Area of an equilateral triangle
=√3/4x²,where x=side
Area=√3/4 *(16)²
=√3/4 *16*4*4
=√3*16*4
=64√3cm²

ans - b

Let the diagonal be x

We know that the length of diagonals of square is congruent

We are given that the product of length of diagonals is 66

So,x^{2}=66

x=\sqrt{66}

Length of diagonal =\sqrt{2}a

a is the side of square

So, \frac{\sqrt{66}}{\sqrt{2}}=a

Area of square = Side^{2} =(\frac{\sqrt{66}}{\sqrt{2}})^2=33 cm^2

Hence the area of rectangle is 33 sq.cm.

 

Area of trapezium =1/2 ×(sum of parallel sides) ×(distance between them )

1/ 2 (20+18) x 15

=285 cm^{2}

The sum of the interior angles of a triangle is 180^o.

The question states that the largest of the angles is 70^o.

The least value for the smallest angle will be when the other two angles are as large as possible.

The largest value that the other two angles can take is 70^o each.

Therefore, the least value of the smallest of the angles = 180 - 70 - 70 = 40^o

Solution: Given: radius of circle=21 cm
Central angle=135°
To find: Perimeter of sector.
Firstly we have to find the length of arc.
So length of arc = theta/ 360°×2pi R
= 135°/360°×2×22/7 × 21
= 27°/72° × 44×3
= 9°/24° × 44×3
= 9/6×44
= 3/2× 44
= 3 × 22= 66 cm
Now, Perimeter of sector= 2×r + length of arc
= 2×21+ 66
= 42+66
= 108 cm

 

The ratio of volume of the two cones is 1: 10

The ratio of the radii of the two cones is 1: 2

we know formula for cone

on simplifying we get to know the value pf h1 and h2

on solving we get 2:5