Find the missing step in the following reasoning in order to find the area of the trapezoid.
| Step | |
| 1 | Using the Pythagoras theorem we have: BF= \sqrt{5^2-3^2}=4 |
| 2 | .... |
| 3 | S=\dfrac{1}{2}\left(AB+DC\right).FC = \dfrac{1}{2}\left(14+5\right).3 |
| 4 | The area of the trapezoid is 28.5. |
To compute the area of the trapezoid, we use the formula in step 3. We can see that the only missing value is AB, so we must find it in step 2 using Step 1.
| Step | |
| 1 | Using the Pythagoras theorem we have: BF= \sqrt{5^2-3^2}=4 |
| 2 | AB=AE + EF + FB = 14 |
| 3 | S=\dfrac{1}{2}\left(AB+DC\right).FC = \dfrac{1}{2}\left(14+5\right).3 |
| 4 | The area of the trapezoid is 28.5. |
Find the missing step in the following reasoning in order to determine a of the parallelogram
| Step | |
| 1 | Since \overline{AB} and \overline{DC} are parallel, \widehat{C} and \widehat{DAB} are corresponding angles. |
| 2 | .... |
| 3 | \widehat{DAB} and a are supplementary. |
| 4 | a=180^\circ - 70^\circ = 110^\circ |
We know that corresponding angles are congruent. Using Step 1, we have:
\widehat{BAD} = 70^\circ
Find the missing step in the following reasoning in order to find a.
| Step | |
| 1 | \widehat{BDC} and the 150 degree angle are supplementary. |
| 2 | .... |
| 3 | 30^\circ+a + 90^\circ + 90^\circ=360^\circ |
| 4 | a=150^\circ |
Since \widehat{BCD} and a 150 degree angle are supplementary, we can write:
\widehat{BCD}= 180^\circ - 150^\circ =30^\circ
Find the missing step in the following reasoning in order to find the perimeter of the kite.
| Step | |
| 1 | The perimeter of the kite equals: P=2AD+2CD |
| 2 | ..... |
| 3 | CD= 2OD= 6 |
| 4 | P=2\left(9\right)+2\left(6\right)=30 |
To find the perimeter of the kite, we need to find AD and DC. Since AD is given, we only need to find DC by using the sine formula:
\dfrac{1}{2}=\sin\left(30^\circ\right)=\dfrac{DO}{CD}
Find the missing step in the following reasoning in order to find the area of the rhombus.
| Step | |
| 1 | The area of the rhombus equals: S=AC \times CD |
| 2 | BD=2OB=10 |
| 3 | .... |
| 4 | AC=2AO=24 |
| 5 | S=24 \times 10 = 240 |
To find the area of the rhombus, we need to find AC. Step 4 implies that we need to prove that AO =12. Therefore, we can write:
Using the Pythagoras theorem we have:
AO=\sqrt{169-25}=12
Find the missing step in the following reasoning in order to find the area of the parallelogram.
| Step | |
| 1 | The area of the parallelogram equals: S=AH \times CD |
| 2 | ..... |
| 3 | DC= DH+CH = 3+4= 7 |
| 4 | S=7 \times 4=28 |
If we find DH, then the area of the parallelogram can be easily computed. So we can write:
By the Pythagoras theorem we have:
DH=\sqrt{25-16}=3
Find the missing step in the following reasoning in order to find the perimeter of the square.
| Step | |
| 1 | The perimeter of the square equals: 4x |
| 2 | .... |
| 3 | 2x^2=9 |
| 4 | x=3\dfrac{\sqrt{2}}{2} |
| 5 | P=4 \times 3\dfrac{\sqrt{2}}{2} = 6\sqrt{2} |
To find the perimeter of the square, we need to determine x. So we can write:
Using the Pythagoras theorem we have: {x^2+x^2}=9