Find the missing step in the following reasoning in order to show that this triangle is isosceles.
| Step | |
| 1 | \widehat{A} +\widehat{B} +\widehat{C} =180^\circ |
| 2 | \widehat{B} =180^\circ-\left(75^\circ+30^\circ\right)=75^\circ |
| 3 | ... |
| 4 | Conclusion : this triangle is isosceles |
A triangle is isosceles if the base angles have the same degree measure.
Here, we know that \widehat{C}=75^\circ and \widehat{B}=75^\circ. Therefore:
In step 3, we can conclude that \widehat{B} = \widehat{C}.
Find the missing step in the following reasoning in order to find the area of this triangle.
| Step | |
| 1 | S=\dfrac{1}{2}.BC.AH |
| 2 | Using the Pythagoras theorem we have: AH=\sqrt{{52}-36}=\sqrt{16}=4 |
| 3 | ... |
| 4 | S=\dfrac{1}{2}\times9\times4=18 |
In Step 1, we need the lengths of AH and BC. The length of AH is given in Step 2. To find the length of BC we only need to determine the length of BH as the length of CH is given. Therefore, we should find BH in Step 3:
In step 3, we can write: Using the Pythagoras theorem we have:
BH=\sqrt{{25}-16}=\sqrt{9}=3
Find the missing step in the following reasoning in order to determine the measure of \widehat{B}
| Step | |
| 1 | Using the Pythagoras theorem we have: AH=\sqrt{25-9}=4 |
| 3 | ... |
| 4 | Conclusion : \widehat{B} = 30^\circ |
To find the measure of \widehat{B}, we first compute \sin\left(\widehat{B}\right). Therefore:
Therefore, in step 2 we can write: \sin\left(\widehat{B}\right) =\dfrac{AH}{AB}= \dfrac{4}{8}= \dfrac{1}{2}
Find the missing step in the following reasoning in order to determine the length of \overline{BC}, given that the triangles are similar.
| Step | |
| 1 | We have \dfrac{DE}{AB}=2, the scale factor is 2. |
| 2 | We have \dfrac{DC}{BC}=\dfrac{CE}{AC}=2 |
| 3 | ... |
| 4 | Conclusion : BC=2 |
In step 2, we can write:
BC=2DC
Find the missing step in the following reasoning in order to determine the measure of \widehat{B}
| Step | |
| 1 | \widehat{a} + \widehat{BAC} +\widehat{b} =180^\circ |
| 2 | ... |
| 3 | \widehat{B}+ \widehat{BAC} + \widehat{C}=180^\circ |
| 4 | \widehat{B}= 180^\circ - \left(50^\circ + 45^\circ\right) = 85^\circ |
In Step 2, we need to find \widehat{BAC} :
\widehat{BAC} =180^\circ-\left(70^\circ+65^\circ\right)=45^\circ.
Find the missing step in the following reasoning in order to show that this triangle is equilateral.
| Step | |
| 1 | Since AB = AC then \widehat{A}= \widehat{B} |
| 2 | We have \widehat{B}= 60^\circ |
| 3 | ... |
| 4 | Conclusion : this triangle is equilateral |
A triangle is an equilateral if all three internal angles are congruent to each other. So in Step 3, we need to conclude that \widehat{C} = 60^\circ.
\widehat{C} = 180^\circ-\left(60^\circ+60^\circ\right)= 60^\circ
Find the missing step in the following reasoning in order to determine the area of ADE
| Step | |
| 1 | \widehat{A} +\widehat{B} +\widehat{C} =180^\circ |
| 2 | \widehat{A} =180^\circ-\left(90^\circ+60^\circ\right)=30^\circ |
| 3 | .... |
| 4 | AE=2\sqrt{6} |
| 5 | Using the Pythagoras theorem we have AD=\sqrt{\left(2\sqrt{6}\right)^2-\sqrt{6}^2}=\sqrt{18} |
| 6 | S=\dfrac{1}{2} \times \sqrt{6} \times \sqrt{18}= \sqrt{27} |
We need to find an equation to determine AE.
\dfrac{DE}{AE}=\sin\left(\widehat{A}\right) =\dfrac{1}{2}