Two circles each with radius 1 are inscribed so that their centers lie along the diagonal of square. Each circle is tangent to two sides of the square and they are tangent to each other. Find the area between the circles and the square.

a) 6+4sqrt2-2pie

b)10-2pie

c)4+2sqrt2 -2pie

d)8sqrt2-2 pie

In a right angle triangle AC = 1, Angle B = 90° and Angle A = 15° Find the area of the triangle ABC.

1/4 square units

1/8 square units

put value of cos15 and find height and base and then area

What's the length of largest equilateral triangle that can be inscribed in an equilateral triangle of side length 1?

(3,-2), (12,4) and (-3,7) are the three vertices of a triangle. Then the distance between the orthocenter and circumcenter of this triangle is-

Sides are √117 √117 √234 which is a isosceles right angle triangle in which mid point of hypotenuse is orthocentre and point where right angle is formed is orthocentre so the distance becomes half of hypotenuse which is (1/2)*√234 gives √117/2

In a circle of radius r units, six points equally spaced are placed on the circumference. ΔABC is formed by joining three points such that ΔABC is neither isosceles nor equilateral. What is the area of ΔABC?

45.

If perimeter is given max area will be of equilateral traingle

A= √3/4*(44/3)

Hence none of these

- A boat, stationed at the North of a lighthouse, is making an angle of 30° with the top of the

lighthouse. Simultaneously, another boat, stationed at the East of the same lighthouse, is making an angle

of 45° with the top of the lighthouse. What will be the shortest distance between these two boats? The

height of the lighthouse is 300 feet. Assume both the boats are of negligible dimensions.

Options:

1) 300 feet

2) 600/√3 feet

3) 300√3 feet

4) 600 feet

5) None of the above

Boat stationed at C forms a 30-60 - 90 triangle (COA), Hence ratio of sides :

AO : OC : AC = 1 : sqrt3 : 2

OC = 300 sqrt 3

Similarly boat stationed at East of the lighthouse forms a 45 - 45 - 90 triangle.

Ratio of the sides 1 :1 : sqrt 2

OB = 300

OA = Height of the light house

C : Boat stationed at North

B : Boat stationed at East

Shortest distance between boats : BC

BC^2 = OC^2 + OB^2

BC^2 = 300^2 +(300sqrt3)^2

BC = 600.

P and Q are mid points, hecne PQ=1/2BC

Now are of ABC is = 1/2 BC* h (considering BC as base)

now distance of A from PQ will be 1/2 h (using similarity)

same way distance of G from PQ will 1/2 h (because GH and PQ are parallel lines)

Hence distance of R from PQ will be 'h'.

So both the triangles ABC and PQR have same height, but base of PQR is 1/2 the base of ABC

Hence ratio of ares is 1/2

Hotel trinity grand has a triangular terrace with sides 20m, 34m and 42m. A square shaped swimming pool is to be constructed such that one of its sides coincides with the largest side of the terrace and two vertices of the swimming pool touch other two sides of the terrace.

Q1. what is the perimeter of the pool?

Q2. What is the sum of the perimeters of the three triangular regions which are not covered by the pool?

Q3. what is the area of the terrace?

find lenght of height on biggest side using area

then use similarity to calculate the length of square