Chapter 5: Understanding Quadrilaterals & Practical Geometry

Solution:-

(a) 40^o

From the given figure TS ∥ BE and also BS is transversal line.

By the rule of alternate interior angles, ∠EBS = ∠BST = 40^o

Then, ∠y = 90^o … [∵diagonal bisect at 90^o]

Consider triangle TSO,

By the rule of exterior angle property of triangle

∠STO + ∠TSO = ∠SOE

x + 40^o = 90^o

x = 90^o – 40^o

x = 50^o

So, the value of y – x is = 90^o – 40^o = 50^o

Solution:-

The closed curve which is also a polygon is figure (a). Because there is no line segments intersect each other.

Solution:-

(d) Each exterior angle = ((n – 2) × 180^o)/n)

Because each exterior angle is equal to 360^o/n

Solution:-

(a) Right angle

The diagonals in the square intersect each other at right angle i.e. 90^o

Therefore, ∠POQ is a right angle.

Solution:-

(a) 30^o, 150^o, 30^o, 150^o

We know that, sum of adjacent angles of a parallelogram = 180^o

Let us assume two angles be x and 5x

Then,

x + 5x = 180^o

6x = 180^o

x = 180^o/6

x = 30^o

Therefore the two angles are x = 30^o

5x = 5 × 30 = 150^o

Solution:-

(b) rectangle

Solution:-

(b) trapezium with PQ || RS

We know that, sum of all interior angle of quadrilaterals is equal to 360^o.

Let us assume the angles be x, 3x, 7x, and 9x

Then,

x + 3x + 7x + 9x = 360^o

20x = 360^o

x = 360/20

x = 18

Therefore the angles are P= x = 18^o

Q = 3x = 3 × 18 = 54^o

R = 7x = 7 × 18 = 126^o

S = 9x = 9 × 18 = 162^o

Therefore, PQ || RS

Solution:-

(a) 70^o

We know that, the adjacent angles in a trapezium are supplementary.

∠R + ∠Q = 180^o

∠R + 110^o = 180^o

∠R = 180^o – 110^o

∠R = 70^o

Solution:-

Now let us assume number of sides of a regular polygon be n.

WKT, sum of all exterior angles of all polygons is equal to 360^o.

Form the question it is given that each exterior angle has a measure of 45^o.

Then,

n = 360^o/Exterior angle

n = 360^o/(180^o – 135^o)

n = 360^o/45^o

n = 8

Solution:-

(c) rhombus