Class 10 Maths: Circles – Extra Questions with Answers

Ans: C

Solution:
2(x+3) – 4x + 1 = 2x + 6 – 4x + 1 = -2x + 7
Answer: -2x + 7

Ans: C

Solution:
Let $O$ be the center of the circle, and let $T$ be the point of tangency on the circle. Then $\triangle OTQ$ is a right triangle with right angle at $T$. We have $OQ = 15$ cm and $OT = 9$ cm. By the Pythagorean theorem, $QT^2 + OT^2 = OQ^2$, so $QT^2 + 9^2 = 15^2$. Then $QT^2 = 15^2 – 9^2 = 225 – 81 = 144$. Thus, $QT = \sqrt{144} = 12$.

Answer: 12 cm

Ans: B

Solution:
The radii to the points of tangency and the tangents form a quadrilateral. The sum of the angles in a quadrilateral is 360 degrees. The angle between the radii and each tangent is 90 degrees. Let x be the angle between the radii. Then, 90 + 90 + 60 + x = 360. So, x = 360 – 240 = 120.
Answer: 120

Ans: A

Solution:
Let the circle’s center be O, the point of tangency be T, and the point B be the external point. We have a right triangle OTB, where OT is the radius (r), BT is the tangent (15 cm), and OB is the distance from B to the center (17 cm). Using the Pythagorean theorem: OT² + BT² = OB². So, r² + 15² = 17². Thus, r² + 225 = 289. Therefore, r² = 64, and r = 8 cm. The diameter is 2r = 2 * 8 = 16 cm.
Answer: 16 cm

Ans: D

Solution:
Since $\angle A = \angle D$ and $\angle A + \angle B = \angle D + \angle E$, it follows that $\angle B = \angle E$. Therefore, triangles ABC and DEF are similar by the Angle-Angle (AA) similarity criterion. The ratio of their corresponding sides is $\frac{BC}{EF} = \frac{9}{12} = \frac{3}{4}$. The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides. Thus, the ratio of the area of triangle ABC to the area of triangle DEF is $(\frac{3}{4})^2 = \frac{9}{16}$.
Answer: $\frac{9}{16}$

Ans: D

Solution:
Using the Pythagorean theorem, radius^2 + tangent^2 = distance^2
radius^2 + 12^2 = 13^2
radius^2 + 144 = 169
radius^2 = 25
radius = 5

Answer: 5 inches

Ans: C

Solution:
Draw radii from the center to the endpoints of the chord (10 cm). Draw a radius to the point of tangency (6 cm), forming a right triangle. The chord is bisected by the radius to the point of tangency. Using the Pythagorean theorem: (chord/2)^2 + 6^2 = 10^2. (chord/2)^2 = 64. chord/2 = 8. chord = 16.
Answer: 16 cm

Ans: B

Solution:
Let the sides be a, b, and c. Given a = 7, b = 15.
By the triangle inequality, 15-7 < c < 15+7, so 8 < c < 22.
For obtuse triangles, we have two cases:
Case 1: c is the longest side (c > 15). Then $7^2 + 15^2 < c^2$, so $49 + 225 < c^2$, which means $274 < c^2$. Since c < 22, we have $\sqrt{274} < c < 22$, so $16.55 < c < 22$. Since c is an integer, c can be 17, 18, 19, 20, 21.
Case 2: 15 is the longest side (c < 15). Then $7^2 + c^2 < 15^2$, so $49 + c^2 < 225$, which means $c^2 < 176$. Thus, $c < \sqrt{176} \approx 13.26$. Since c > 8, c can be 9, 10, 11, 12, 13.
Total number of possible triangles = 5 + 5 = 10.

Answer: 10

Ans: C

Solution:
Using the Pythagorean theorem on triangles ABD and CBD, let AD = x, then CD = 17 – x.
Also, let BD = h.
In triangle ABD: x^2 + h^2 = 15^2 = 225
In triangle CBD: (17-x)^2 + h^2 = 10^2 = 100
Subtracting the second equation from the first:
x^2 – (17-x)^2 = 125
x^2 – (289 – 34x + x^2) = 125
34x – 289 = 125
34x = 414
x = 414/34 = 207/17
Answer: 207/17

Ans: C

Solution:
Answer: C, D, A, B