Calculate the area of trapezoid.
$begingroup$
In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $angle BAD=theta,$
$BD=h$,
$angle ABD=90^circ$
$angle CBD=90^circ-theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$cos(90-theta)=frac{h^2+2^2-2^2}{2times 2times h}=frac{h}{4}$
$sintheta=frac{h}{4}..(1)$
In right triangle $ABD,sin theta=frac{h}{sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2sqrt3$
Area of $ABCD=frac{1}{2}times 2times h+frac{1}{2}times 2times 2times sin2theta=3sqrt3$
But the answer given is $2sqrt2(sqrt{5}+1)$
geometry
asked Dec 18 '18 at 14:46
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $angle BAD=theta,$
$BD=h$,
$angle ABD=90^circ$
$angle CBD=90^circ-theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$cos(90-theta)=frac{h^2+2^2-2^2}{2times 2times h}=frac{h}{4}$
$sintheta=frac{h}{4}..(1)$
In right triangle $ABD,sin theta=frac{h}{sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2sqrt3$
Area of $ABCD=frac{1}{2}times 2times h+frac{1}{2}times 2times 2times sin2theta=3sqrt3$
But the answer given is $2sqrt2(sqrt{5}+1)$
geometry
asked Dec 18 '18 at 14:46
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $angle BAD=theta,$
$BD=h$,
$angle ABD=90^circ$
$angle CBD=90^circ-theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$cos(90-theta)=frac{h^2+2^2-2^2}{2times 2times h}=frac{h}{4}$
$sintheta=frac{h}{4}..(1)$
In right triangle $ABD,sin theta=frac{h}{sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2sqrt3$
Area of $ABCD=frac{1}{2}times 2times h+frac{1}{2}times 2times 2times sin2theta=3sqrt3$
But the answer given is $2sqrt2(sqrt{5}+1)$
geometry
asked Dec 18 '18 at 14:46
user984325user984325
246112
$endgroup$
In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $angle BAD=theta,$
$BD=h$,
$angle ABD=90^circ$
$angle CBD=90^circ-theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$cos(90-theta)=frac{h^2+2^2-2^2}{2times 2times h}=frac{h}{4}$
$sintheta=frac{h}{4}..(1)$
In right triangle $ABD,sin theta=frac{h}{sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2sqrt3$
Area of $ABCD=frac{1}{2}times 2times h+frac{1}{2}times 2times 2times sin2theta=3sqrt3$
But the answer given is $2sqrt2(sqrt{5}+1)$
geometry
geometry
asked Dec 18 '18 at 14:46
user984325user984325
246112
asked Dec 18 '18 at 14:46
user984325user984325
246112
asked Dec 18 '18 at 14:46
user984325user984325
246112
asked Dec 18 '18 at 14:46
user984325user984325
246112
asked Dec 18 '18 at 14:46
user984325user984325
246112
246112
add a comment |
add a comment |
2 Answers
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$begingroup$
I've got $3sqrt 3$ using a slightly different method. $angle BAD=theta$, $angle ABD =90^circ$ means $angle CBD=angle BDA=90^circ-theta$. Since the trapezoid is isosceles, $angle CDA=theta$, and you can get $angle CDB=2theta -90^circ$. Since $BC=AB=CD$ you get $angle CDB=angle CDB$ or $$2theta-90^circ=90^circ-theta$$ so $theta=60^circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2ABcos 60^circ=4$, and the height $h=ABsin60^circ=sqrt 3$. Therefore the area is $$frac12 (BC+AD)cdot h=frac12 6sqrt3=3sqrt 3$$
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
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$begingroup$
you are correct. The answer should be $3sqrt{3}$
answered Dec 18 '18 at 14:51
william122william122
52412
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
I've got $3sqrt 3$ using a slightly different method. $angle BAD=theta$, $angle ABD =90^circ$ means $angle CBD=angle BDA=90^circ-theta$. Since the trapezoid is isosceles, $angle CDA=theta$, and you can get $angle CDB=2theta -90^circ$. Since $BC=AB=CD$ you get $angle CDB=angle CDB$ or $$2theta-90^circ=90^circ-theta$$ so $theta=60^circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2ABcos 60^circ=4$, and the height $h=ABsin60^circ=sqrt 3$. Therefore the area is $$frac12 (BC+AD)cdot h=frac12 6sqrt3=3sqrt 3$$
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
I've got $3sqrt 3$ using a slightly different method. $angle BAD=theta$, $angle ABD =90^circ$ means $angle CBD=angle BDA=90^circ-theta$. Since the trapezoid is isosceles, $angle CDA=theta$, and you can get $angle CDB=2theta -90^circ$. Since $BC=AB=CD$ you get $angle CDB=angle CDB$ or $$2theta-90^circ=90^circ-theta$$ so $theta=60^circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2ABcos 60^circ=4$, and the height $h=ABsin60^circ=sqrt 3$. Therefore the area is $$frac12 (BC+AD)cdot h=frac12 6sqrt3=3sqrt 3$$
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
I've got $3sqrt 3$ using a slightly different method. $angle BAD=theta$, $angle ABD =90^circ$ means $angle CBD=angle BDA=90^circ-theta$. Since the trapezoid is isosceles, $angle CDA=theta$, and you can get $angle CDB=2theta -90^circ$. Since $BC=AB=CD$ you get $angle CDB=angle CDB$ or $$2theta-90^circ=90^circ-theta$$ so $theta=60^circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2ABcos 60^circ=4$, and the height $h=ABsin60^circ=sqrt 3$. Therefore the area is $$frac12 (BC+AD)cdot h=frac12 6sqrt3=3sqrt 3$$
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
$endgroup$
I've got $3sqrt 3$ using a slightly different method. $angle BAD=theta$, $angle ABD =90^circ$ means $angle CBD=angle BDA=90^circ-theta$. Since the trapezoid is isosceles, $angle CDA=theta$, and you can get $angle CDB=2theta -90^circ$. Since $BC=AB=CD$ you get $angle CDB=angle CDB$ or $$2theta-90^circ=90^circ-theta$$ so $theta=60^circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2ABcos 60^circ=4$, and the height $h=ABsin60^circ=sqrt 3$. Therefore the area is $$frac12 (BC+AD)cdot h=frac12 6sqrt3=3sqrt 3$$
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
answered Dec 18 '18 at 15:15
AndreiAndrei
11.7k21026
11.7k21026
add a comment |
add a comment |
$begingroup$
you are correct. The answer should be $3sqrt{3}$
answered Dec 18 '18 at 14:51
william122william122
52412
$endgroup$
add a comment |
$begingroup$
you are correct. The answer should be $3sqrt{3}$
answered Dec 18 '18 at 14:51
william122william122
52412
$endgroup$
add a comment |
$begingroup$
you are correct. The answer should be $3sqrt{3}$
answered Dec 18 '18 at 14:51
william122william122
52412
$endgroup$
you are correct. The answer should be $3sqrt{3}$
answered Dec 18 '18 at 14:51
william122william122
52412
answered Dec 18 '18 at 14:51
william122william122
52412
answered Dec 18 '18 at 14:51
william122william122
52412
answered Dec 18 '18 at 14:51
william122william122
52412
52412
add a comment |
add a comment |
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