Geometry - Study Mode

[#216] For an equilateral triangle, the ratio of the in-radius and the outer-radius is
Correct Answer

(A) 1 : 2

Explanation

Solution: Let the side of equilateral ΔABC be a & r = in-radius & R = outer radius $$eqalign{
& r = frac{a}{{2sqrt 3 }},,,,R = frac{a}{{sqrt 3 }} cr
& r:R = frac{a}{{2sqrt 3 }}:frac{a}{{sqrt 3 }} = 1:2 cr} $$

[#217] If in the following figure (not to the scale), ∠ACB = 135° and the radius of the circle is 2√2 cm, then the length of the chord AB is
Correct Answer

(C) 4 cm

Explanation

Solution: $$eqalign{
& { ext{Hence }}A{B^2} = {left( {2sqrt 2 }
ight)^2} + {left( {2sqrt 2 }
ight)^2} cr
& A{B^2} = 8 + 8 cr
& A{B^2} = 16 cr
& AB = sqrt {16} cr
& AB = 4 cr} $$

[#218] A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the circumference is
Correct Answer

(D) 30°

Explanation

Solution: ∵ OA = AB = OB (given) ∴ ΔAOB is equilateral triangle So, ∠AOB = ∠OAB = ∠ABO = 60° ∵ ∠ACB = $$frac{1}{2}$$ ∠AOB = $$frac{1}{2}$$ × 60° = 30°

[#219] AB is a common tangent to both the circles in the given figure. Find the distance (correct to two decimal places) between the centres of the two circles.
Correct Answer

(B) 23.58 units

Explanation

Solution: $$eqalign{
& Rightarrow { ext{By figure,}} cr
& Delta CAE sim Delta DBE cr
& frac{{CA}}{{BD}} = frac{{AE}}{{BE}} cr
& frac{5}{x} = frac{8}{{12}} cr
& x = 7.5 = left( {frac{{15}}{2}}
ight) cr
& Rightarrow { ext{Length of T}}{ ext{.L}} = sqrt {{d^2} - {{left( {{R_1} + {R_2}}
ight)}^2}} cr
& {left( {20}
ight)^2} = {d^2} - {left( {5 + 7.5}
ight)^2} cr
& 400 = {d^2} - 156.25 cr
& {d^2} = 400 + 156.25 cr
& {d^2} = 556.25 cr
& d = sqrt {556.25} cr
& d = 23.58 cr} $$

[#220] If two circles do not touch or intersect each other and one does not lie inside the other, then find the number of common tangents.
Correct Answer

(B) 4

Explanation

Solution: 4 tangents