Problem 1: Horizontal distance
Both points have \(y=5\), so the segment is horizontal. \(|9-2|=7\).
The distance is 7 units.
Problem 2: Vertical distance
Both points have \(x=-3\), so the segment is vertical. \(|6-(-4)|=10\).
The distance is 10 units.
Problem 3: Draw the legs
Horizontal leg: \(4-1=3\). Vertical leg: \(5-1=4\). Then \(d^{2}=3^{2}+4^{2}=9+16=25\).
\(AB=\)5 units.
Problem 4: Distance from the origin
The legs are 5 and 12. \(d^{2}=5^{2}+12^{2}=25+144=169\).
\(d=\)13 units.
Problem 5: Two points in Quadrant I
Legs: \(10-2=8\) and \(9-3=6\). \(d^{2}=64+36=100\).
\(d=\)10 units.
Problem 6: Across quadrants
Legs: \(4-(-2)=6\) and \(7-(-1)=8\). \(d^{2}=36+64=100\).
\(d=\)10 units.
Problem 7: Moving down and right
Legs: \(5-(-7)=12\) and \(|-2-3|=5\). \(d^{2}=144+25=169\).
\(d=\)13 units.
Problem 8: Distance that is not whole
Legs: \(5-1=4\) and \(6-1=5\). \(d^{2}=16+25=41\), so \(d=\sqrt{41}\).
\(\sqrt{41}\approx 6.40\), so the distance is about 6.4 units.
Problem 9: Larger legs
Legs: \(2-(-3)=5\) and \(|-6-4|=10\). \(d^{2}=25+100=125\).
\(d=\sqrt{125}\approx 11.18\), so the distance is about 11.2 units.
Problem 10: Use the distance formula
\(d=\sqrt{(7-(-8))^{2}+(5-(-3))^{2}}=\sqrt{15^{2}+8^{2}}=\sqrt{225+64}=\sqrt{289}\).
\(d=\)17 units.
Problem 11: Find the error
Liam added the legs. The distance is the hypotenuse, so the legs must be squared, added, and then the square root is taken.
\(d^{2}=3^{2}+4^{2}=25\), so \(d=5\). The distance is 5 units, not 7.
Problem 12: Which point is closer?
\(OP=\sqrt{5^{2}+(-6)^{2}}=\sqrt{61}\approx 7.8\). \(OQ=\sqrt{(-7)^{2}+3^{2}}=\sqrt{58}\approx 7.6\).
Since 58 is less than 61, point \(Q\) is closer to the origin.
Problem 13: Missing coordinate
The horizontal leg is \(7-1=6\). So \(6^{2}+(y-2)^{2}=10^{2}\), which gives \((y-2)^{2}=100-36=64\).
Then \(y-2=8\), so \(y=10\).
Problem 14: Perimeter of a grid triangle
Bottom side: 6. Right side: 8. Slanted side: \(\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\).
Perimeter: \(6+8+10=\)24 units.
Problem 15: Perimeter of a rhombus
Each side has legs 4 and 3, so each side is \(\sqrt{4^{2}+3^{2}}=\sqrt{25}=5\).
All four sides are 5, so the perimeter is \(4\cdot 5=\)20 units.
Problem 16: Perimeter with a slanted side
Horizontal side: \(2-(-4)=6\). Vertical side: \(3-0=3\). Slanted side: \(\sqrt{6^{2}+3^{2}}=\sqrt{45}\approx 6.71\).
Perimeter: \(6+3+6.71\approx\)15.7 units.
Problem 17: Is the triangle isosceles?
\(AB=|5-1|=4\). \(AC=\sqrt{2^{2}+4^{2}}=\sqrt{20}\). \(BC=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20}\).
Two sides have the same length \(\sqrt{20}\approx 4.5\), so the triangle is isosceles.
Problem 18: Is the triangle a right triangle?
\(AB^{2}=3^{2}+4^{2}=25\). \(BC^{2}=4^{2}+(-3)^{2}=25\). \(AC^{2}=7^{2}+1^{2}=50\).
Check the converse: \(AB^{2}+BC^{2}=25+25=50=AC^{2}\), so the triangle is a right triangle with the right angle at \(B\).
Problem 19: Library to park
Legs: \(8-2=6\) miles and \(11-3=8\) miles. \(d^{2}=36+64=100\).
They are 10 miles apart.
Problem 20: Drone flight
Legs: \(8-(-4)=12\) and \(2-(-3)=5\). \(d^{2}=144+25=169\).
The drone flies 13 yards.
Problem 21: Home to school
Legs: \(6-(-3)=9\) and \(|-10-2|=12\). \(d^{2}=81+144=225\), so \(d=15\) units.
Each unit is 0.1 mile: \(15\times 0.1=\)1.5 miles.
Problem 22: Fence for a garden
Sides: \(9\) ft, \(12\) ft and \(\sqrt{9^{2}+12^{2}}=\sqrt{225}=15\) ft. Perimeter: \(9+12+15=36\) ft.
Cost: \(36\times 3.50=\)\$126.
Problem 23: Jogging loop
Two sides are 5 blocks. The slanted side is \(\sqrt{5^{2}+5^{2}}=\sqrt{50}\approx 7.07\).
Loop: \(5+5+7.07\approx\)17.1 blocks.
Problem 24: Video game character
Legs: \(6-(-6)=12\) and \(3-(-2)=5\). \(d^{2}=144+25=169\), so \(d=13\) units.
Time: \(13: 2.6=\)5 seconds.
See every chapter on the Grade 8 math page.