Answer:
251.2
I used 3.14 as pi
Step by step explanation:
SA= 2(pi*r²) + (pi*d)h
SA= 2 ( * 4² ) + (pi*8)*6
<u>you could also do this:</u>
SA=2πrh + 2πr²
What is the length of the 9 centimeter ribbon in millimeters?
2 answers:
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The answer is 18m^2. if u find the area of each section and add it up together at the end then u get 18
Answer:
Transitive property of equality is not a justification for the proof.
Step-by-step explanation:
We draw a right angle ΔACB. CD is perpendicular to AB.
Let AC = a , BC = b , AB = c and CD = h
Now in ΔABC and ΔACD
∠C = ∠D and ∠A = ∠A
from AA similarity postulate
ΔABC similar to ΔACD.
Hence,
=
= c × x ·····················(1)
Now in ΔABC and ΔCBD
∠C = ∠D and ∠B = ∠B
from AA similarity postulates
ΔABC similar to ΔCBD
Hence,
= c × y······················(2)
Add equation (1) and (2)
+
= cx + cy
+
= c(x+y)
+
=
[because x+y=c]
Transitive property is not useful for this proof.
If AM=BM, then point M is a middle point of side AB.
1. Consider triangle AML. You know that AM=ML, then this triangle is isosceles and AL is its base. The angles adjacent to the base of isosceles triangle are conruent, this means that
2. Consider lines ML and AC. The angle bisector AL is transversal. Since alternate interior angles you have that lines ML and AC are parallel. This means that ML is a middle line of triangle and 2ML=AC. Also you know that AC=2AL. This gives you that ML=AL. Now ML=AL and ML=AM gives you that triangle AML is equilateral.
3. In equilateral triangle AML all angles are congruent and have measures 60°. Thus, m∠AML=m∠MLA=m∠LAM=60°.
4. AL is angle bisector, then m∠MAL=m∠LAC=60° and m∠BAC=m∠MAL+m∠LAC=120°.
5. Consider ΔBML, it is isosceles, because BM=ML and m∠BML=180°-m∠AML=180°-60°=120°. Then,
6. Consider triangle ABC. In this triangle m∠A=120°, m∠B=30°, then
m∠C=180°-m∠A-m∠B=180°-120°-30°=30°.
Answer: m∠A=120°, m∠B=m∠C=30°.