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Solving Literal Equations with Geometry Formulas

Authored by Lisaira Daniels

Mathematics

9th - 10th Grade

CCSS covered

Used 42+ times

Solving Literal Equations with Geometry Formulas
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10 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Area of a square is  A=s2A=s^2  . Which equation will find the side of a square?

 s=2As=2A  

 s=A2s=A^2  

 s=As=\sqrt{A}  

 s=2As=\sqrt{2A}  

Tags

CCSS.HSA.CED.A.4

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 Area of a rectangle is  A=lwA=lw . Which equation will find the width of the rectangle?

 w=Alw=Al  

 w=Alw=\frac{A}{l}  

 w=lAw=\frac{l}{A}  

 l=Awl=\frac{A}{w}  

Tags

CCSS.HSA.CED.A.4

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

  Area of a trapezoid is  A=12h(b1+b2)A=\frac{1}{2}h\left(b_1+b_2\right) . What is the equation to find the height of a trapezoid?

 h=2A(b1 +b2)h=\frac{2A}{\left(b_{1\ }+b_2\right)}  

 h=A2(b1+b2)h=\frac{A}{2\left(b_1+b_2\right)}  

 h=(b1+b2)2Ah=\frac{\left(b_1+b_2\right)}{2A}  

 h=(b1+b2)2Ah=\frac{\left(b_1+b_2\right)}{2A}  

Tags

CCSS.HSA.CED.A.4

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

   Area of a triangle is  A=12bh.A=\frac{1}{2}bh. Which equation will find the base of the triangle?

 b=2Ahb=\frac{2A}{h}  

 h=2Abh=\frac{2A}{b}  

 b=h2Ab=\frac{h}{2A}  

 b=A2hb=\frac{A}{2h}  

Tags

CCSS.HSA.CED.A.4

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Area of a circle is A=πr2.A=\pi r^2. Which equation will find the radius of the circle?

 r=Aπr=\sqrt{A\pi}  

 r=Aπr=\sqrt{\frac{A}{\pi}}  

 r=A2πr=A^2\pi  

 r=πAr=\pi\sqrt{A}  

Tags

CCSS.HSA.CED.A.4

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Volume of a rectangular solid is V=lwh.V=lwh.  Which equation will find the height of the rectangular solid?

 h=Vlwh=Vlw  

 h=Vlwh=\frac{V}{lw}  

 h=lwVh=\frac{lw}{V}  

 l=Vwhl=\frac{V}{wh}  

Tags

CCSS.HSA.CED.A.4

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Volume of a right circular cylinder is  V=πr2h.V=\pi r^2h.  Which equation will find the height of a right circular cylinder?

 h=Vπr2h=\frac{V\pi}{r^2}  

 h=r2Vπh=\frac{r^2}{V\pi}  

 h=Vπr2h=\frac{V}{\pi r^2}  

 h=πr2Vh=\frac{\pi r^2}{V}  

Tags

CCSS.HSA.CED.A.4

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