The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
<h3>How to prove an Isosceles Triangle?</h3>
Let ABC be an isosceles triangle such that AB = AC.
Let AD be the bisector of ∠A.
We want to prove that BD=DC
In △ABD & △ACD
AB = AC(Thus, △ABC is an isosceles triangle)
∠BAD =∠CAD(Because AD is the bisector of ∠A)
AD = AD(Common sides)
By SAS Congruency, we have;
△ABD ≅ △ACD
By corresponding parts of congruent triangles, we can say that; BD=DC
Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
Read more about Isosceles Triangle at; brainly.com/question/1475130
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This should be correct I am sorry if it is not.
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Answer:
here
Step-by-step explanation:
it is basically a sector , use the area formula i.e.
[(theta)/360]*πr^2
Answer:
Find the LCD of the first two, then the LCD of that and the third one.
Step-by-step explanation:
You can do it by finding the LCD of two of the denominators, then the LCD of that and the third denominator.
Or, you can factor each of the denominators and find their LCM by multiplying the unique factors to their highest powers.
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<u>Example:</u>
1/21 + 1/35 + 1/45
The LCD of 1/21 and 1/35 is (21·35)/5 = 105. The LCD of 1/105 and 1/45 is ...
(105·45)/15 = 315
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Using factoring ...
- 21 = 3·7
- 35 = 5·7
- 45 = 3²·5
LCD = 3²·5·7 = 315
Answer:
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Step-by-step explanation:
