NCERT Solutions for Exercise 8.2 Class 9 Maths Chapter 8 - Quadrilaterals

# NCERT Solutions for Exercise 8.2 Class 9 Maths Chapter 8 - Quadrilaterals

Edited By Vishal kumar | Updated on Jul 18, 2022 02:45 PM IST

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2- This chapter takes you into the study of quadrilaterals, four-sided shapes. 9th class maths exercise 8.2 answers provide practical problems and solutions to help you understand their properties and types. Access these free PDF solutions to improve your understanding, prepare for exams, and strengthen your math skills.

The NCERT Solutions for Class 9 Maths Exercise 8.2 deal with concepts that are mostly related to triangles rather than quadrilateral.

The question that is derived using the midpoint theorem from Class 9 Maths Chapter 8 exercise 8.2 are mostly revolving around triangles and their areas.

The two theorems that are present in NCERT solutions for Class 9 Maths chapter 8 exercise 8.2 are:

• The line segment joining the midpoints of the two sides of a triangle is parallel to the third side of the triangle (also known as the base of the triangle, generally).

• The line segment that is drawn through the midpoint of one side of a triangle, and moves parallel to the third side of the triangle (base) bisects the other side of the triangle into two equal halves.

The figure below shows a triangle ABC and the line EF passes through the midpoints of the sides AB and AC. Since it divides the sides into two AE=EB and AB=2AE

Along with Class 9 Maths, chapter 8 exercise 8.2 the following exercise is also present.

## Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.2

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove :

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$

Proof: In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove : $\small PQ=SR$

Proof : In $\triangle$ ACD,

S is mid point of DA. (Given)

R is mid point of DC. (Given)

By mid point theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is mid point of AB. (Given)

Q is mid point of BC. (Given)

By mid point theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2,we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$

Given : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove : PQRS is a parallelogram.

Proof : In PQRS,

Since,

$\small PQ\parallel SR$ and $\small PQ=SR$ .

So,PQRS is a parallelogram.

Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In $\triangle$ ACD,

S is midpoint of DA. (Given)

R is midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is midpoint of AB. (Given)

Q is mid point of BC. (Given)

By mid point theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2,we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So,the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$ BCD,

Q is mid point of BC. (Given)

R is mid point of DC. (Given)

By mid point theorem,

$\small QR\parallel BD$

So, QN || LM ...........5

LQ || MN ..........6 (Since, PQ || AC)

From 5 and 6, we get

LMPQ is a parallelogram.

Hence, $\small \angle$ LMN= $\small \angle$ LQN (opposite angles of the parallelogram)

But, $\small \angle$ LMN= 90 (Diagonals of a rhombus are perpendicular)

so, $\small \angle$ LQN=90

Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof :

In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By midpoint theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2, we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$ BCD,

Q is the midpoint of BC. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small QR\parallel BD$ and $\small QR=\frac{1}{2}BD$ ...................5

AC = BD.......................6(diagonals )

From 2, 5 and 6, we get

PQ=QR

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

Given: ABCD is a trapezium in which $\small AB\parallel DC$ , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. $\small 8.30$ ).

To prove: F is the mid-point of BC.

In $\triangle$ ABD,

E is the midpoint of AD. (Given)

EG || AB (Given)

By converse of midpoint theorem,

G is the midpoint of BD.

In $\triangle$ BCD,

G is mid point of BD. (Proved above)

FG || DC (Given)

By converse of midpoint theorem,

F is the midpoint of BC.

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

AB=CD (Given)

$\frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AE=CF$ (E and F are midpoints of AB and CD)

AE=CF (Given)

AE || CF (Opposite sides of a parallelogram)

Hence, AECF is a parallelogram.

In $\triangle$ DCQ,

F is the midpoint of DC. (given )

FP || CQ (AECF is a parallelogram)

By converse of midpoint theorem,

P is the mid point of DQ.

DP= PQ....................1

Similarly,

In $\triangle$ ABP,

E is the midpoint of AB. (given )

EQ || AP (AECF is a parallelogram)

By converse of midpoint theorem,

Q is the midpoint of PB.

OQ= QB....................2

From 1 and 2, we have

DP = PQ = QB.

Hence, the line segments AF and EC trisect the diagonal BD.

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.

To prove: the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Proof: In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By midpoint theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2, we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.

Thus, SQ and PR bisect each other.

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove :D is mid point of AC.

Proof: In $\triangle$ ABC,

M is mid point of AB. (Given)

DM || BC (Given)

By converse of mid point theorem,

D is the mid point of AC.

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.

To prove : $\small MD\perp AC$

Proof : $\angle$ ADM = $\angle$ ACB (Corresponding angles)

$\angle$ ADM= $90 \degree$ . ( $\angle$ ACB = $90 \degree$ )

Hence, $\small MD\perp AC$ .

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove : $\small CM=MA=\frac{1}{2}AB$

Proof : In $\triangle$ ABC,

M is the midpoint of AB. (Given)

DM || BC (Given)

By converse of midpoint theorem,

D is the midpoint of AC i.e. AD = DC.

In $\triangle$ AMD and $\triangle$ CMD,

$\angle$ ADM = $\angle$ CDM (Each right angle)

DM = DM (Common)

$\triangle$ AMD $\cong$ $\triangle$ CMD (By SAS)

AM = CM (CPCT)

But , $\small AM=\frac{1}{2}AB$

Hence, $\small CM=MA=\frac{1}{2}AB$ .

## More About NCERT Solutions for Class 9 Maths Exercise 8.2

NCERT solutions Class 9 Maths exercise 8.2 includes some examples and questions that blends the concepts of midpoints theorem from triangles and then it applies the question to quadrilaterals

Since a quadrilateral is generally a combination of two triangles this becomes a possibility to apply the midpoint theorem in quadrilaterals like a trapezium, or a parallelogram.

The figure shown below shows a trapezium ABCD and then we can see two triangles ABD and BCD thus we have two triangles and the line EF is parallel to the sides AB and CD and passes through the midpoints of AD and BC thus we can apply the midpoint theorem in such scenarios.

What is the basic condition required to make a quadrilateral?

A quadrilateral is formed when we have four non-colinear points in a plane and they are connected by straight lines in such a way that they form a closed four-sided figure.

## Benefits of NCERT Solutions for Class 9 Maths Exercise 8.2

• Exercise 8.2 Class 9 Maths, is based on mid-point theorem and its application.

• From Class 9 Maths chapter 8 exercise 8.2

• Understanding the concepts from Class 9 Maths chapter 8 exercise 8.2 will permit us to comprehend the ideas identified with the various kinds of area and side issue of similarity of figures, ideas which are on the way in Class 10 Maths NCERT syllabus.

## Key Features of Exercise 8.2 Class 9 Maths

1. Practical Problems: 9th class maths exercise 8.2 answers presents practical problems related to quadrilaterals, allowing students to apply their knowledge of their properties and classifications.

2. Step-by-Step Solutions: The class 9 maths ex 8.2 provides step-by-step solutions for each problem, making it easier for students to follow and understand the problem-solving process.

3. Enhanced Understanding: It aids in enhancing students' understanding of quadrilaterals by offering a variety of problems and solutions to reinforce concepts.

4. Exam Preparation: The ex 8.2 class 9 helps students prepare for exams by offering a range of problems that align with the curriculum and potential exam questions.

5. Practice Material: It serves as valuable practice material for students to improve their problem-solving skills and confidence in working with quadrilaterals.

6. Free Access: The class 9 ex 8.2 and its solutions are readily accessible for free, allowing students to study and practice at their convenience.

Also, See

## Subject Wise NCERT Exemplar Solutions

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 8.2?

NCERT solutions for Class 9 Maths exercise 8.2 is based mid-point theorem and its application.

2. What is the sum of all the angles of a quadrilateral and a triangle?

The sum of all the angles of a quadrilateral that has four sides is 360 and the sum of the angles of a triangle is 180 degrees. A triangle has three sides

3. What is midpoint theorem?

The midpoint hypothesis expresses that assuming a line segment is drawn joining the midpoints of any different sides of a triangle it is parallels corresponding to the third side and is equivalent to half of the length of the third side.

4. A line is drawn parallel to the base of a triangle passes through one of the mid-points of the side of the triangle. Find the point where it touches the third side?

It will also pass through the mid-point of the other side.

5. A line parallel to the base passes through the mid-points of the side of triangle. What will be the area of that small triangle?

The area of the smaller triangle formed will be one-fourth of the area of the original triangle.

6. A line parallel to the base passes through the mid-points of the side of triangle. What is the relation between the original triangle and the new smaller triangle formed?

The new smaller triangle will always be similar to the original triangle if it is formed by the line parallel to the base that passes through the mid-points of the side of the triangle.

7. How many triangles are formed when all the diagonals are drawn inside a quadrilateral?

A total of four triangles are formed when all the diagonals are drawn inside a quadrilateral.

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