Chapter 5: Understanding Quadrilaterals & Practical Geometry

Solution:-

(a) 72^o, 108^o

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

Let us assume two angles be 2x and 3x

Then,

2x + 3x = 180^o

5x = 180^o

x = 180^o/5

x = 36^o

Therefore the two angles are 2x = 2 × 36 = 72^o

3x = 3 × 36 = 108^o

Solution:-

(d) 0^o

We know that opposite angles ∠P and ∠R are equal in parallelogram.

So, ∠P – ∠R = 0^o

Solution:-

(a) 180^o

Solution:-

(b) 150^o

ABCD is a parallelogram.

From the question it is given that, ∠EBF = 30^o

WKT, sum of interior angles of a quadrilateral = 360^o

Then,

∠EBF + ∠BED + ∠EDF + ∠DFB = 360^o

∠EDF = 360^o – (90^o + 90^o + 30^o)

∠EDF = 150^o which is an obtuse angle.

Solution:-

(a) 60^o

From the given figure,

∠BAD = 30^o

∠BCD = 30^o … [∵opposite angles of parallelogram are equal]

Now, let us consider the triangle CBD

From angle sum property, ∠DBC + ∠BCD + ∠CDB = 180^o

90^o + 30^o + ∠CDB = 180^o

120^o + ∠CDB = 180^o

∠CDB = 180^o – 120^o

∠CDB = 60^o

∴∠BEC = 60^o, because opposite angles of parallelogram are equal.

Solution:-

(c) 26 cm

PQRS is a rectangle,

Where SR = 24 cm, QR = 10 cm

Now, consider the triangle QRS

From the rule of Pythagoras theorem,

QS² = SR² + QR²

QS² = 24² + 10²

QS² = 576 + 100

QS² = 676

QS = √676

QS = 26 cm

Solution:-

(a) rectangle

Solution:-

(a) 140^o, 40^o, 20^o, 160^o

We know that sum of interior angles of quadrilaterals is 360^o.

So, 140^o + 40^o + 20^o + 160^o = 360^o

In option (d) has angle sum equal to 360^o, but one angle is 180^o if it is considered then the quadrilateral becomes a triangle.

Solution:-

(c) equal to 360^o

We know that sum of angles of concave and convex quadrilateral is equal to 360^o.

Solution:-

(a) 22^o

We know that, Sum of exterior angles of a polygon is equal to 360⁰

If we divide 360⁰ by any one of the angles must be a whole number since it gives the number of sides.

Then, 360^o divide by 22 it gives fraction. So 22^o can never be the measure of exterior angle of a regular polygon.