Selina Concise Class 9 Maths Chapter 14 Rectilinear Figures Exercise 14B Solutions
EXERCISE – 14B
Exercise – 14.2
(1) State, ‘true’ or ‘False’
(i) The diagonals of a rectangle bisect each other
=> True :
We can see diagram,
The diagonals of a rectangle bisect each other.
(ii) The diagonals of a quadrilateral bisect each other.
=> False :
(iii) The diagonals of a parallelogram bisect each other at right angle.
=> False:
The diagonals of a parallelogram bisect each other but not at right angle.
(iv) Each diagonal of a rhombus bisects it.
=> True: (Photo)
(v) The quadrilateral, whose four sides are equal, is a square.
=> False:
All quadrilateral is not a square.
(vi) Every rhombus is a parallelogram.
=> True:
A rhombus has opposite sides equal or parallel, so every rhombus is a parallelogram.
(vii) Every parallelogram is a rhombus.
=> False:
(viii) Diagonals of a rhombus are equal.
=> False :
Diagonals of a rhombus are not equal.
(ix) If two adjacent sides of a parallelogram are equal, it is a rhombus.
=> True :
(x) If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square,
=> False : (Photo)
(2) In the figure, given below, AM bisects angles A and DM bisects angle D of parallelogram ABCD. Prove that: ∠AMD = 90°
=> Given :
AM bisect angle A and DM bisects angle D of parallelogram ABCD.
To prove :
∠AMD = 90°
Proof :
∠A + ∠D = 180° (∵ co-interior angles)
2x + 2y = 180°
2 (x + y) = 180°
x + y = 180°/2
x + y = 90°
In △AMD,
∠MAD + ∠MDA + ∠AMD = 180°
(Sum of all interior angles = 180°)
(x + y ) + ∠AMD = 180°
90 + ∠AMD = 180°
∠AMD = 180 – 90
∠AMD = 90° [Hence proved]
(3) In the following figure, AE and BC are equal and parallel and the three sides AB, CD and DE are equal to one another. If angle A is 102°. Find angles AEC and BCD.
Solution :
AECB is a parallelogram (∵ AE = BC and AE || BC)
∠BAE + ∠AEC = 180° (co-interior angles)
102° + ∠AEC = 180°
∠AEC = 180° – 102°
∠AEC = 78°
Also,
∠BAE = ∠BCE = 102°
AB = EC (because AECB is a parallelogram)
Then,
∠ECD = 60° (because ECD is an equilateral triangle)
∴ ∠BCD = ∠BCE + ∠ECD
= 102° + 60°
∠BCD = 162°
(4) In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP show that
(i) ∠POC = 22 1/2°
(ii) ∠BDC = 2∠POC
(iii) ∠BOP = 2∠COP
Solution :
Given, OB = BP
To show:
(i) ∠POC = 22 1/2°
(ii) ∠BDC = 2∠POC
(iii) ∠BOP = 3∠COP
Proof :
In △BOP,
Sum of all interior angles = 180°
∠OBP + ∠BOP + ∠BPO = 180°
45° + x + x = 180°
45 + 2x = 180°
2x = 180 – 45
2x = 135
x = 135/2
x = 67.5
Now,
∠POC = ∠BOC – ∠BOP
= 90° – x
= 90 – 67.5
∠POC = 22.5
(ii) ∠BDC = 2∠POC
= 2 × 22.5
= 45°
(iii) ∠BOP = 3∠COP
= 3 × 22.5
= 67.5°
(5) The given figure shows a square ABCD and an equilateral triangle ABP calculate
(i) ∠AOB
(ii) ∠BPC
(iii) ∠PCD
(iv) Reflex ∠APC
Given, ABP is an equilateral triangle. (Photo)
Find :
(i) ∠AOB
(ii) ∠BPC
(iii) ∠PCD
(iv) ∠Reflex ∠APC
Solution : In △AOB,
(i) Sum of all interior angles = 180°
∠AOB + ∠OAB + ∠OBA = 180°
∠AOB + 60° + 45° = 180°
∠AOB + 105° = 180°
∠AOB = 180° – 105°
∠AOB = 75°
(ii) In △BPC,
∠BPC + ∠BCP + ∠PBC = 180°
x + x + 30° = 180°
2x + 3o° = 180°
2x = 180° – 30°
2x = 150°
x = 150/2
x = 75°
(iii) Now,
∠BCD = ∠PCD + ∠BCP
90° = ∠PCD + 75°
90° – 75° = ∠PCD
15° = ∠PCD
(iv) ∠APC = ∠APB + ∠BPC
= 60° + 75°
∠APC = 135°
reflex ∠APC = 360° – 135°
= 225°
Here is your solution of Selina Concise Class 9 Maths Chapter 14 Rectilinear Figures Exercise 14B
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.
For more solutions, See below ⇓