Let AB extended intersect DC extended at point E
<span>We now have right triangle BEC with E = 90 degrees </span>
<span>For triangle BEC: </span>
<span>Exterior angle at E = 90 </span>
<span>Exterior angle at C = 148 (given) </span>
<span>Exterior angle of all polygons add up to 360 degrees </span>
<span>Exterior angle at B = 360−148−90 = 122 </span>
<span>So in quadrilateral ABCD </span>
<span>B = 122 </span>
<span>D = 360−44−148−122 = 46</span>
The answer is 310.93, rounded to the nearest hundredth. Hope this helps!
Our inequality looks like this:
2(x+6)≤52
Using the Distributive Property, we have
2*x + 2*6 ≤52
2x+12≤52
Cancel the 12 by subtracting from both sides:
2x+12-12≤52-12
2x≤40
Divide both sides by 2:
2x/2 ≤ 40/2
x≤20
x cannot be any more than 20 to satisfy this inequality.
Answer:
Yes it is a right angled triangle
Step-by-step explanation:
If its a right triangle it would satisfy the Pythagoras theorem.
29.3^2 = 858.49
28.5^2 = 812.25
6.8^2 = 46.24
Adding the last 2 values:
812.25 + 46.24
= 858.49
So YES.
