The triangle PQR is a right triangle (since 13^2 = 11^2 + 20^2, which satisfies the Pythagorean theorem). Therefore, the diameter of the semicircle is the altitude from the right angle, which splits the hypotenuse into two segments.

Let's denote the points where the semicircle touches PR and QR as S and T respectively. Then, the triangles PQS and TQR are similar (since they are both right triangles and share the angle at Q). Therefore, the ratio of corresponding sides is the same, i.e., PQ/QR = QS/QT.

Substituting the given lengths, we get 20/13 = QS/(20 - QS). Solving this equation gives QS = 80/7. Therefore, QT = 20 - QS = 60/7.

Since O is the midpoint of ST, OQ - OP = (QT - QS)/2 = (60/7 - 80/7)/2 = -20/14 = -10/7.

Question

The triangle PQR is a right triangle (since 13^2 = 11^2 + 20^2, which satisfies the Pythagorean theorem). Therefore, the diameter of the semicircle is the altitude from the right angle, which splits the hypotenuse into two segments.

Let's denote the points where the semicircle touches PR and QR as S and T respectively. Then, the triangles PQS and TQR are similar (since they are both right triangles and share the angle at Q). Therefore, the ratio of corresponding sides is the same, i.e., PQ/QR = QS/QT.

Substituting the given lengths, we get 20/13 = QS/(20 - QS). Solving this equation gives QS = 80/7. Therefore, QT = 20 - QS = 60/7.

Since O is the midpoint of ST, OQ - OP = (QT - QS)/2 = (60/7 - 80/7)/2 = -20/14 = -10/7.

Knowee AI · Accepted Answer

The triangle PQR is a right triangle (since 13^2 = 11^2 + 20^2, which satisfies the Pythagorean theorem). Therefore, the diameter of the semicircle is the altitude from the right angle, which splits the hypotenuse into two segments.

Let's denote the points where the semicircle touches PR and QR as S and T respectively. Then, the triangles PQS and TQR are similar (since they are both right triangles and share the angle at Q). Therefore, the ratio of corresponding sides is the same, i.e., PQ/QR = QS/QT.

Substituting the given lengths, we get 20/13 = QS/(20 - QS). Solving this equation gives QS = 80/7. Therefore, QT = 20 - QS = 60/7.

Since O is the midpoint of ST, OQ - OP = (QT - QS)/2 = (60/7 - 80/7)/2 = -20/14 = -10/7.

In triangle PQR, PQ = 20, PR = 11 and QR = 13. A semicircle is drawn inside PQR such that it touches PR and QR, and its diameter lies on PQ. If the semicircle is centred at O, what is the value of OQ – OP?Cannot be determined

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