Q1. The area of an equilateral triangle is 400√3 sq. m. Its perimeter is:
(a) 120 m
(b) 150 m
(c) 90 m
(d) 135 m
Q2. From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are √3 cm, 2√3 cm and 5√3 cm. The perimeter (in cm) of the triangle is
(a) 64
(b) 32
(c) 48
(d) 24
Q3. The perimeter of a triangle is 30 cm and its area is 30 cm^2. If the largest side measures 13 cm, What is the length of the smallest side of the triangle?
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Q4. The length of the perpendiculars drawn from any point in the interior of an equilateral triangle to the respective sides are p_1,p_2 and p_3. The length of each side of the triangle is
(a) 2/√3 (p1+p2+p3 )
(b) 1/3 (p1+p2+p3 )
(c) 1/√(3 ) (p1+p2+p3 )
(d) 4/√3 (p1+p2+p3 )
Q5. Find the diameter of a wheel that makes 113 revolutions to go 2 km 26 decameters. (Take ? = 22/7 )
(a) 4 4/13 m
(b) 6 4/11 m
(c) 12 4/11 m
(d) 12 8/11 m
Q6. A circle is inscribed in a square whose length of the diagonal is 12√2 cm. An equilateral triangle is inscribed in that circle. The length of the side of the triangle is
(a) 5√3 cm
(b) 8√3 cm
(c) 6√3 cm
(d) 7√3 cm
Q9. The area of a circle is increased by 22 cm, if its radius is increased by 1 cm. The original radius of the circle is.
(a) 6 cm
(b) 3.2 cm
(c) 3 cm
(d) 3.5 cm
Q10. The circumference of a circle is 11 cm and the angle of a sector of the circle is 60°. The area of the sector is (use ? = 22/7 )
(a) 1 29/48 cm^2
(b) 2 29/48 cm^2
(c) 1 27/48 cm^2
(d) 2 27/48 cm^2
Solutions
S1. Ans.(a)
Sol. √3/4 a^2 = 400√3
a^2 = 1600
a=40
Perimeter = 40 × 3
= 120 m
S2. Ans.(c)
Sol. a/h = 2/√3
a = 2/√3 (p1+p2+p3 )
a=2/√3 (√3 + 2√3 + 5√3)
a=(2/√3) × (8√3)
a = 16
Perimeter = 16 × 3
= 48 m
S4. Ans.(a)
Sol. (a/h) = (2/√3)
a=(2/√3) h
a=(2/√3) (p1+p2+p3 )
S5. Ans.(b)
Sol. 2?Rh = (2 × 1000) + (26 × 10)
2 × (22/7) × R × 113 = 2000 + 260
2 × (22/7) × R × 113 = 2260
2R = 70/11
2R = 6 4/11 m
Diameter = 6 4/11 m
S9. Ans.(c)
Sol. π(r+1)^2-πr^2 = 22
22/7 (r^2+1+2r-r^2 )=22
2r + 1 = 7
2r = 6
r = 3
S10. Ans.(a)
Sol. 2?r = 11
r = 7/4 cm
area of sector = (πr^2 θ)/360
=(22/7)×(7/4)×(7/4)×(60/360)
=77/48
=1 29/48 cm^2
