Area of Shapes 3
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Question 1 of 6
1. Question
Find the area of the Triangle- `\text(Area )=` (6.975) `cm^2`
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Area of a Triangle Formula
`\text(Area )=1/2 times``\text(base)``times``\text(height)`Identify the known lengths
`\text(base)=4.5``\text(height)=3.1`Solve for the area using the Area of a Triangle formula`\text(Area)` `=` `1/2 times``\text(base)``times``\text(height)` Area of a Triangle formula `=` `1/2 times ``4.5``times``3.1` Plug in the known lengths `=` `6.975 cm^2` The given measurements are in metres, so the area is measured as square metres`\text(Area)=6.975 cm^2` -
Question 2 of 6
2. Question
Find the area of the Square- `\text(Area )=` (25) `cm^2`
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Area of a Square Formula
`\text(Area )=``\text(side)``times``\text(side)`In a regular square, all sides are equalIdentify the known lengths
`\text(side)=5`Solve for the area using the Area of a Square formula`\text(Area)` `=` `\text(side)``times``\text(side)` Area of a Square formula `=` `5``times``5` Plug in the known lengths `=` `25` `\text(Area)=25 cm^2` -
Question 3 of 6
3. Question
Find the area of the Trapezium- `\text(Area ) =` (49.77) `cm^2`
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Area of a Trapezium Formula
`\text(Area)=1/2 times``\text(height)``times (``\text(base)_1``+``\text(base)_2``)`Identify the known lengths
`\text(height)=7.9``\text(base)_1=3.1``\text(base)_2=9.5`Solve for the area using the Area of a Trapezium formula`\text(Area)` `=` `1/2 times``\text(height)``times (``\text(base)_1``+``\text(base)_2``)` Area of a Trapezium formula `=` `1/2 times``7.9``times (``3.1``+``9.5``)` Plug in the known lengths `=` `49.77 cm^2` The given measurements are in centimetres, so the area is measured as square centimetres`\text(Area)=49.77 cm^2` -
Question 4 of 6
4. Question
The area of the trapezium below is: `226.8 m^2`
Find the missing length `? m`The answer should be in metres.Round your answer to one decimal place.- `\text(?) =` (10.5)`m`
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Incorrect
Area of a Trapezium Formula
`\text(Area)``=1/2 times``\text(height)``times (``\text(base)_1``+``\text(base)_2``)`Identify the known lengths
`\text(Area)=226.8 m^2``\text(base)_1=? m``\text(base)_2=21.9 m``\text(height)=14 m`Substitute the known values to the Area of a Trapezium formula$$\color{#9a00c7}{\text{Area}}$$ `=` `1/2 times``\text(height)``times (``\text(base)_1``+``\text(base)_2``)` Area of a Trapezium formula $$\color{#9a00c7}{226.8}$$ `=` `1/2 times``14``times (``\text(base)_1``+``21.9``)` Plug in the known lengths `226.8` `=` `(14(\text(base)_1 + 21.9))/2` Change to quotient form Finally, get `\text(base)_1` on one side of the equation to get its value`226.8` `=` `(14(\text(base)_1 + 21.9))/2` `226.8``times2` `=` `(14(\text(base)_1 + 21.9))/2``times2` Multiply both sides by `2` `226.8``times2` `=` `(2xx[14(\text(base)_1 + 21.9)])/2` `453.6` `=` `14(\text(base)_1 + 21.9)` The coefficient `2/2` cancels out `453.6``divide14` `=` `14(\text(base)_1 + 21.9)``divide14` Divide both sides by `14` `453.6``divide14` `=` `14``(\text(base)_1 + 21.9)``divide14` `times14divide14` cancels out `32.4` `=` `\text(base)_1 + 21.9` `32.4` `-21.9` `=` `\text(base)_1 + 21.9` `-21.9` Subtract `21.9` from both sides `32.4` `-21.9` `=` `\text(base)_1``+21.9` `-21.9` `21.9-21.9` cancels out `10.5` `=` `\text(base)_1` `\text(base)_1` `=` `10.5 m` `\text(?)` `=` `10.5 m` `\text(?)=10.5 m` -
Question 5 of 6
5. Question
Find the area of the Kite- `\text(Area )=` (10.5) `mm^2`
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Area of a Kite Formula
`\text(Area)= 1/2 times``\text(base)``times``\text(height)`Identify the known lengths
`\text(base)=3``\text(height)=7`Solve for the area using the Area of a Kite formula`\text(Area)` `=` `1/2 times``\text(base)``times``\text(height)` Area of a Kite formula `=` `1/2 times ``3``times``7` Plug in the known lengths `=` `10.5 mm^2` The given measurements are in millimetres, so the area is measured as square millimetres`\text(Area)=10.5 mm^2` -
Question 6 of 6
6. Question
Find the missing measurement of the Parallelogram- `h =` (33) `km`
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Area of a Parallelogram Formula
`\text(Area)``=``\text(base)``xx \text(height)`Identify the known lengths
`\text(Area)=1650 km^2``\text(base)=50 km`To find the missing measurement, `h`, solve using the Parallelogram formula and get `h` by itself.Substitute the known values to the Area of a Parallelogram formula`\text(Area)` `=` `\text(base)``times \text(height)` `1650` `=` `50``timesh` `1650` `=` `50h` Finally, divide both sides of the equation by `50` to get the value of `h``1650``divide50` `=` `50h``divide50` `1650``divide50` `=` `50``h``divide50` `times50divide50` cancels out `33` `=` `h` `h` `=` `33 km` `h=33 km`