer & worked solution
  1. A. 152\dfrac{15}{2}
  2. B. 99
  3. C. 212\dfrac{21}{2}✓ correct
  4. D. 1515

Area=12 ∣1(5−2)+4(2−1)+7(1−5)∣=12 ∣3+4−28∣=12⋅21=212\text{Area} = \dfrac{1}{2}\,|1(5 - 2) + 4(2 - 1) + 7(1 - 5)| = \dfrac{1}{2}\,|3 + 4 - 28| = \dfrac{1}{2} \cdot 21 = \dfrac{21}{2}.

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