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Elitmus Geometry Questions & Answers

Elitmus Geometry Questions & Answers

Elitmus Geometry Questions & Answers

1.The perimeter of a regular hexagon ABCDEF is 42.find the area of ABDEF polygon?
A)49
B)49 root 3
C)36
D)36 root3/2

Correct Answer : B

Explanation:
none of the answers is correct .there is some problem with the question,i also think they have asked for quad ABED ,then only 49 root 3 could be the answer.

2.Find the regular polygon whose interior angle is 150 more than exterior angle. note the total sum of interior and exterior angle should be 180 at one point?
A)12
B)16
C)32
D)24

Correct Answer : D

Explanation:
sum of interior angle+ exterior angle=180 exterior angle + exterior angle+150=180 2*exterior angle=30 exterior angle=15 now,
Sum of exterior angles=360 so,n*15=360 n=360/15=24

3.A square of 2 unit is given and four circles are made on its corners having an equal radius of each. and another circle of equAl radius is made on the center of the square. find out the area of the square which is not covered by circles?
A)0.43
B)0.86
C)0.74
D)0.37

Correct Answer : B

Explanation:
science there is given there are 4 circles on the corner and one in the center which means,
the diameter of center’s circle+sum of the radius of two circles = diagonal of a square 2r+r+r = 2root2 r=1/root2 therefor, required area = area of the square – 4*pi/4*r^2 – pi*r^2 (science corner’s sectors make an angle of 90degree) =4 – pi/2 – pi/2 =4 – pi =4 – 3.14 =0.86
square unit.

4.In the quadrilateral PQRS d=7 cm, h1=4.2 cm and h2=2.1 cm. Find the area?
A)22.05 cm^2
B)44.1 cm^2
C)66.15 cm^2
D)11.025 cm^2

Correct Answer : A

Explanation:
area of trapezium= (1/2)(s1+s2)(h) => 0.5 * (4.2+2.1)*7 =>0.5*6.3*7 =22.05 cm^2

5.ABC is the scalene triangle. Its angle A is kept constant, rest sides are tripled in length such that AC’=3AC and AB’=3AB. how much is the area increased in triangle AB’C’ from ABC?
A)800%
B)900%
C)850%
D)950%

Correct Answer : A

Explanation:
Scalene Triangle- A triangle with all sides of different lengths and different angle.
Let us assume triangle(ABC) having sides 3,4,5 and angle A = 90degree.
Hence Ar(ABC) = 1/2 * 3* 4 = 6.
Similarly, assume triangle AB’C’ where AB’ = 3AB, AC’ = 3AC’ (A/Q) Hence Ar(AB’C) = 1/2 * 9 * 12 = 54.
Difference = 54-6 = 48. % CHANGE = 48 * 100/ 6 = 800%

6.The side of a square is 10 cm. after joining the midpoint of all sides makes an another inner square and this process goes to infinite.Find the sum of perimeter of all squares.
A)80+40(square root of 2)
B)80
C)r=1 / root 2
D)160

Correct Answer : A

Explanation:
perimeter of largest square= 4 *(10)=40
perimeter of 2nd largest square = 4*(square root of 2)*5=20
square root of 2=40/(square root of 2)
—— 3rd ——————-
=4*5=20= 40/(square root of 2)^2 so on…………….. so ,
sum of all perimeter is = 40+40/(square root of 2) + 40/(square root of 2)^2 + ….. = 40{1+1/(square root of 2)+1/(square root of 2)^2+……}
this is G.P series with ratio is 1/(square root of 2) so, sum = 40*{1/(1-(1/(square root of 2)))} = 80+40(square root of 2)

7.PR is tangent to a circle at point P. Q is another point on the circle such that PQ is the diameter of a circle and RQ cuts circle at M. if the radius of the circle is 4 units and PR=6units. find the ratio of triangle PMR to PQR.
A)11/20
B)3/5
C)13/20
D)18/25

Correct Answer :B

Explanation:
PR^2+PQ^2=QR^2 8^2+6^2=QR^2 QR=10
NOW TO FIND THE RATIO OF PERIMETER OF TWO TRIANGLE TAKE THE OF UNCOMMON ANGLE
i.e, angle QPR so COS (QPR)= COS (PR/QR) =6/10 =3/5

8.There is a triangle with sides of any positive integer having area A1. Three squares are drawn on each side. The area of squares is S1, S2, S3. What is the value of T if T= A1+S1+S2+S3?
A)can be a fractional value
B)even or odd
C)even
D)odd

Correct Answer :A

Explanation:
consider a triangle with sides (4,4,4)
then the area of the three squares will be an integer.
and area of the triangle will be {(a)sq* root(3)}/2.
and root(3) will be a fractional no. therefore, t1+s1+s2+s3= fractional no

9.A quadrilateral pqrs is inscribed in a circle of center o. PQ is parallel to RS and pq= 3rs, also QR=PS. perimeter(pqrs)= perimeter of a square whose area is 36sqm, then find the approx area of pqrs?
A)6root(3)
B)12root(3)
C)9root(3)
D)18root(3)

Correct Answer ūüėÄ

Explanation:
AS THE AREA OF SQUARE IS 36… THEN ONE SIDE WILL BE 6…AND THE PERIMETER OF SQUARE WOULD BE 24=PERIMETER OF PQRS……(1)
AND BY GIVEN CONDITIONS WE KNOW THAT PQRS IS A CASE OF ISOCLESS TRAPEZIUM..AS PQ IS PARALLEL TO RSAND PS=QR..
NOW BY (1)…. PQ+RS+QR+PS=24 3RS+QR+RS+QR=24 2RS+QR=12 NOW BY HIT AND TRIAL PQ=9 QR=PS=6 SR=3
NOW WE CAN FIND THE HEIGHT OF TRAPEZIUM PQRS WHICH WILL BE 3ROOT3….
AREA OF TRAPEZIUM= 1/2 *SUM OF PARALLEL SIDES *HEIGHT…. =1/2*(9+3)*3ROOT3 = 18ROOT3.

10.A triangle of side 10,17,21.a square in the triangle whose two corners touch two sides of the triangle & one side of the square on the biggest side of triangle side. find out side of square?
A)5.4
B)5.8
C)6.7
D)7.2

Correct Answer :B

Explanation:
A=sqrt[(s)(s-a)(s-b)(s-c)] A=sqrt[24*14*7*3] A=sqrt[7056] A=84 now if you name the triangle PQR with PQ=10 QR=21 PR=17 and then draw a perpendicular from P to the side QR and name it PM now you can find the height of the triangle PQR by 1/2 * base * height= Area(by heroes formula) 1/2 * 21 * h =84 h= 8 if u imagine the two edges of square touching the sides of triangle let that be A on side PQ and B on side PR and the perpendicular PM cutting AB @ C now let the side of square be = x cm total height of PQR = 8
then height of triangle PAB is PC=8-x the the triangle PAB is similar to triangle PQR then AB/QR=PC/PM x/21=8-x/8 8x=168-21x 29x=168 x=168/29 x=5.79 ~ 5.8 cm

 

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