Answer :
To find the height and base of the triangle given the conditions, we can follow these steps:
1. Define Variables and Relationships:
- Let [tex]\( h \)[/tex] be the height of the triangle in centimeters.
- According to the problem, the base [tex]\( b \)[/tex] of the triangle is twelve more than twice its height. Therefore, [tex]\( b = 2h + 12 \)[/tex].
2. Express the Area in Terms of the Height:
- The formula for the area [tex]\( A \)[/tex] of a triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
- Given that the area is 45 square centimeters, we can set up the equation:
[tex]\[ \frac{1}{2} \times b \times h = 45 \][/tex]
3. Substitute the Expression for the Base into the Area Equation:
- Substitute [tex]\( b = 2h + 12 \)[/tex] into the area equation:
[tex]\[ \frac{1}{2} \times (2h + 12) \times h = 45 \][/tex]
4. Simplify and Solve for [tex]\( h \)[/tex]:
- Distribute and solve for [tex]\( h \)[/tex]:
[tex]\[ \frac{1}{2} \times (2h^2 + 12h) = 45 \][/tex]
[tex]\[ h^2 + 6h = 45 \][/tex]
[tex]\[ h^2 + 6h - 45 = 0 \][/tex]
- This is a quadratic equation in the form [tex]\( ah^2 + bh + c = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -45 \)[/tex].
5. Use the Quadratic Formula to Solve for [tex]\( h \)[/tex]:
- The quadratic formula is given by:
[tex]\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
- Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -45 \)[/tex]:
[tex]\[ h = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \][/tex]
[tex]\[ h = \frac{-6 \pm \sqrt{36 + 180}}{2} \][/tex]
[tex]\[ h = \frac{-6 \pm \sqrt{216}}{2} \][/tex]
[tex]\[ h = \frac{-6 \pm 6\sqrt{6}}{2} \][/tex]
[tex]\[ h = -3 \pm 3\sqrt{6} \][/tex]
- Since height cannot be negative, we take the positive value:
[tex]\[ h = 3\sqrt{6} - 3 \][/tex]
6. Calculate the Height:
- The simplified value of [tex]\( 3\sqrt{6} - 3 \)[/tex] in approximate form:
[tex]\[ h \approx 4.348 \][/tex]
7. Calculate the Base:
- Using [tex]\( b = 2h + 12 \)[/tex]:
[tex]\[ b = 2(4.348) + 12 \][/tex]
[tex]\[ b = 8.696 + 12 \][/tex]
[tex]\[ b \approx 20.696 \][/tex]
Therefore, the height of the triangle is approximately [tex]\( 4.348 \)[/tex] cm and the base of the triangle is approximately [tex]\( 20.696 \)[/tex] cm.
1. Define Variables and Relationships:
- Let [tex]\( h \)[/tex] be the height of the triangle in centimeters.
- According to the problem, the base [tex]\( b \)[/tex] of the triangle is twelve more than twice its height. Therefore, [tex]\( b = 2h + 12 \)[/tex].
2. Express the Area in Terms of the Height:
- The formula for the area [tex]\( A \)[/tex] of a triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
- Given that the area is 45 square centimeters, we can set up the equation:
[tex]\[ \frac{1}{2} \times b \times h = 45 \][/tex]
3. Substitute the Expression for the Base into the Area Equation:
- Substitute [tex]\( b = 2h + 12 \)[/tex] into the area equation:
[tex]\[ \frac{1}{2} \times (2h + 12) \times h = 45 \][/tex]
4. Simplify and Solve for [tex]\( h \)[/tex]:
- Distribute and solve for [tex]\( h \)[/tex]:
[tex]\[ \frac{1}{2} \times (2h^2 + 12h) = 45 \][/tex]
[tex]\[ h^2 + 6h = 45 \][/tex]
[tex]\[ h^2 + 6h - 45 = 0 \][/tex]
- This is a quadratic equation in the form [tex]\( ah^2 + bh + c = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -45 \)[/tex].
5. Use the Quadratic Formula to Solve for [tex]\( h \)[/tex]:
- The quadratic formula is given by:
[tex]\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
- Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -45 \)[/tex]:
[tex]\[ h = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \][/tex]
[tex]\[ h = \frac{-6 \pm \sqrt{36 + 180}}{2} \][/tex]
[tex]\[ h = \frac{-6 \pm \sqrt{216}}{2} \][/tex]
[tex]\[ h = \frac{-6 \pm 6\sqrt{6}}{2} \][/tex]
[tex]\[ h = -3 \pm 3\sqrt{6} \][/tex]
- Since height cannot be negative, we take the positive value:
[tex]\[ h = 3\sqrt{6} - 3 \][/tex]
6. Calculate the Height:
- The simplified value of [tex]\( 3\sqrt{6} - 3 \)[/tex] in approximate form:
[tex]\[ h \approx 4.348 \][/tex]
7. Calculate the Base:
- Using [tex]\( b = 2h + 12 \)[/tex]:
[tex]\[ b = 2(4.348) + 12 \][/tex]
[tex]\[ b = 8.696 + 12 \][/tex]
[tex]\[ b \approx 20.696 \][/tex]
Therefore, the height of the triangle is approximately [tex]\( 4.348 \)[/tex] cm and the base of the triangle is approximately [tex]\( 20.696 \)[/tex] cm.