Answer:
From the given figure we can see that :---
MN||PO, MN=PO and MP=NO
Therefore, angle M=angle N (adjacent sides of the trapezoid)
angle M=angle N= (6×18)+20=108+20 =128°
angle O =(180°-128°)=52°
<h3>Hence, angle O is <u>52°</u><u>.</u></h3>
Answer:
7. ∠CBD = 100°
8. ∠CBD = ∠BCE = 100°; ∠CED = ∠BDE = 80°
Step-by-step explanation:
7. We presume the angles at A are congruent, so that each is 180°/9 = 20°.
Then the congruent base angles of isosceles triangle ABC will be ...
∠B = ∠C = (180° -20°)/2 = 80°
The angle of interest, ∠CBD is the supplement of ∠ABC, so is ...
∠CBD = 180° -80°
∠CBD = 100°
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8. In the isosceles trapezoid, base angles are congruent, and angles on the same end are supplementary:
∠CBD = ∠BCE = 100°
∠CED = ∠BDE = 80°
Since the length is 15, and the width would be 15, you need to first do 15 x 15.
15 x 15 = 225
Multiply the 6 faces.
225x6
Your final answer would be 1,350.
X=width of the lake
y=length of the lake.
we set out this system of equiations
x*y=9/20
x=y/5
we solve by sustitution method
(y/5)*y=9/20
y²/5=9/20
y²=(5*9)/ 20
y²=2.25
y=√2.25=1.5
x=y/5
x=1.5/5=0.3
Solution:
width of the lake=0.3 miles
lenght of the lake=1.5 miles
