An isosceles, right-angled triangle has a hypotenuse of 10cm. Calculate the length of the two shorter sides. (Hint: Call both shorter sides 'x').
Can someone please help me figure this one out? With working out. Thanks!!! xX
2007-02-24 15:39:00 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Very easy. asqrd + bsqrd = csqrd therefore 2x sqrd = 10sqrd
therefore 2xsqrd = 100 => xsqrd = 50, and x = 7.07.
2007-02-24 15:51:22 · answer #1 · answered by DuckyWucky 3 · 0⤊ 1⤋
Ok, this actually easier than it looks at first, you just have to Pythagorize it. The hypotenuse is 10, so c^2 is 100. The a^2 + b^2 = 100. Since the two are equal, you can change this to
2a^2 = 100
Divide both sides by two and square root to find a (or x), which is the length of each of the shorter sides. (Square root of 50)
2007-02-24 23:50:48 · answer #2 · answered by Greg Z 3 · 0⤊ 0⤋
If it is an isosceles right triangle you can use pythagorean theorem or the fact the the hypotenuse is equal to the leg times sqrt2
x(sqrt2) = 10
divide by sqrt2
x = 10/(sqrt2)
Rationalize the den by multiplying by (sqrt2)/(sqrt2)
Solution is 5sqrt2
2007-02-24 23:48:00 · answer #3 · answered by lizzie 3 · 0⤊ 0⤋
x^2+x^2=10^2
Hence 2 x^2=100
So x=Square root of 100/2=Square root of 50=7.1(Ans)
2007-02-24 23:48:53 · answer #4 · answered by Yako 2 · 0⤊ 0⤋