Quiz 1

Explanation:
6*a=42
therefore a=7
we are asked to find the area of ABDEF leaving the node C.
hence we will have only 4 equilateral triangles instead of 6.
therefore, required area= 4* (root3/4) *a*a
req. area= 49 root3

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

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

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

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% Ans :-)..;

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) ans..

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 (ANSWER)

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

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 3√3....
AREA OF TRAPEZIUM= 1/2 *SUM OF PARALLEL SIDES *HEIGHT....
=1/2*(9+3)*3√3
= 18√3.

By Heron\'s Formula, the area, A, of a triangle with sides a, b, c is given by A = square root[s(s − a)(s − b)(s − c)], where s = ½(a + b + c) is the semi-perimeter of the triangle.

Then s = ½(10 + 17 + 21) = 24, and 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 drop a perpendicular of length h onto the side of length 21.
We also have A = ½ × base × perpendicular height.
Hence A = 21h/2 = 84, from which h = 8.
Notice that the triangle above the square is similar to the whole triangle. (This follows because its base, the top of the square, is parallel to the base of the whole triangle.)

Let the square have side of length d.

Considering the ratio of altitude to base in each triangle, we have 8/21 = (8 − d)/d = 8/d − 1.

Therefore the length of the side of the square is 168/29=5.79=5.8