Geometry - Study Mode

[#146] ABCD is a quadrilateral such that ∠D = 90°. A circle with centre O touches the sides AB, BC, CD and DA at P, Q, R and S, respectively. If BC = 40 cm, BP = 28 cm and CD = 25 cm, then what is the radius of the circle?
Correct Answer

(B) 13 cm

Explanation

Solution: BC = 40, BP = 28, CD 25 (given) BQ = BP = 28 (Tangent from an external point to a circle are equal) CQ = 40 - 28 = 12 RC = CQ = 12 (Tangent from an external point to a circle are equal) DR = 25 - 12 = 13 Join the line OR & OS (Both are radius) ∠DSO = ∠DRO = 90° In quadrilateral DROS ∠ROS = 360° - (90° + 90° + 90°) = 90° Hence, DROS is a square. Hence radius of circle = 13 ∠SDR = 90° (Given) ∴ ∠ROS = 90° Hence DROS is a square.

[#147] In ΔABC, Perimeter of ΔABD = Perimeter of ΔBCD and AB = 60 cm, BC = 80 cm and AC = 100 cm. Then find BD.
Correct Answer

(C) 24√5 cm

Explanation

Solution: Let AD = x and BD = y Perimeter of ΔABD = Perimeter of ΔBCD 60 + x + y = 80 + y + 100 - x 2x = 120 x = 60 CD = 100 - 60 = 40 Let ∠ACB = θ In ΔABC, cosθ $$ = frac{{80}}{{100}} = frac{4}{5}$$ In ΔCBD, by using cosine formula we get $$eqalign{
& cos heta = frac{{{ ext{B}}{{ ext{C}}^2} + { ext{C}}{{ ext{D}}^2} - { ext{B}}{{ ext{D}}^2}}}{{2{ ext{BC}}.{ ext{CD}}}} cr
& frac{4}{5} = frac{{{{80}^2} + {{40}^2} - {{ ext{y}}^2}}}{{2 imes 40 imes 80}} cr
& frac{4}{5} imes 80 imes 80 = 6400 + 1600 - {{ ext{y}}^2} cr
& 5120 = 8000 - {{ ext{y}}^2} cr
& {{ ext{y}}^2} = 2880 cr
& { ext{y}} = 24sqrt 5 { ext{ cm}} cr} $$

[#148] The lengths of the two sides forming the right angle of a right-angled triangle are 21 cm and 20 cm. What is the radius of the circle circumscribing the triangle?
Correct Answer

(A) 14.5 cm

Explanation

Solution: $$eqalign{
& AC = sqrt {{{21}^2} + {{20}^2}} cr
& = sqrt {441 + 400} cr
& = sqrt {841} cr
& = 29 cr
& R = frac{{AC}}{2} = frac{{29}}{2} = 14.5{ ext{ cm}} cr} $$

[#149] Two sides of a triangle are of length 3 cm and 8 cm. If the length of the third side is '$$x$$' cm, then:
Correct Answer

(B) 5 < $$x$$ < 11

Explanation

Solution: 8 - 3 < $$x$$ < 8 + 3 5 < x < 11

[#150] In a triangle PQR, ∠PQR = 90°, PQ = 10 cm and PR = 26 cm, then what is the value (in cm) of inradius of incircle?
Correct Answer

(B) 4

Explanation

Solution: $$eqalign{
& QR = sqrt {{{left( {26}
ight)}^2} - {{left( {10}
ight)}^2}} cr
& = sqrt {676 - 100} cr
& = sqrt {576} cr
& = 24 cr
& { ext{Incirde radius in a right angled triangle }}left( r
ight) cr
& = frac{{{ ext{P}} + { ext{B}} - { ext{H}}}}{2} cr
& = frac{{10 + 24 - 26}}{2} cr
& = frac{8}{2} cr
& = 4{ ext{ cm}} cr} $$