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Ncert Solution | Ncert Solution for class 9th | Ncert Solution for Class 9 Maths | NCERT SOLUTIONS MATHS CHAPTER 8 QUADRILATERALS
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# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below:

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
∴ 3x + 5x + 9x + 13x = 360°
[Angle sum property of a quadrilateral]
⇒ 30x = 360°
⇒ x = 360?30 = 12°
∴ 3x = 3 x 12° = 36°
5x = 5 x 12° = 60°
9x = 9 x 12° = 108°
13a = 13 x 12° = 156°
⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.

# Some questions are asked by class 9 students related to Maths

Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
(i) SR || AC and SR = 12 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:
(i) In ?ACD, We have
∴ S is the mid-point of AD and R is the mid-point of CD.
SR = 12AC and SR || AC …(1)
[By mid-point theorem]

(ii) In ?ABC, P is the mid-point of AB and Q is the mid-point of BC.
PQ = 12AC and PQ || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 12AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
PQ = SR and PQ || SR [Proved]
∴ PQRS is a parallelogram.

Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC.

In ?ABC, P and Q are the mid-points of AB and BC respectively.
∴ PQ = 12AC and PQ || AC …(1)
[By mid-point theorem]
In ?ADC, R and S are the mid-points of CD and DA respectively.
∴ SR = 12AC and SR || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 12AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
∴ PQRS is a parallelogram. …….(3)
Now, in ?ERC and ?EQC,
∠1 = ∠2
[ ? The diagonals of a rhombus bisect the opposite angles]
CR = CQ [ ?CD2 = BC2]
CE = CE [Common]
∴ ?ERC ≅ ?EQC [By SAS congruency]
⇒ ∠3 = ∠4 …(4) [By C.P.C.T.]
But ∠3 + ∠4 = 180° ……(5) [Linear pair]
From (4) and (5), we get
⇒ ∠3 = ∠4 = 90°
Now, ∠RQP = 180° – ∠b [ Y Co-interior angles for PQ || AC and EQ is transversal]
But ∠5 = ∠3
[ ? Vertically opposite angles are equal]
∴ ∠5 = 90°
So, ∠RQP = 180° – ∠5 = 90°
∴ One angle of parallelogram PQRS is 90°.
Thus, PQRS is a rectangle.