The hypotenuse of a right triangle is 10 inches. The length of the two legs are given by two consecutive even?
6,8,10 is multiple of 3,4,5 (x)^2+ (x+2)^2 = (10)^2 x^2+x^2+4x+4=100 2x^2+4x+4-100=0 2x^2+4x-96=0 2(x^2+2x-48)=0 (x^2+2x-48)=0 (x-6)(x+8)=0 x=6 or x=-8 x+2=8 reject -8 as length can't be negative check: 6sq+8sq=?10sq 36+64=?100 100=yes100 so, it works. solution set: {6,8}.
The hypotenuse of a right triangle is 58 inches. The lengths of the two legs are given by two consecutive even integers. find the 2 legs.?
One leg ............ x Other leg .......... x + 2 Hypotenuse .... 58 a² + b² = c² x² + ( x + 2 )² = 58² x² + x² + 4x + 4 = 3364 Combine like terms 2x² + 4x + 4 = 3362 Divide all the terms on both sides by 2 x² + 2x + 2 = 1682 Subtract 1682 from both sides to set the equation equal to 0 x² + 2x - 1680 = 0 Bah. It would be a waste of study time to factor this. "Guess" -- instead of factoring x ( x + 2 ) = 1680 Thinking: "The numbers "x" and "x+2" are so close to each other that 1680 is nearly a perfect square. So I wonder what the square root of 1680 is -- because it will be near "x" and "x + 2" Finding the square root of 1680 √1680 ≅ 41 So try 40 and 42 as the numbers because no one is going to actually factor that. (You can also use an on-line factoring calculator) How much is 40² plus 42² 1600 + 1764 3364 BINGO!! A lucky guess! Answer: The two legs are 40 inches and 42 inches Check a² + b² = c² 40² + 42² should equal 58² 1600 + 1764 should equal 3364 3364 does equal 3364 Check When the numbers are big, use an on-line factoring calculator https://www.google.com/#q=factoring+calc...
The hypotenuse of a right triangle is 17. The lengths of the two legs are whole numbers. What are they?
8²+15²=17²
The hypotenuse of a right triangle is 10 inches. The length of the two legs are given by two consecutive even?
x^2 + y^2 = 10^2 x^2 + y^2 = 100 y = x + 2 x^2 + (x + 2)^2 = 100 x^2 + x^2 + 4x + 4 = 100 2x^2 + 4x + 4 = 100 2*(x^2 + 2x + 2) = 100 x^2 + 2x + 2 = 50 x^2 + 2x = 48 x^2 + 2x - 48 = 0 a = 1 b = 2 c = -48 [-b +- sqrt{b^2 - 4*a*c}]/2*a [-2 +- sqrt{4 - 4*1*(-48)}]/2 [-2 +- sqrt{196}]/2 [-2 +- 14]/2 12/2 or -16/2 6 or -8 Lenghths are positive so x = 6 Therefore, the two legs are 6 and 8 Check: 6^ + 8^2 = 36 + 64 = 100 = 10^2 Correct.
b=7 and c=25if A=90º then [math]a=\sqrt{b^2+c^2}=25.96[/math], link: TrianCalif B=90º then imposible B must be ≤ 16º 15' 36.7369499", link: TrianCalif C=90º then [math]a=\sqrt{c^2-b^2}=24[/math], link: TrianCal
The hypotenuse of a right triangle is 20 inches. One of the legs is 4 inches more than the other.?
Name the one leg x so that the other will be x+4 Pythagoras' theorem: 20² = x² + (x+4)² Expand/simplify: 400 = x² + x² +16+4x+4x 400 = 2x² + 8x + 16 2x² + 8x - 384 = 0 Divide by 2: x² + 4x - 192 = 0 Factorise: ( x -12 )( x +16 ) = 0 so x = 12 or -16 the 16 given cannot be negative because we are talking about actual measurements so make it positive. So here you go 12 and 16
Let the side AB=8 cm and BC = 15 cm and the hypotenuse is AC[math]We know that since ABC is a right angled triangle,[/math][math]AB^2+BC^2=AC^2[/math][math]8^2+15^2=AC^2[/math][math]64+225=AC^2[/math][math]AC^2=289[/math][math]AC=\sqrt{289}=17[/math][math]Hypotenuse = 17 cm[/math]
If the non-hypotenuse sides of a right angled triangle have lengths of 5 and 12 units, (the hypotenuse length)^2 = 25 + 144 = 169 = 13^2, so the length of the hypotenuse is 13 units.It is worth remembering a few of the Pythagorean triples such as: 5, 12, 13 (as found above) and 3, 4, 5 which is the Pythagorean triple for the right angled triangle with the shortest sides.By enlarging these two triangles by various scale factors, other Pythagorean triples may be found, eg, 3, 4, 5, enlarged by scale factor 2 results in the right angled 6, 8, 10 triangle. Enlarged by scale factor 3, the 3, 4, 5, triangle becomes a 9, 12, 15 triangle. Similarly, a 5, 12, 13 triangle can be enlarged to 10, 24, 26, and 15, 36, 39, triangles.
From Pythagoras theorem for any right angled triangle:perpendicular^2 + base^2 = hypotenuse^2Hence, length of the other leg = sqrt(5^2 - 2^2)= sqrt(21)
The hypotenuse of a right triangle is 3 inches longer than the longer leg. The shorter leg is 3 inches shorter than the longer leg. What are the lengths of the sides of the triangle? What are the variables?Using Pythagorus’ theorem, for a right triangle, c = (a^2 + b^2)^0.5, let the long leg be x, the short leg is x – 3, and the hypotenus is x + 3, so:(x + 3)^2 = x^2 + (x – 3)^2 = 2x^2 – 6x + 9or:6x + 9 = x^2 – 6x + 9;so:6x = x^2 – 6x, or 12x = x^2, or x =- 12;So the hypotenus is x + 3 or 15 inches, the long arm is x or 12 inches, and the short arm is x – 3 or 9 inches.Since (12 – 3)^2 + (12)^2 = (12 + 3)^2 or9^2 + 12 ^2 = 15^2 = 144 = 81 = 225;the conditions that you specified and Pythagoras’ theorem are fulfilled.