Geometry - Study Mode

[#56] In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is
Correct Answer

(D) 70°

Explanation

Solution: According to the figure. OM = OY = ON ∴ In ΔOMY ∠OMY = ∠OYM = 15° ∴ ∠MOY = 180° - 15° - 15° ∠MOY = 150° In ΔONY ∠ONY = ∠OYN = 50° ∴ ∠NOY = 180° - 50° - 50° ∠NOY = 80° ∴ ∠MON = 150° - 80° ∠MON = 70°

[#57] In the given figure, two identical circles of radius 4 cm touch each other. A and B are the centres of the two circles. If RQ is a tangent to the circle, then what is the length (in cm) of RQ?
Correct Answer

(C) 4√2

Explanation

Solution: n 2 = 8 × 16 n 2 = 128 n = 8$$sqrt 2 $$ b = (8$$sqrt 2 $$ + a) a 2 + (16) 2 = b 2 a 2 + (16) 2 = (8$$sqrt 2 $$ + a) 2 a 2 + 256 = 128 + a 2 + 16$$sqrt 2 $$ a 128 = 16$$sqrt 2 $$ a a = $$frac{8}{{sqrt 2 }}$$ a = 4$$sqrt 2 $$

[#58] Two equal circles intersect so that there centres, and the point at which they intersect from a square of side 1 cm. The area (in sq. cm) of the portion that is common to the circles is
Correct Answer

(B) $$frac{pi }{2} - 1$$

Explanation

Solution: Now, Area of arc AC 1 B $$ = pi {r^2}.frac{{90}}{{360}} = frac{pi }{4}{left( 1
ight)^2} = frac{pi }{4}$$ And area of arc AC 2 B = $$frac{pi }{4}$$ Area of square = (side) 2 = 1 Area of common portion = area of arc (AC 1 B + AC 2 B) - Area of square $$eqalign{
& = frac{pi }{4} + frac{pi }{4} - 1 cr
& = frac{pi }{2} - 1,{ ext{sq}}{ ext{. m}} cr} $$

[#59] ABC is a triangle in which ∠ABC = 90°. BD is perpendicular to AC. Which of the following is TRUE? I. Triangle BAD is similar to triangle CBD. II. Triangle BAD is similar to triangle CAB. III. Triangle CBD, is similar to triangle CAB.
Correct Answer

(D) All I, II and III

Explanation

Solution: In ΔBAD and ΔBDC ∠BDA = ∠BDC xa0 (Each 90°) If ∠C = 30° Then ∠A = 60° Also ∠ABD = 30° ∠BCD = ∠ABD ΔBAD ∼ ΔBDC . . . . . (i) In ΔBAD and ΔCAB ∠BDA = ∠ABC xa0 (Each 90°) ∠BAD = ∠BAC xa0 (common) ΔBAD ∼ ΔCAB . . . . . (ii) Similarly ΔCBD ∼ ΔCAB . . . . . (iii) All conditions are true.

[#60] ABC is a right angled triangle, right angled at A. A circle is inscribed in it. The lengths of two sides containing the right angle are 48 cm and 14 cm. The radius of the inscribed circle is:
Correct Answer

(B) 6 cm

Explanation

Solution: BC 2 = 48 2 + 14 2 BC 2 = 2304 + 196 BC 2 = 2500 BC = 50 cm In radius r = $$frac{{14 + 48 - 50}}{2}$$ xa0 = 6 cm