The figure shows a circle with tangents CP and CQ at points P and Q respectively. The circle is to be…

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The figure shows a circle with tangents CP and CQ at points P and Q respectively. The circle is to be inscribed in a triangle ABC, where point B lies on CQ produced and CBA = 90^. a) Locate point O, t

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10. The figure shows a circle with tangents CP and CQ at points P and Q respectively. The circle is to be inscribed in a triangle ABC, where point B lies on CQ produced and ∠ CBA = 90°. a) Locate point O, the centre of the circle. Step 1: The center of an inscribed circle is equidistant from the sides of the triangle. Since CP and CQ are tangents from C, the center O must lie on the angle bisector of ∠ PCQ. Step 2: Since the circle is tangent to CQ at Q, the radius OQ is perpendicular to CQ. Therefore, draw a line perpendicular to CQ at point Q. This line passes through O. Step 3: The intersection of the angle bisector of ∠ PCQ and the perpendicular line from Q is the center O. b) Complete triangle ABC. Step 1: Measure the radius of the circle, which is the length OQ. Step 2: Since ∠ CBA = 90° and the circle is tangent to BC at Q, the distance BQ must be equal to the radius OQ. Mark point B on the line CQ such that BQ = OQ and B is to the left of Q (as B lies on CQ produced). Step 3: From point B, draw a line perpendicular to CQ. This line represents the side AB of the triangle. Step 4: Extend the line CP. The intersection of line AB and line CP is point A. Step 5: Connect points A, B, and C to complete triangle ABC. 11. The deviations of the masses of 10 students from an assumed mean are: −10, −5, −2, 1, 4, 5, 7, 8, 9, 13. The mass of the heaviest student was 58 kg. Calculate the mean mass of the students. Step 1: Calculate the sum of the deviations (∑ d). ∑ d = −10 + (−5) + (−2) + 1 + 4 + 5 + 7 + 8 + 9 + 13; ∑ d = −17 + 47; ∑ d = 30 Step 2: Determine the assumed mean (A). The heaviest student has the largest positive deviation, which is 13. Let xheaviest be the mass of the heaviest student. The deviation dheaviest = xheaviest − A. Given xheaviest = 58 kg and dheaviest = 13. 13 = 58 − A; A = 58 − 13; A = 45 kg; Step 3: Calculate the mean mass (x) using the formula x = A + (∑ d)/(n). Here, n = 10 (number of students). x = 45 + 30/10; x = 45 + 3; x = 48 kg The mean mass of the students is: 48 kg

The figure shows a circle with tangents CP and CQ at points P and Q respectively. The circle is to be inscribed in a triangle ABC, where point B lies on CQ produced and CBA = 90^. a) Locate point O, t

Answer

Step-by-step solution

Part a· 3 steps

The center of an inscribed circle is equidistant from the sides of the triangle.

Since the circle is tangent to CQ at Q, the radius OQ is perpendicular to CQ.

The intersection of the angle bisector of $\angle PCQ$ and the perpendicular line from Q is the center O.

Part b· 8 steps
$48 \text{ kg}$

Measure the radius of the circle, which is the length OQ.

Since $\angle CBA = 90^\circ$ and the circle is tangent to BC at Q, the distance BQ must be equal to the radius OQ.

From point B, draw a line perpendicular to CQ.

Extend the line CP.

Connect points A, B, and C to complete triangle ABC.

Calculate the sum of the deviations ( $\sum d$ ).

Determine the assumed mean ( $A$ ).

Calculate the mean mass ( $\bar{x}$ ) using the formula $\bar{x} = A + \frac{\sum d}{n}$ .

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The figure shows a circle with tangents CP and CQ at points P and Q respectively. The circle is to be inscribed in a triangle ABC, where point B lies on CQ produced and CBA = 90^. a) Locate point O, t

Answer

Step-by-step solution

Part a· 3 steps

The center of an inscribed circle is equidistant from the sides of the triangle.

Since the circle is tangent to CQ at Q, the radius OQ is perpendicular to CQ.

The intersection of the angle bisector of ∠PCQ\angle PCQ∠PCQ and the perpendicular line from Q is the center O.

Part b· 8 steps48 kg48 \text{ kg}48 kg

Measure the radius of the circle, which is the length OQ.

Since ∠CBA=90∘\angle CBA = 90^\circ∠CBA=90∘ and the circle is tangent to BC at Q, the distance BQ must be equal to the radius OQ.

From point B, draw a line perpendicular to CQ.

Extend the line CP.

Connect points A, B, and C to complete triangle ABC.

Calculate the sum of the deviations (∑d\sum d∑d).

Determine the assumed mean (AAA).

Calculate the mean mass (xˉ\bar{x}xˉ) using the formula xˉ=A+∑dn\bar{x} = A + \frac{\sum d}{n}xˉ=A+n∑d​.

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The intersection of the angle bisector of $\angle PCQ$ and the perpendicular line from Q is the center O.

Part b· 8 steps
$48 \text{ kg}$

Since $\angle CBA = 90^\circ$ and the circle is tangent to BC at Q, the distance BQ must be equal to the radius OQ.

Calculate the sum of the deviations ( $\sum d$ ).

Determine the assumed mean ( $A$ ).

Calculate the mean mass ( $\bar{x}$ ) using the formula $\bar{x} = A + \frac{\sum d}{n}$ .