Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.
Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.
Therefore, the angle BKQ is equal to 180-50-45=85°.
Of course angle BKP=180-85=95°.
Hope this helps :)
Answer:
2.14
Step-by-step explanation:
Basically its like regular subtraction
4.6
-
2.46
________
2.14
Answer:
A ≈ 35.3 units²
Step-by-step explanation:
Calculate the radius CB using Pythagoras' identity in the right triangle.
CB² + AB² = AC²
CB² + 6² = 9²
CB² + 36 = 81 ( subtract 36 from both sides )
CB² = 45 = r²
Then area of quarter circle is
A = × πr² = × π × 45 ≈ 35.3
Answer: 23/18
Step-by-step explanation: