Mixing
Working out the value of $x$ on two triangle with the same area in the form $a+sqrt b$
$begingroup$
Here are two triangles T1 and T2.
The lengths of the sides are in centimeters.
The area of triangle T1 is equal to the area of triangle T2.
Work out the value of x, giving your answer in the form $a+ sqrt b$ where $a$ and $b$ are integers.
I'm in maths set 5 and this is the final question so i have no idea how to solve it. Any help would be greatly appreciated.
algebra-precalculus geometry
edited Jan 25 at 16:14
André 3000
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
$endgroup$
add a comment |
$begingroup$
Here are two triangles T1 and T2.
The lengths of the sides are in centimeters.
The area of triangle T1 is equal to the area of triangle T2.
Work out the value of x, giving your answer in the form $a+ sqrt b$ where $a$ and $b$ are integers.
I'm in maths set 5 and this is the final question so i have no idea how to solve it. Any help would be greatly appreciated.
algebra-precalculus geometry
edited Jan 25 at 16:14
André 3000
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
$endgroup$
add a comment |
$begingroup$
Here are two triangles T1 and T2.
The lengths of the sides are in centimeters.
The area of triangle T1 is equal to the area of triangle T2.
Work out the value of x, giving your answer in the form $a+ sqrt b$ where $a$ and $b$ are integers.
I'm in maths set 5 and this is the final question so i have no idea how to solve it. Any help would be greatly appreciated.
algebra-precalculus geometry
edited Jan 25 at 16:14
André 3000
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
$endgroup$
Here are two triangles T1 and T2.
The lengths of the sides are in centimeters.
The area of triangle T1 is equal to the area of triangle T2.
Work out the value of x, giving your answer in the form $a+ sqrt b$ where $a$ and $b$ are integers.
I'm in maths set 5 and this is the final question so i have no idea how to solve it. Any help would be greatly appreciated.
algebra-precalculus geometry
algebra-precalculus geometry
edited Jan 25 at 16:14
André 3000
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
edited Jan 25 at 16:14
André 3000
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
edited Jan 25 at 16:14
André 3000
12.8k22243
edited Jan 25 at 16:14
André 3000
12.8k22243
edited Jan 25 at 16:14
André 3000
12.8k22243
12.8k22243
asked Apr 28 '14 at 17:40
user146422user146422
61
asked Apr 28 '14 at 17:40
user146422user146422
61
asked Apr 28 '14 at 17:40
user146422user146422
61
61
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
My monitor is fuzzy, and I cannot read upside down. So I will assume that the legs of the right triangle are $1+x$ and $x-2$. You can use the same strategy if the expressions are different.
The right triangle has area $frac{1}{2}(1+x)(x-2)$.
As to the isosceles triangle, it has area $frac{1}{2}x^2sin(30^circ)$.
Set these areas equal to each other. We get
$$frac{(1+x)(x-2)}{2}=frac{x^2}{4}.$$
Multiply through by $4$. We get $2x^2-2x-4=x^2$, that is, $x^2-2x-4=0$. Solve, using the Quadratic Formula, and discarding the negative root. We get
$x=frac{2+sqrt{20}}{2}$, which can be simplified to the desired form.
edited Jan 25 at 12:29
Namaste
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
My monitor is fuzzy, and I cannot read upside down. So I will assume that the legs of the right triangle are $1+x$ and $x-2$. You can use the same strategy if the expressions are different.
The right triangle has area $frac{1}{2}(1+x)(x-2)$.
As to the isosceles triangle, it has area $frac{1}{2}x^2sin(30^circ)$.
Set these areas equal to each other. We get
$$frac{(1+x)(x-2)}{2}=frac{x^2}{4}.$$
Multiply through by $4$. We get $2x^2-2x-4=x^2$, that is, $x^2-2x-4=0$. Solve, using the Quadratic Formula, and discarding the negative root. We get
$x=frac{2+sqrt{20}}{2}$, which can be simplified to the desired form.
edited Jan 25 at 12:29
Namaste
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
$endgroup$
add a comment |
$begingroup$
My monitor is fuzzy, and I cannot read upside down. So I will assume that the legs of the right triangle are $1+x$ and $x-2$. You can use the same strategy if the expressions are different.
The right triangle has area $frac{1}{2}(1+x)(x-2)$.
As to the isosceles triangle, it has area $frac{1}{2}x^2sin(30^circ)$.
Set these areas equal to each other. We get
$$frac{(1+x)(x-2)}{2}=frac{x^2}{4}.$$
Multiply through by $4$. We get $2x^2-2x-4=x^2$, that is, $x^2-2x-4=0$. Solve, using the Quadratic Formula, and discarding the negative root. We get
$x=frac{2+sqrt{20}}{2}$, which can be simplified to the desired form.
edited Jan 25 at 12:29
Namaste
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
$endgroup$
add a comment |
$begingroup$
My monitor is fuzzy, and I cannot read upside down. So I will assume that the legs of the right triangle are $1+x$ and $x-2$. You can use the same strategy if the expressions are different.
The right triangle has area $frac{1}{2}(1+x)(x-2)$.
As to the isosceles triangle, it has area $frac{1}{2}x^2sin(30^circ)$.
Set these areas equal to each other. We get
$$frac{(1+x)(x-2)}{2}=frac{x^2}{4}.$$
Multiply through by $4$. We get $2x^2-2x-4=x^2$, that is, $x^2-2x-4=0$. Solve, using the Quadratic Formula, and discarding the negative root. We get
$x=frac{2+sqrt{20}}{2}$, which can be simplified to the desired form.
edited Jan 25 at 12:29
Namaste
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
$endgroup$
My monitor is fuzzy, and I cannot read upside down. So I will assume that the legs of the right triangle are $1+x$ and $x-2$. You can use the same strategy if the expressions are different.
The right triangle has area $frac{1}{2}(1+x)(x-2)$.
As to the isosceles triangle, it has area $frac{1}{2}x^2sin(30^circ)$.
Set these areas equal to each other. We get
$$frac{(1+x)(x-2)}{2}=frac{x^2}{4}.$$
Multiply through by $4$. We get $2x^2-2x-4=x^2$, that is, $x^2-2x-4=0$. Solve, using the Quadratic Formula, and discarding the negative root. We get
$x=frac{2+sqrt{20}}{2}$, which can be simplified to the desired form.
edited Jan 25 at 12:29
Namaste
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
edited Jan 25 at 12:29
Namaste
1
edited Jan 25 at 12:29
Namaste
1
edited Jan 25 at 12:29
Namaste
1
1
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
answered Apr 28 '14 at 19:27
André NicolasAndré Nicolas
454k36432818
454k36432818
add a comment |
add a comment |
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