Volume of a triangular pyramid is the product of its base area, A, and height, h, multiplied by 13. To find the value of x, use the Rational Zero Theorem.
5 inches, 9 inches, and 28 inches
Practice makes perfect
Volume of a triangular pyramid is the product of its base area, A, and height, h, multiplied by 13. With this, an equation for the volume of the triangular pyramid can be written.
V=1/3Ah
Notice that its base is right triangle. Area of a right triangle is the product of its legs divided by 2. Let's find it!
A=x(2x-1)/2Now that we found the area of its base, we will substitute the values in the equation and write it in standard form.
Next, we will find x using the Rational Zero Theorem. Any rational zero will have the form ± p q, where p is a factor of the constant term, - 1260, and q is a factor of the leading coefficient, 10.
p: & ± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 7, ± 9, ± 10,
& ± 12, ± 14, ± 15, ± 18, ± 20, ± 21, ± 28,
& ± 30, ± 35, ± 36, ± 42, ± 45, ± 60, ± 63,
& ± 70, ± 84, ± 90, ± 105, ± 126, ± 140, ± 180,
& ± 210, ± 252, ± 315, ± 420, ± 630, ± 1260
q: & ± 1, ± 2, ± 5, ± 10
Since length can only be positive, we only need to check the possible positive zeros. Let's examine the changes in signs of the coefficients to state the number of positive real zeros.
There is only one sign change for the coefficients of the polynomial. According to Descartes' Rule of Signs, the function has one positive real zero. Now, we can check the possible combinations of p q.
x
10x^3+x^2-3x-1260
10x^3+x^2-3x-1260=0
1/1= 1
10( 1)^3+ 1^2-3( 1)-1260
-1252≠ 0 *
2/1= 2
10( 2)^3+ 2^2-3( 2)-1260
-1182≠ 0 *
3/1= 3
10( 3)^3+ 3^2-3( 3)-1260
-990≠ 0 *
4/1= 4
10( 4)^3+ 4^2-3( 4)-1260
-616≠ 0 *
5/1= 5
10( 5)^3+ 5^2-3( 5)-1260
0 = 0 ✓
Since x=5 satisfied the equation and there is only one positive real zero, we can say that the value of x is 5 inches. With this, we can find the dimensions of the solid.
x:& 5=5
2x-1:& 2(5)-1=9
5x+3:& 5(5)+3=28
