Maths-

General

Easy

Question

# In the figure, ABC is an isosceles triangle with AB = AC. If D is the midpoint of BC, prove that:

(i) ΔABD⩭ ΔACD

(ii) 90°

Hint:

### Find the congruence rule and from that find the angles.

## The correct answer is: 90°

### (i) In ΔABD and ΔACD, we have

AB = AC (isosceles triangle)

AD = AD (common side)

BD = DC (Since D is the midpoint of BC)

Hence ΔABD⩭ ΔACD by SSS congruence rule.

(ii) Since ΔABD⩭ ΔACD by SSS congruence rule

We have,

(linear pair)

### Related Questions to study

Maths-

### State the property of congruence and write symbolically the congruency in the correct order.

(i)

(ii)

(i) ASA congruence rule (Angle - Side - Angle)

Two angles are said to be congruent if two angles and the included side of one

triangle is equal to two angles and the included side of another triangle.

Here,

and AB = PQ

Comparing these correspondence between sides and angles, we have

ΔABC ⩭ ΔPQR

(ii) RHS congruence rule (Right angle - Hypotenuse - Side)

Two right angles are congruent if the hypotenuse and one side of first triangle are

respectively equal to the hypotenuse and one side of the second triangle.

Here,

, XY = LN= 3 cm and YZ = NM = 5cm

Comparing these correspondence between sides and eagles, we have

ΔXYZ ⩭ ΔLNM

Two angles are said to be congruent if two angles and the included side of one

triangle is equal to two angles and the included side of another triangle.

Here,

and AB = PQ

Comparing these correspondence between sides and angles, we have

ΔABC ⩭ ΔPQR

(ii) RHS congruence rule (Right angle - Hypotenuse - Side)

Two right angles are congruent if the hypotenuse and one side of first triangle are

respectively equal to the hypotenuse and one side of the second triangle.

Here,

, XY = LN= 3 cm and YZ = NM = 5cm

Comparing these correspondence between sides and eagles, we have

ΔXYZ ⩭ ΔLNM

### State the property of congruence and write symbolically the congruency in the correct order.

(i)

(ii)

Maths-General

(i) ASA congruence rule (Angle - Side - Angle)

Two angles are said to be congruent if two angles and the included side of one

triangle is equal to two angles and the included side of another triangle.

Here,

and AB = PQ

Comparing these correspondence between sides and angles, we have

ΔABC ⩭ ΔPQR

(ii) RHS congruence rule (Right angle - Hypotenuse - Side)

Two right angles are congruent if the hypotenuse and one side of first triangle are

respectively equal to the hypotenuse and one side of the second triangle.

Here,

, XY = LN= 3 cm and YZ = NM = 5cm

Comparing these correspondence between sides and eagles, we have

ΔXYZ ⩭ ΔLNM

Two angles are said to be congruent if two angles and the included side of one

triangle is equal to two angles and the included side of another triangle.

Here,

and AB = PQ

Comparing these correspondence between sides and angles, we have

ΔABC ⩭ ΔPQR

(ii) RHS congruence rule (Right angle - Hypotenuse - Side)

Two right angles are congruent if the hypotenuse and one side of first triangle are

respectively equal to the hypotenuse and one side of the second triangle.

Here,

, XY = LN= 3 cm and YZ = NM = 5cm

Comparing these correspondence between sides and eagles, we have

ΔXYZ ⩭ ΔLNM

Maths-

### In the figure, ray AZ bisects as well as .

(i) State the three pairs of equal parts in triangles BAC and DAC.

(ii) Is ΔBAC ⩭ ΔDAC ? Give reasons.

(iii) Is AB = AD? Justify your answer.

(iv) Is CD = CB? Give reasons.

(i) From the figure, we have

AC = AC (common side)

(ii) By, ASA congruence criterion, the triangles are congruent.

That is,

AC = AC

(iii) Since the triangles are congruent, AB = AD because both sides originate from

equal angles in 2 different triangles.

(iv) Since the triangles are congruent, CD = CB because both sides originate from

equal angles in 2 different triangles.

AC = AC (common side)

(ii) By, ASA congruence criterion, the triangles are congruent.

That is,

AC = AC

(iii) Since the triangles are congruent, AB = AD because both sides originate from

equal angles in 2 different triangles.

(iv) Since the triangles are congruent, CD = CB because both sides originate from

equal angles in 2 different triangles.

### In the figure, ray AZ bisects as well as .

(i) State the three pairs of equal parts in triangles BAC and DAC.

(ii) Is ΔBAC ⩭ ΔDAC ? Give reasons.

(iii) Is AB = AD? Justify your answer.

(iv) Is CD = CB? Give reasons.

Maths-General

(i) From the figure, we have

AC = AC (common side)

(ii) By, ASA congruence criterion, the triangles are congruent.

That is,

AC = AC

(iii) Since the triangles are congruent, AB = AD because both sides originate from

equal angles in 2 different triangles.

(iv) Since the triangles are congruent, CD = CB because both sides originate from

equal angles in 2 different triangles.

AC = AC (common side)

(ii) By, ASA congruence criterion, the triangles are congruent.

That is,

AC = AC

(iii) Since the triangles are congruent, AB = AD because both sides originate from

equal angles in 2 different triangles.

(iv) Since the triangles are congruent, CD = CB because both sides originate from

equal angles in 2 different triangles.

Maths-

### If line segments AB and CD bisect each other at O, then using SAS congruence rule prove that ΔAOC ⩭ ΔBOD.

From the figure, we have

AO = OB

And, CO = OD

To prove: (By SAS congruence rule)

AO = OB (given)

CO = OD (given)

(vertically opposite angles)

by SAS congruence rule.

AO = OB

And, CO = OD

To prove: (By SAS congruence rule)

AO = OB (given)

CO = OD (given)

(vertically opposite angles)

by SAS congruence rule.

### If line segments AB and CD bisect each other at O, then using SAS congruence rule prove that ΔAOC ⩭ ΔBOD.

Maths-General

From the figure, we have

AO = OB

And, CO = OD

To prove: (By SAS congruence rule)

AO = OB (given)

CO = OD (given)

(vertically opposite angles)

by SAS congruence rule.

AO = OB

And, CO = OD

To prove: (By SAS congruence rule)

AO = OB (given)

CO = OD (given)

(vertically opposite angles)

by SAS congruence rule.

Maths-

### What least number must be multiplied to 6912 so that the product becomes a perfect Cube?

Explanation :-

6912 = 2 × 3456

= 2 × 2 × 1728

= 2 × 2 × 2 × 864

= 2 × 2 × 2 × 2 × 432

= 2 × 2 × 2 × 2 × 2 × 216

= 2 × 2 × 2 × 2 × 2 × 2 ×108

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

=

we know is perfect square and to make perfect square multiply it with 3

We need to multiply 6912 with 3 to make

6912 = 2 × 3456

= 2 × 2 × 1728

= 2 × 2 × 2 × 864

= 2 × 2 × 2 × 2 × 432

= 2 × 2 × 2 × 2 × 2 × 216

= 2 × 2 × 2 × 2 × 2 × 2 ×108

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

=

we know is perfect square and to make perfect square multiply it with 3

We need to multiply 6912 with 3 to make

### What least number must be multiplied to 6912 so that the product becomes a perfect Cube?

Maths-General

Explanation :-

6912 = 2 × 3456

= 2 × 2 × 1728

= 2 × 2 × 2 × 864

= 2 × 2 × 2 × 2 × 432

= 2 × 2 × 2 × 2 × 2 × 216

= 2 × 2 × 2 × 2 × 2 × 2 ×108

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

=

we know is perfect square and to make perfect square multiply it with 3

We need to multiply 6912 with 3 to make

6912 = 2 × 3456

= 2 × 2 × 1728

= 2 × 2 × 2 × 864

= 2 × 2 × 2 × 2 × 432

= 2 × 2 × 2 × 2 × 2 × 216

= 2 × 2 × 2 × 2 × 2 × 2 ×108

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

=

we know is perfect square and to make perfect square multiply it with 3

We need to multiply 6912 with 3 to make

Maths-

### Rohit, Peter and Santosh walk around a circular park. They take 𝑎𝑛𝑑 to complete one round. What is the total time taken by them to complete a round in minutes.

Time taken by Rohith to walk around the circular path =

Time taken by Peter to walk around the circular path =

Time taken by Santhosh to walk around the circular path =

Total time taken =

On taking the LCM of denominators 3,5 and 12, we get LCM as 60.

So, and

and

On adding the terms , we get

To convert it into minutes, multiply by 60.

minutes.

They took 69 minutes to complete a round in a circular park.

Time taken by Peter to walk around the circular path =

Time taken by Santhosh to walk around the circular path =

Total time taken =

On taking the LCM of denominators 3,5 and 12, we get LCM as 60.

So, and

and

On adding the terms , we get

To convert it into minutes, multiply by 60.

minutes.

They took 69 minutes to complete a round in a circular park.

### Rohit, Peter and Santosh walk around a circular park. They take 𝑎𝑛𝑑 to complete one round. What is the total time taken by them to complete a round in minutes.

Maths-General

Time taken by Rohith to walk around the circular path =

Time taken by Peter to walk around the circular path =

Time taken by Santhosh to walk around the circular path =

Total time taken =

On taking the LCM of denominators 3,5 and 12, we get LCM as 60.

So, and

and

On adding the terms , we get

To convert it into minutes, multiply by 60.

minutes.

They took 69 minutes to complete a round in a circular park.

Time taken by Peter to walk around the circular path =

Time taken by Santhosh to walk around the circular path =

Total time taken =

On taking the LCM of denominators 3,5 and 12, we get LCM as 60.

So, and

and

On adding the terms , we get

To convert it into minutes, multiply by 60.

minutes.

They took 69 minutes to complete a round in a circular park.

Maths-

### Determine y so that

To find

So here, and

can be written as

On adding on both the sides, we have

On taking the LCM of denominators 12 and 3, we get LCM as 12.

So,

On adding, we get

Hence

So here, and

can be written as

On adding on both the sides, we have

On taking the LCM of denominators 12 and 3, we get LCM as 12.

So,

On adding, we get

Hence

### Determine y so that

Maths-General

To find

So here, and

can be written as

On adding on both the sides, we have

On taking the LCM of denominators 12 and 3, we get LCM as 12.

So,

On adding, we get

Hence

So here, and

can be written as

On adding on both the sides, we have

On taking the LCM of denominators 12 and 3, we get LCM as 12.

So,

On adding, we get

Hence

Maths-

### Find the simplest form of

To find

Determine the LCM of the denominators of rational numbers inside the brackets.

LCM of 2, 3 and 6 = 6

So, and

On adding the terms inside the bracket, we have

Divide with , we get

Hence

Determine the LCM of the denominators of rational numbers inside the brackets.

LCM of 2, 3 and 6 = 6

So, and

On adding the terms inside the bracket, we have

Divide with , we get

Hence

### Find the simplest form of

Maths-General

To find

Determine the LCM of the denominators of rational numbers inside the brackets.

LCM of 2, 3 and 6 = 6

So, and

On adding the terms inside the bracket, we have

Divide with , we get

Hence

Determine the LCM of the denominators of rational numbers inside the brackets.

LCM of 2, 3 and 6 = 6

So, and

On adding the terms inside the bracket, we have

Divide with , we get

Hence

Maths-

### What is the result of

To find

Taking the LCM of 39 and 26, we have 78 as LCM.

So, and

and

So,

Taking the LCM of 39 and 26, we have 78 as LCM.

So, and

and

So,

### What is the result of

Maths-General

To find

Taking the LCM of 39 and 26, we have 78 as LCM.

So, and

and

So,

Taking the LCM of 39 and 26, we have 78 as LCM.

So, and

and

So,

Maths-

### What is the difference between the greatest and least numbers and

Given numbers are and

Greatest number among the given numbers =

Least number among the given numbers =

Difference of greatest and least number =

Greatest number among the given numbers =

Least number among the given numbers =

Difference of greatest and least number =

### What is the difference between the greatest and least numbers and

Maths-General

Given numbers are and

Greatest number among the given numbers =

Least number among the given numbers =

Difference of greatest and least number =

Greatest number among the given numbers =

Least number among the given numbers =

Difference of greatest and least number =

Maths-

### Which is the correct descending order of

Given numbers are

Taking the LCM of 5,4 and 20. we get the LCM as 20.

So, and

and

Now we have

Hence fractions in descending order are

That is,

Hence option B is the correct order.

Taking the LCM of 5,4 and 20. we get the LCM as 20.

So, and

and

Now we have

Hence fractions in descending order are

That is,

Hence option B is the correct order.

### Which is the correct descending order of

Maths-General

Given numbers are

Taking the LCM of 5,4 and 20. we get the LCM as 20.

So, and

and

Now we have

Hence fractions in descending order are

That is,

Hence option B is the correct order.

Taking the LCM of 5,4 and 20. we get the LCM as 20.

So, and

and

Now we have

Hence fractions in descending order are

That is,

Hence option B is the correct order.

Maths-

### The sum of two rational numbers is -5. If one of the numbers is What is the other number?

Given that one rational number is and let the other one be x.

Their sum = -5, that is

Their sum = -5, that is

### The sum of two rational numbers is -5. If one of the numbers is What is the other number?

Maths-General

Given that one rational number is and let the other one be x.

Their sum = -5, that is

Their sum = -5, that is

Maths-

### In the given figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7cm, find the area of the shaded region

It is given that OA = 7 cm

i.e. Radius of bigger circle = 7 cm

Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

Now, OD is radius of bigger circle and diameter of smaller circle

i.e. radius of smaller circle = half of radius of bigger circle

r

Area of smaller circle = (r

= 3.14 × (3.5)

Area of semi-circle OACB = × r

= × 3.14 × 7

AB = diameter of bigger circle = 2 × 7 = 14 cm

OC = radius of bigger circle = 7 cm

Area of triangle ABC = × b × h = × AB × OC

= × 14 × 7 = 49 cm

Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

= 38.465 cm

= 66.395 cm

i.e. Radius of bigger circle = 7 cm

Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

Now, OD is radius of bigger circle and diameter of smaller circle

i.e. radius of smaller circle = half of radius of bigger circle

r

_{1}= 3.5 cmArea of smaller circle = (r

_{1})^{2}= 3.14 × (3.5)

^{2}= 38.465 cm^{2}Area of semi-circle OACB = × r

^{2}= × 3.14 × 7

^{2}= 76.93 cm^{2}AB = diameter of bigger circle = 2 × 7 = 14 cm

OC = radius of bigger circle = 7 cm

Area of triangle ABC = × b × h = × AB × OC

= × 14 × 7 = 49 cm

^{2}Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

= 38.465 cm

^{2}+ 76.93 cm^{2}- 49 cm^{2}= 66.395 cm

^{2}### In the given figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7cm, find the area of the shaded region

Maths-General

It is given that OA = 7 cm

i.e. Radius of bigger circle = 7 cm

Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

Now, OD is radius of bigger circle and diameter of smaller circle

i.e. radius of smaller circle = half of radius of bigger circle

r

Area of smaller circle = (r

= 3.14 × (3.5)

Area of semi-circle OACB = × r

= × 3.14 × 7

AB = diameter of bigger circle = 2 × 7 = 14 cm

OC = radius of bigger circle = 7 cm

Area of triangle ABC = × b × h = × AB × OC

= × 14 × 7 = 49 cm

Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

= 38.465 cm

= 66.395 cm

i.e. Radius of bigger circle = 7 cm

Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

Now, OD is radius of bigger circle and diameter of smaller circle

i.e. radius of smaller circle = half of radius of bigger circle

r

_{1}= 3.5 cmArea of smaller circle = (r

_{1})^{2}= 3.14 × (3.5)

^{2}= 38.465 cm^{2}Area of semi-circle OACB = × r

^{2}= × 3.14 × 7

^{2}= 76.93 cm^{2}AB = diameter of bigger circle = 2 × 7 = 14 cm

OC = radius of bigger circle = 7 cm

Area of triangle ABC = × b × h = × AB × OC

= × 14 × 7 = 49 cm

^{2}Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC

= 38.465 cm

^{2}+ 76.93 cm^{2}- 49 cm^{2}= 66.395 cm

^{2}Maths-

### What is the number that is to be subtracted from 𝑡𝑜 𝑔𝑒𝑡

Given rational number = and let the other rational number to be subtracted from

= x

On subtraction, we get the result to be

That is, we have

On rearranging the equation, we get

On taking LCM of 12 and 8 , we have 24 as the LCM.

= x

On subtraction, we get the result to be

That is, we have

On rearranging the equation, we get

On taking LCM of 12 and 8 , we have 24 as the LCM.

### What is the number that is to be subtracted from 𝑡𝑜 𝑔𝑒𝑡

Maths-General

Given rational number = and let the other rational number to be subtracted from

= x

On subtraction, we get the result to be

That is, we have

On rearranging the equation, we get

On taking LCM of 12 and 8 , we have 24 as the LCM.

= x

On subtraction, we get the result to be

That is, we have

On rearranging the equation, we get

On taking LCM of 12 and 8 , we have 24 as the LCM.

Maths-

### The circle touch each other internally. The sum of their area is 116 pie cm^{2} and the distance between their centres is 6 cm. Find the radii of the circles.

It is given that the distance between their centres is 6 cm

r

The sum of their area of two circles = 116 cm

(r

[ (r

(r

2 (r

2 (r

Using splitting the middle term,

We get, (r

(r

Since radius is always positive, r

r

r

_{2}- r_{1 }= 6 r_{2}= 6 + r_{1}The sum of their area of two circles = 116 cm

^{2}(r

_{1})^{2}+ (r_{2})^{2}= 116[ (r

_{1})^{2}+ (6 + r_{1})^{2}] = 116(r

_{1})^{2}+ 36 + (r_{1})^{2}+ 12 r = 1162 (r

_{1})^{2}+ 12 r + 36 – 116 = 02 (r

_{1})^{2}+ 12 r – 80 = 0 (r_{1})^{2}+ 6 r – 40 = 0Using splitting the middle term,

We get, (r

_{1})^{2}+ 10 r – 4r – 40 = 0(r

_{1 }+ 10 )( r_{1}– 4) = 0 r_{1}= 4 , - 10Since radius is always positive, r

_{1}= 4 cmr

_{2}= 4 + 6 = 10 cm### The circle touch each other internally. The sum of their area is 116 pie cm^{2} and the distance between their centres is 6 cm. Find the radii of the circles.

Maths-General

It is given that the distance between their centres is 6 cm

r

The sum of their area of two circles = 116 cm

(r

[ (r

(r

2 (r

2 (r

Using splitting the middle term,

We get, (r

(r

Since radius is always positive, r

r

r

_{2}- r_{1 }= 6 r_{2}= 6 + r_{1}The sum of their area of two circles = 116 cm

^{2}(r

_{1})^{2}+ (r_{2})^{2}= 116[ (r

_{1})^{2}+ (6 + r_{1})^{2}] = 116(r

_{1})^{2}+ 36 + (r_{1})^{2}+ 12 r = 1162 (r

_{1})^{2}+ 12 r + 36 – 116 = 02 (r

_{1})^{2}+ 12 r – 80 = 0 (r_{1})^{2}+ 6 r – 40 = 0Using splitting the middle term,

We get, (r

_{1})^{2}+ 10 r – 4r – 40 = 0(r

_{1 }+ 10 )( r_{1}– 4) = 0 r_{1}= 4 , - 10Since radius is always positive, r

_{1}= 4 cmr

_{2}= 4 + 6 = 10 cmMaths-

### A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

It is given that diameter of the wheel = 77 cm

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

### A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

Maths-General

It is given that diameter of the wheel = 77 cm

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions