Geometry I
1)How many different regular polygons can be formed with the interior angle exceeding the exterior angle nad the sum of the interior angles not exceeding 180 degree?
a) 7 b) 8 c) 9 d) NOT
2) The ratio of the sides of triangle ABC is 1:2:4 .What is the ratio of the altitude drawn onto these sides?
a) 4:2:1 b) 1:2:4 c) 1:4: 16 d)NOT
3) A quadrilateral is inscribed in a circle.If an angle is inscribed in each of the segments outside the quadrilateral,then what is the sum of the four angles?
a) 270 b) 360 c) 540 d) 720
4) In a triangle PQR ,PQ=6cm,QR=8cm and PR=10.Then the length of the median bisecting the shortest side is (approx.)
a) 10 b) 8.5 c)9 d) NOT
5)A square is inscribed in acircle which is inscribed in an equliateral triangle .If one side of the triangle is "a". Find the area of the square.
a)a^/3 b) a^2/6 c) 3a^2/8 d) a^2/12
6)A cylindriacl quadrilateral is such that two of its adjacent angles are divisible by 6 and 10 respectively.One of the remaining angle will be divided by-
a)3 b) 4 c) 8 d) NOT
7) In the above question if one of the remaining angle be divided by 16,then its measure can be:
a) 128 b) 48 c) 132 d) 112
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ans d. NOT
Its not possible to create a triangle with the side ratio = 1 : 2 : 4
n/a
4) In a triangle PQR ,PQ=6cm,QR=8cm and PR=10.Then the length of the median bisecting the shortest side is (approx.)
a) 10 b) 8.5 c)9 d) NOT
ans:
The length is = sqrt (64 - 9) = sqrt(54)
so ans is D
How cum u got that??
"To get the BEST, be the BEST"
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i think answer to question 4 is 8.5. it is sqrt(64+9)= sqrt(73)=8.5
triangle PQR is right angled at Q, so it should be sqrt(64+9) not sqrt(64-9)
Soln.:
As each angle has 180 deg.
Total angle:180 x 4 - 180 x 2=360 deg.
So option (b) is correct....
Giv ur best to d world.
Nd d best will cm back to u...
Regards,
Dipanjan.....
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Soln.:
It is a right angle triangle..Whose hypotuse is 10cm.,Smallest side 6 cm.
Draw d diagram...
Median=sqrt(9+64)=sqrt(73)=8.5(approx)
So option(b) is correct...
Giv ur best to d world.
Nd d best will cm back to u...
Regards,
Dipanjan....
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Soln. Plz draw d picture at first as per as my instruction otherwised u cant maintain d link...
Draw the triangle ABC..From point B draw d angle bisector nd allow it to cut at d point D of BC.(As we know incentre is d cuttin poin of the angle bisectors)
As it is an equilateral triangle dn angle bisector shud be MEDIAN( As,AB/AC=BD/DC=>BD=DC[As,AB+AC])
So,BD=CD=a/2.
We know:
AB x AC - BD x DC = BD^2
=>BD=Sqrt(3a^2/4)=Sqrt(3)/2 * a......(i)
If incentre is O.
Dn OD=1/3 x BD[Properties of MEDIAN]
=>OD=Sqrt(3)/6 * a
As it is d radius of d circle
Sqrt(2) x s = Sqrt(3)/6 x 2 x a[s is d length of d square]
=>Area=s^2=a^2/6
So Option (b) is correct...
If u have any doubts dn olz ask me...
Giv ur best to d world.
Nd d best will cm back to u...
Regards,
Dipanjan.....
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Hey the solution is 8.544(sqrt(8^2 + 3^2)) ....pls do confirm the correct answere.......
Can anybody explain me 5th & 6th questions and solutions for both the questions
it'll b really helpful if someone could post the answers....
n i dont get the meanin f da first question....
cheers,
ani
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