How Cheenta works to ensure student success?

Explore the Back-StoryThe area of the curve enclosed by is :

(A) 16

(B) 12

(C) 8

(D) 4

Area of is equivalent to area of

So it is a rhombus

Hence area will be,

In a rectangle , point lies on such that and point lies on such that 2. Lines and intersect at and respectively. If , are relatively prime positive integers, then the minimum value of is :

(i)

.(ii)

A regular polygon has 100 sides each of length. A another regular polygon has 200 sided each of length 2. When the area of the larger polygon is divided by the area of the smaller polygon, the quotient is closest to the integer

(A) 2

(B) 4

(C) 8

(D) 16

Area of polygon 1

Area of polygon 2

In a rectangle , points are taken on the sides respectively such that the lengths and are integers and is rectangle. The largest possible area of PQRS is

Area of

is a point inside an equilateral triangle . The perpendicular distance to the sides of the triangle are in the ratio . If where are co-prime positive integers, then equals

Let

Similarly

Area of quadrilateral

Area of

In and . Point lies on and bisects . Point lies on and bisects . If the bisectors intersect at , then the ratio

bisect

Length of

so

and also bisect

Ans.

In quadrilateral and . If is an integer then

Difference of two side is less than third side

in

also in

so

A box of dimension units is used to keep smaller cuboidal boxes so that no space is left between the boxes. If the box is packed with 100 such smaller boxes of the same size, then dimension of the smaller box is

Dimension of bigger box

No. of smaller box

Volume of smaller box

Small rectangular sheets of length units and breadth units are available. These sheets are assembled and pasted in a big cardboard sheet, edge to edge and made into a square. The minimum number of such sheet required is

Area of square

No. of rectangle

Therefore

is a square. is one fourth of the way from to and is one fourth of the way from to C. is the centre of the square. Side of the square is . Then the area of the shaded region in the figure in is

Draw perpendicular from to \& on .

So required area Area of Area of Area of quadrilateral .

Therefore

is a rectangle with and are midpoints of and respectively and is the mid-point of . The ratio of the area of to area of is

Let length of rectangle

breadth of rectangle

So area of trapezium

In quadrilateral and . If is an integer then

Extend to such that

So by SAS rule

CPCT

So of of

add on both sides

So (1)

In

In the given figure, is a right angled triangle with are points on , respectively such that and . Then, (in degree).

Let

So

So

The area of a sector and the length of the arc of the sector are equal in numerical value. Then the radius of the circle is

Area of sector = length of are of sector

In , the medians through and are perpendicular. Then is equal to

Let

In

In

In

Equation (i) + (ii)

Option (D).

In a quadrilateral . Then

(A) is a cyclic quadrilateral

(B) ABCD has an in-circle

(C) ABCD has both circum-circle and in-circle

(D) It has neither a circum-circle nor an in-circle

For incircle

In circle is possible

for cyclic quadrilateral (circumcircle) theorem should be followed.

It is not a cyclic quadrilateral

Option B is correct

In a rhombus of side length 5 , the length of one of the diagonals is at least 6 , and the length of the other diagonal is at most 6 . What is the maximum value of the sum of the diagonals ?

(A)

(B) 14

(C)

(D) 12

Let diagonal are and

We have to find ?

By option

from here we get it is not possible.

By option (B)

it is possible maximum value which is greater by other two options.

The number of acute angled triangles whose vertices are chosen from the vertices of a rectangular box is

(A) 6

(B) 8

(C) 12

(D) 24

From each surface diagonals there are 4 triangles are possible which are equilateral of side

each.

But in this manner each triangle is counted thrice therefore

Option (B).

Circles A, B and C are externally tangent to each other and internally tangent to circle D. Circles A and are congruent. Circle has radius 1 unit and passes through the centre of circle D. Then the radius of circle is units.

In

In CAN

In shows below, is a point on and a point on such that is a point in the interior of is a point on and is a point on such that = BG. Also, . The measure of in degree is

As

Therefore

The area of the curve enclosed by is :

(A) 16

(B) 12

(C) 8

(D) 4

Area of is equivalent to area of

So it is a rhombus

Hence area will be,

In a rectangle , point lies on such that and point lies on such that 2. Lines and intersect at and respectively. If , are relatively prime positive integers, then the minimum value of is :

(i)

.(ii)

A regular polygon has 100 sides each of length. A another regular polygon has 200 sided each of length 2. When the area of the larger polygon is divided by the area of the smaller polygon, the quotient is closest to the integer

(A) 2

(B) 4

(C) 8

(D) 16

Area of polygon 1

Area of polygon 2

In a rectangle , points are taken on the sides respectively such that the lengths and are integers and is rectangle. The largest possible area of PQRS is

Area of

is a point inside an equilateral triangle . The perpendicular distance to the sides of the triangle are in the ratio . If where are co-prime positive integers, then equals

Let

Similarly

Area of quadrilateral

Area of

In and . Point lies on and bisects . Point lies on and bisects . If the bisectors intersect at , then the ratio

bisect

Length of

so

and also bisect

Ans.

In quadrilateral and . If is an integer then

Difference of two side is less than third side

in

also in

so

A box of dimension units is used to keep smaller cuboidal boxes so that no space is left between the boxes. If the box is packed with 100 such smaller boxes of the same size, then dimension of the smaller box is

Dimension of bigger box

No. of smaller box

Volume of smaller box

Small rectangular sheets of length units and breadth units are available. These sheets are assembled and pasted in a big cardboard sheet, edge to edge and made into a square. The minimum number of such sheet required is

Area of square

No. of rectangle

Therefore

is a square. is one fourth of the way from to and is one fourth of the way from to C. is the centre of the square. Side of the square is . Then the area of the shaded region in the figure in is

Draw perpendicular from to \& on .

So required area Area of Area of Area of quadrilateral .

Therefore

is a rectangle with and are midpoints of and respectively and is the mid-point of . The ratio of the area of to area of is

Let length of rectangle

breadth of rectangle

So area of trapezium

In quadrilateral and . If is an integer then

Extend to such that

So by SAS rule

CPCT

So of of

add on both sides

So (1)

In

In the given figure, is a right angled triangle with are points on , respectively such that and . Then, (in degree).

Let

So

So

The area of a sector and the length of the arc of the sector are equal in numerical value. Then the radius of the circle is

Area of sector = length of are of sector

In , the medians through and are perpendicular. Then is equal to

Let

In

In

In

Equation (i) + (ii)

Option (D).

In a quadrilateral . Then

(A) is a cyclic quadrilateral

(B) ABCD has an in-circle

(C) ABCD has both circum-circle and in-circle

(D) It has neither a circum-circle nor an in-circle

For incircle

In circle is possible

for cyclic quadrilateral (circumcircle) theorem should be followed.

It is not a cyclic quadrilateral

Option B is correct

In a rhombus of side length 5 , the length of one of the diagonals is at least 6 , and the length of the other diagonal is at most 6 . What is the maximum value of the sum of the diagonals ?

(A)

(B) 14

(C)

(D) 12

Let diagonal are and

We have to find ?

By option

from here we get it is not possible.

By option (B)

it is possible maximum value which is greater by other two options.

The number of acute angled triangles whose vertices are chosen from the vertices of a rectangular box is

(A) 6

(B) 8

(C) 12

(D) 24

From each surface diagonals there are 4 triangles are possible which are equilateral of side

each.

But in this manner each triangle is counted thrice therefore

Option (B).

Circles A, B and C are externally tangent to each other and internally tangent to circle D. Circles A and are congruent. Circle has radius 1 unit and passes through the centre of circle D. Then the radius of circle is units.

In

In CAN

In shows below, is a point on and a point on such that is a point in the interior of is a point on and is a point on such that = BG. Also, . The measure of in degree is

As

Therefore

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Online Live Classroom Programs

Online Self Paced Programs [*New]

Past Papers

More