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Sw Kkk 6 years, 10 months ago
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Sia ? 6 years, 6 months ago
Steps of construction:-

- Draw a line segment BC at 8 cm.
- At B draw an angle of 90°
- With centre B and radius 6 cm draw an arc which intersect line of angle at A.
- Join AC.
- At B, draw an angle CBX of any measure.
- Starting from B, cut 4 equal parts on BX such that
BX1 = X1X2 = X2X3 = X3X4 - Join X4C
- Through X3, Draw X3Q||X4C
- Through Q, Draw QP||CA
{tex}\therefore \triangle P B Q \sim \triangle A B C{/tex}
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Gaurav Seth 6 years, 10 months ago
let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ <r<3
so the remainders may be 0,1 and 2
so n may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3
CASE 2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3
CASE 3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3
Posted by Anup Raj 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
We know that, the lengths of tangents drawn from an external point to a circle are equal.
∴ TP = TQ
In ΔTPQ,
TP = TQ
⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)
∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))
⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP ⊥ PT,
∴ ∠OPT = 90º
⇒ ∠OPQ + ∠TPQ = 90º
⇒ ∠OPQ = 90º – ∠TPQ
⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)
From (1) and (2), we get
∠PTQ = 2∠OPQ
Posted by Vishal Vishu K. Mathur 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Given:
Length (l)= 18 cm
Breadth (b)= 15 cm
Height (h)= 4.5 cm
Volume of metallic cuboid = l× b×h
= 18 X 15 X 4.5
=1215 cm³
Let the total no. of small cubes be x.
volume of a cube = side ³ = a³
= 3³ = 27cm³
volume of a cube= 27 cm³
No.of cubes = volume of cuboid/ volume of cube
= 1215/27 = 45
Hence, 45 cubes can be made from the given metallic cuboid.
Posted by Varsha Rani?? 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
The tangents at any point of a circle is perpendicular to radius through the point of contact


Posted by Khushi Rustogi 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago

- Draw a line segment AB of length 5cm.
- Taking B as point Construct an angle of measure 60o using a compass.
- Name the angle as angle ABX.
- Draw a line EF at a height of 3 cm such that it is parallel to the line segment AB. It must intersect ray BX at point C. Now join AC.
- Draw CD perpendicular to AB. CD is altitude of triangle ABC having height 3cm.
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Sw Kkk 6 years, 10 months ago
1Thank You