1.Find five rational numbers between 3/5 and 4/5.
We have to find five rational numbers between 3/5 and 4/5.
So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1)
Now,
3/5 = (3/5) × (6/6) = 18/30
4/5 = (4/5) × (6/6) = 24/30
Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30
5.The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is ₹ 49,57,200 at the rate of ₹ 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
Solution:
4.The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.
Solution:
3.The area of an equilateral triangle is 144√3 cm2; find its perimeter.
Solution:
2.The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m2; find its base and height.
Solution:
1.Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
(i) area of the triangle
(ii) height of the triangle corresponding to 8 cm side.
5. For any angle θ, state the value of:
Sin2 θ + cos2 θ.
Sin2 ? + cos2 ? = Sin2 ? + 1 – sin2 ?.
= 1
3.If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.
sin x = cos y = sin (90o – y)
If x and y are acute angles
x = 90o – y
which implies,
x + y = 90o
hence x and y are complementary angles.
2. If sin x = cos x and x is acute, state the value of x.
The angle, x is acute and hence we have, 0 < x
We know that
Cos2x + sin2 x = 1
Since cos x = sin x
Above equation will become
2 sin2 x = 1
Sin x = 1/v2
Therefore, x = 45o
Prove that:
3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
Given 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
LHS = 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
= 3 (2/v3)2 – 2 (v3)2 + (v2)2
= 4 – 6 + 2
= 0
= RHS
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