Q1. If the orthocenter and the centroid of a triangle are the same, then the triangle is:
(a) Obtuse Triangle
(b) Right angled
(c) Equilateral
(d) Scalene Triangle
Q2. The sides of a triangle have lengths 3, 4, and 5. What kind of triangle is it?
(a) acute triangle
(b) right triangle
(c) obtuse triangle
(d) none of these
Q3. G is the centroid of the equilateral ∆ ABC. If AB = 10 cm. Then length of AG is
(a) (2√3)/3 cm
(b) (10√3)/3 cm
(c) 5√2 cm
(d) 8√7 cm
Q4. ABC is an isosceles triangle with AB = AC. A circle through B touching AC at the middle point intersects AB at P. Then AP: AB is
(a) 4: 1
(b) 3: 7
(c) 5: 4
(d) 1: 4
Q5. If ∆ABC is similar to ∆DEF such that BC = 3cm, EF = 4cm and area of ∆ABC = 54 cm2, then the area of ∆DEF is:
(a) 66 cm2
(b) 78 cm2
(c) 96 cm2
(d) 54 cm2
Q8. A fruit seller buys mangoes at the rate of 15 for 12 and sells them at the rate of 15 per dozen. Find his gain percentage.
(a) 25.65%
(b) 32.25%
(c) 51.35%
(d) 56.25%
Q9. Summer started cycling along the boundaries of a square field from corner point A. After half an hour he reached the corner point C, diagonally opposite of A. If his speed was 8 kmph, what is the area of the field in sq. km?
(a) 64
(b) 8
(c) 4
(d) None of these
Q10. A man riding a cycle at 12 km/hr can reach a village in 4(1/2) hours. If he is delayed by 1(1/2 ) hours at the start, then in order to reach his destination in time, he should ride with a speed of:
(a) 15 km/hr
(b) 16 km/hr
(c) 18 km/hr
(d) 20 km/hr
Solutions:
S8. Ans.(d)
Sol.
C.P. of one mango = 12/15 = 80 paise ; S.P. of one mango = 15/12=Rs 1.25
Gain = 45 paise , gain = 45/80 × 100 = 56.25 %
S9. Ans.(c)
Sol.
Distance covered in half an hour = 4 km.
2× side of the square = 4 km
side of square = 2 km
So Area of the field = 4 sq.km.
S10. Ans.(c)
Sol.
Distance = (Speed × Time)= (12 × 9/12) km = 54 km
Now, distance = 54 km and time taken = 3 hrs.
so speed = distance/time = 54/3 = 18 km/hr.
