Geometry - Study Mode

[#151] The circles of same radius 13 cm intersect each other at A and B. If AB = 10 cm, then the distance between their centres
Correct Answer

(C) 24 cm

Explanation

Solution: In right angle ΔAOC, OC = $$sqrt {{{13}^2} - {5^2}} $$ xa0 = 12 OO' = 12 + 12 = 24 cm

[#152] In ΔABC, D is a point on side BC such that ∠ADC = ∠BAC. If CA = 12 cm, CD = 8 cm, then CB (in cm) = ?
Correct Answer

(A) 18

Explanation

Solution: In ΔABC, ∠BAC = ∠ADC, (given) In, ΔABC & ΔADC, ∠BAC = ∠ADC (given) ∠C = ∠C (common) $$eqalign{
& herefore Delta { ext{ABC}} sim Delta { ext{DAC}},,,left( {{ ext{A}} - { ext{A}},{ ext{theorem}}}
ight) cr
& frac{{{ ext{BC}}}}{{{ ext{AC}}}} = frac{{{ ext{AC}}}}{{{ ext{DC}}}} cr
& Rightarrow frac{{{ ext{BC}}}}{{12}} = frac{{12}}{8} cr
& Rightarrow { ext{BC}} = frac{{144}}{8} = 18{ ext{ cm}} cr} $$

[#153] If in the following figure (not to scale), ∠DAB + ∠CBA = 90°, BC = AD, AB = 20 cm, CD = 10 cm
then the area of the quadrilateral ABCD is:
Correct Answer

(D) 75 cm 2

Explanation

Solution: $${ ext{Area of ABCD}} = frac{1}{2} imes 30 imes 5 = 75{ ext{ c}}{{ ext{m}}^2}$$

[#154] A circle touches the side BC of ΔABC at D and AB and AC are produced to E and F, respectively. If AB = 10 cm, AC = 8.6 cm and BC = 6.4 cm, then BE = ?
Correct Answer

(D) 2.5 cm

Explanation

Solution: 10 + x = 8.6 + y xa0 (AE = AF) x - y = -1.4 . . . . . . (i) x + y = 6.4 . . . . . . (ii) 2x = 5.0 x = 2.5 BE = 2.5

[#155] In an isosceles triangle ABC, AB = AC, XY || BC. If ∠A = 30°, the ∠BXY = ?
Correct Answer

(D) 105°

Explanation

Solution: $$eqalign{
& { ext{If }}angle A = {30^ circ } cr
& { ext{Then }}angle ABC = angle ACB = frac{{{{180}^ circ } - {{30}^ circ }}}{2} = {75^ circ } cr
& angle BXY = {180^ circ } - angle ABC cr
& = {180^ circ } - {75^ circ } cr
& = {105^ circ } cr} $$