NCERT Solutions for Class 8 Maths Chapter 5 - Understanding Quadrilaterals & Practical Geometry

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Answer:

Sum of angles of quadrilateral = 360º
Let fourth angle be xº
∴ 75º + 75º + 75º + x = 360º  (∵ three angles = 75º each)
⇒ 225º + x = 360º
⇒ x = 135º

Hence, the correct answer is option B. 
 

Answer:

We know, Diagonals of squares bisect each other. But diagonals of kite, trapezium and quadrilateral do not bisect each other.

Hence, the correct answer is option A.

Answer:

In rectangle, all angles are equal, i.e., each 90º.

Hence, the correct answer is option A.

Answer:

Diagonals of kite are perpendicular to each other.

Hence, the correct answer is option B.

Answer:

For rectangle, the diagonals are equal.

Hence, the correct answer is option D.

Answer:

All the properties are satisfy by figure R.

As, figure R has

(i) all its side congruent.

(ii) all angles are right angles.

(iii) opposite sides are parallel.

Hence, the correct answer is option C.

Answer:

We know, Rhombus has two pair of congruent adjacent sides.

Hence, the correct answer is option B.

Answer:

Given: Only one pair of sides are parallel. This property is satisfy by figure P.

Hence, the correct answer is option A. 

Answer:

Given properties are not satisfy by figure P only.

As figure P does not have

(i) all angles right angles.

(ii) parallel opposite sides.

Hence, the correct answer is option A.

Answer:

We know, In trapezium, a pair of opposite sides are parallel.

Hence, the correct answer is option A.

 

Answer:

In a parallelogram, opposite sides are parallel.

Hence, the correct answer is option A.

Answer:

We know, sum of angles of quadrilateral is 360º
Also, an obtuse angle is more than 90º
but less than 180º
Thus, almost 3 angles of quadrilateral can be obtuse.

Hence, the correct answer is option C.

Answer:

We can make (n – 2) non-overlapping triangles in a n-gon by joining the vertices.

Hence, the correct answer is option B.

Answer:

We know that,

Sum of interior angles of a polygon = (n − 2) × 180º

∴ Sum of interior angles of a pentagon = (5 − 2) × 180º = 540º

Hence, the correct answer is option C.

Answer:

We know that,
Sum of angles of a polygon = (n – 2) × 180º

Here n = 6 (for hexagon)

Sum of angles of a hexagon = (6 – 2) × 180º = 720º

Hence, the correct answer is option D.  
                        

Answer:

In parallelogram , sum of adjacent angles = 180º
$$\begin{gathered} \left(5x–5\right)º+\left(10x+35\right)º=180º \\ \Rightarrow 15x=150º \\ \Rightarrow x=10º \end{gathered}$$

∴ Required angles are
$$\begin{gathered} \left(5x–5\right)º=\left(5\times 10 –5\right)º=45º \\ \mathrm{And} \left(10x+35\right)°=\left(10\times 10 +35\right)º=135º \end{gathered}$$

Now required ratio$=\frac{45}{135}=\frac{1}{3}=1:3$

Hence, the correct answer is option A.
 

Answer:

A quadrilateral, where all sides are equal, opposite angles are equal and diagonals bisect each other at right angle is rhombus.

Hence, the correct answer is option A.

Answer:

In rectangle, opposite sides are equal and angles are equal.

Hence, the correct answer is option A.

Answer:

In square, all sides, angles and diagonals are equal.

Hence, the correct answer is option A.

Answer:

Number of diagonal in a polygon having n sides = $\frac{n\left(n-3\right)}{2}$
Now, for hexagon (n = 6)

∴ Number of diagonal$=\frac{6\left(6-3\right)}{2}=9$

​Hence, the correct answer is option A.

Answer:

Parallelogram, in which adjacent sides are equal is rhombus.

Hence, the correct answer is option C.

Answer:

In rectangle, diagonals are equal and bisect each other.

Hence, the correct answer is option B.

Answer:

Sum of exterior angle of a triangle = 360º.

Hence, the correct answer is option B.

Answer:

In square, all sides and all angle are equal.
Thus, square is a equiangular and equilateral polygon.

Hence, the correct answer is option A.

Answer:

Rhombus has all properties of parallelogram and kite.

Hence, the correct answer is option B.

Answer:

Let angles be x, 2n, 3x, and 4x

Sum of angles of quadrilateral = 360º

$$\begin{gathered} \therefore x+2x+3x+4x=360º \\ 10x=360º \\ x=36º \end{gathered}$$

Thus, smallest angle is 36º

Hence, the correct answer is option C.

Answer:

We know, sum of supplementary angles = 180º
Then, 
$$\begin{gathered} \angle D+\angle A=180º \\ \angle D+55º=180º \\ \angle D=180º-55º \\ \angle D=125º \end{gathered}$$

Hence, the correct answer is option D.

Answer:

We know, sum of angles of quadrilateral  = 360º
Let fourth angle be xº
Given: 3 angles are 80º each
3 × 80º + xº = 360º
x + 240º = 360º
x = 120º

Hence, the correct answer is option B.

Answer:

Sum of exterior angle of polygon = 360º
Each angle = 45º (given)

∴ Number of sides$=\frac{360º}{45º}=8$

Hence, the correct answer is option is A.

Answer:

In parallelogram,
sum of supplementary angles = 180º
Here, ∠P and ∠Q are supplementary
$$\begin{gathered} \therefore \angle P+\angle Q=180º \\ 60º+\angle Q=180º \\ \angle Q=120º \end{gathered}$$

Now, 
$$\begin{gathered} \angle P=\angle R=60º \left(\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal}\right) \\ \angle Q=\angle S=120º \left(\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal}\right) \end{gathered}$$

Hence, the correct answer is option B.

Answer:

Let the angles be 2x and 3x
In parallelogram,
sum of supplementary angles = 180º
2x + 3x = 180º
⇒ 5x = 180º
⇒ x = 36º
Thus, angles are 72º, and 108º.

Hence, the correct answer is option A. 
 

Answer:

In parallelogram, opposite angles are equal.

$\therefore \angle P–\angle R=0º$

Hence, the correct answer is option D.

Answer:

In parallelogram
Sum of adjacent angles = 180º

Hence, the correct answer is option A.
 

Answer:

∠EBF = 30º (given)
Sum of angles of quadrilateral = 360º
∴ ∠EBF + ∠BED + ∠EDF + ∠DFB = 360º
⇒ ∠EDF = 360º – (90 + 90 + 30º)
⇒ ∠EDF = 150º

Hence, the correct answer is option B.

Answer:

In parallelograms ABCD,

∠BCD = ∠A = 30º (∵ Opposite angles)

In â–³DBC, using angle sum property, we get
∠DBC + ∠BCD + ∠BDC = 180º
⇒ ∠BDC = 180º – 120º = 60º

In parallelograms BDCE

∠BEC = ∠BDC = 60º (∵ Opposite angles)

Hence, the correct answer is option A.

Answer:

Given: Ab = 24 cm, Bc = 10 cm
From phythagoras theorem

$$\begin{gathered} {\left(AD\right)}^2={\left(AD\right)}^2+{\left(BC\right)}^2 \\ \Rightarrow {\left(AD\right)}^2={\left(24\right)}^2+{\left(10\right)}^2 \\ \Rightarrow {\left(AD\right)}^2=576 \\ \Rightarrow AD=26 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option C.

Answer:

If the adjacent angles of a parallelogram are equal, then since, adjacent angles are supplementary. Therefore, each angle will be 90º.
Thus, parallelogram will be rectangle.

Hence, the correct answer is option A.

Answer:

Sum of interior angle of quadrilateral = 360º
So, 140º + 40º + 20º + 160º = 360º

Hence, the correct answer is option A.

Answer:

Sum of interior angles of any polygon (concave or convex) = (n – 2) × 360º
Thus, sum of concave quadrilateral = 360º

Hence, the correct answer is option C.

Answer:

Sum of exterior angle of polygon = 360º

∴ Measure of each angle = $\frac{360}{n}$

Hence, n = number of sides.
Thus, measure of each exterior angle will always divide 360º connectively.
Thus, 22º can never be the measure of as exterior angle.

Hence, the correct answer is option A.

Answer:

Here, TS || BE
∠SBE = ∠BST = 40º
⇒ y = 90º                   (Diameters of a rhombus are intersect each other) 

In âˆ†TSD
∠STO + ∠OST = ∠EOS   (exterior angle property)
⇒ x + 40º = 90º
⇒ x = 50º
So, y – x = 90º – 50º = 40º

Hence, the correct answer is option A.
 

Answer:

A polygon is a simple closed curve made up of line segments.

Hence, the correct answer is option A.

Answer:

$\mathrm{Each} \mathrm{exterior} \mathrm{angle}=\frac{360°}{n}$

Formula in option D is not correct.

Hence, the correct answer is option D.

Answer:

Diagonal in square intersect each other at 90º.

∴ ∠POQ = 90º

Hence, the correct answer is option A.

Answer:

Let the angles be x and 5x 
x + 5x = 180º      (Supplementary angles)
⇒ 6x = 180º
⇒ x = 30º
⇒ 5x = 150º

Thus, angles are 30º, 150º, 30º, 150º

Hence, the correct answer is option A.

Answer:

In a parallelogram, if one angle is 90º then all angles will be 90º.

Thus, parallelogram is rectangle.

Hence, the correct answer is option B.

Answer:

Let the angle be x, 3x, 7x and 9x, then sum of angles of quadrilateral  = 360º
⇒ x + 3x + 7x + 9x = 360º
⇒ 20x = 360º
⇒ x = 18º
Thus, angles are 18º, 54º, 126º, and 162º

Since ∠P + ∠S = 180º , ∠Q + ∠R = 180º
Thus, PQ || RS.

Hence, the correct answer is option B.
 

Answer:

Given PQ || SR
∠R + ∠Q = 180º      (Supplementary angles)
⇒ ∠R + 110º = 180º
⇒ ∠R = 70º

Hence, the correct answer is option A.

Answer:

Sum of interior angles of polygon = (n – 2) × 180º
Since, polygon is interior,

$$\begin{gathered} \therefore \left(n-2\right)\times 180º=135º\times n \\ \Rightarrow 180º\times n-360º = 135º\times n \\ \Rightarrow 45º\times n=360º \\ \Rightarrow n=8 \end{gathered}$$

Hence, the correct answer is option C.

Answer:

In rhombus, diagonals bisect both the angles.

Hence, the correct answer is option C.

Answer:

To construct a parallelogram, we require the measurement of any two non-parallel sides and measure of an angle.
Thus, a minimum of 3 measurement are required.

Hence, the correct answer is option B.

Answer:

Since, in rectangle opposite sides are equal and parallel.
So, we need the measurement of only two adjacent sides.
Thus, we required only two measurement. 

Hence, the correct answer is option C.

Answer:

Pair of opposite side = HO and PE, PO and HE.

Answer:

In figure Rope, ∠R and ∠O, ∠O and ∠P, ∠P and ∠E, ∠E and ∠R are adjacent angles.

Answer:

In figure WXYZ, ∠W and ∠Y, ∠X and ∠Z are opposite angles.

Answer:

Hence, the diagonals of the quadrilateral DEFG are DF and EG.

Answer:

Sum of all angles of quadrilateral is 360º.

Answer:

Measure of each exterior angle of a polygon of n side is $\frac{360°}{n}$.

∴ Measure of each exterior angle$=\frac{360°}{5}=72°$

Hence, the measure of each exterior angle of a regular pentagon is 72°.

Answer:

We know, sum of interior angle of regular polygon = (n – 2) × 180º.

Here, n = 6

∴ Sum of angles = (6 – 2) × 180º = 720º

Hence, the sum of the angles of a hexagon is 720º.

Answer:

Measure of each exterior angle of a polygon of n sides = $\frac{360°}{n}$

Here n = 18

∴ Measure of exterior angle$=\frac{360°}{18}=20°$

Hence, the measure of each exterior angle of a regular polygon of 18 sides is 20°.

Answer:

We know, measure of each exterior of regular polygon = $\frac{360°}{n}$
Each exterior angle = 36º    (given)

$\therefore 36º=\frac{360°}{n}$

$\begin{array}{rcl}& \Rightarrow & n=\frac{360º}{36º}=10\end{array}$

Hence, the number of sides of a regular polygon, where each exterior angle has a measure of 36°, is 10.

Answer:

The closed curve entirely made up of line segments is called polygon.

Hence, the another name for this shape is polygon.

Answer:

Quadrilateral that is not a parallelogram but has exactly two opposite angles is kite.

Answer:

Sum of angles of regular pentagon = (n – 2) × 180º

= (5 – 2) × 180º 

= 540º

Measure of each interior angle = $\frac{540º}{5}$ = 108º

Hence, the measure of each angle of a regular pentagon is 108º.

Answer:

Three sided regular polygon is known as triangle.

Answer:

Number of diagonal is regular polygon = $\frac{n\left(n–3\right)}{2}$

Here, n = 6
Number of diagonals = $\frac{6\left(6-3\right)}{2}$ = 9

Hence, the number of diagonals in a hexagon is 9.

Answer:

Polygon is closed curve made up of only line segments.

Answer:

In regular polygon, all sides are equal and all angles are equal.
 

Answer:

Sum of interior angles of regular polygon = (2n – 4) × 90º

Hence, he sum of interior angles of a polygon of n sides is (2n – 4) right angles.

Answer:

The sum of all exterior angles of a polygon is 360º.

Answer:

Square is a regular quadrilateral.

Answer:

In parallelogram, pair of opposite sides are parallel.

Hence, a quadrilateral in which a pair of opposite sides is parallel is parallelogram.

Answer:

In square and rhombus, all sides are equal.

Hence, if all sides of a quadrilateral are equal, it is a rhombus/square.

Answer:

In a rhombus, diagonal intersect at right angles.

Answer:

We know to construct a unique quadrilateral, we require 5 measurement, i.e., four sides and one angle or three sides and two included angles or two adjacent sides and three angles are given.

Hence, 5 measurements can determine a quadrilateral uniquely.

Answer:

Two

A quadrilateral can be construct uniquely if its three sides and two angles are given.

Answer:

All
In rhombus, all sides are equal.

Answer:

One
In concave polygon, atleast one interior angle is more than 180º.

Answer:

Opposite
Diagonal of a quadrilateral is a line segment that joins two opposite vertices.

Answer:

Sum of exterior angle of regular polygon = $\frac{360°}{n}$

$$\begin{gathered} 72°=\frac{360º}{n} \\ \Rightarrow n=5 \end{gathered}$$

Hence, number of sides is 5.

Answer:

Parallelogram.
If diagonal of  quadrilateral bisect each other, than it is parallelogram.

Answer:

Perimeter of parallelogram = 2 (length + breadth)
                                         =  2 (5 + 9)
                                         = 28 cm.

Hence, the perimeter is 28 cm.

Answer:

9
A nonagon has 9 sides.

Answer:

Equal
Diagonal of a rectangle are equal.

Answer:

Decagon
Polygon having 10 sides is called Decagon.

Answer:

Square
Rectangle whose adjacent sides are equal is called square.

Answer:

Since diagonals of rectangle are equal.

Hence, if one diagonal of a rectangle is 6 cm long, length of the other diagonal is also 6 cm.

Answer:

Sum of adjacent angles of parallelogram is 180º.

Hence, adjacent angles of a parallelogram are supplementary.
.

Answer:

In Kite one diagonal bisects the other.
Hence, if only one diagonal of a quadrilateral bisects the other, then the quadrilateral is known as kite.

Answer:

Since, AB || Cd, ∠A and ∠D are supplementary
∴ ∠A + ∠D = 180º
⇒ 100º + ∠D = 180º
⇒ ∠D = 80º
Hence, in trapezium ABCD with AB||CD, if ∠A = 100°, then ∠D = 80º.

Answer:

A quadrilateral is a polygon in which the sum of both interior and exterior angles are equal.

Hence, the polygon in which sum of all exterior angles is equal to the sum of interior angles is called a quadrilateral. 

Answer:

False.
In a trapezium, all angles are not equal.

Answer:

True.
Every square is a rectangle because it is quadiladeral with all 4 right angles.

Answer:

False.
Because a kite is a quadilateral with two distinct pairs of adjacent congruent sides.

Answer:

True
A rectangle is a parallelogram with four right angles.

Answer:

False.
All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent. Diagonals of a rhombus always bisect each other at right angles.

Answer:

False
Sum of all the angles of quadrilateral is 360º.

Answer:

True.
Quadrilateral is a polygon which has 4 side and two diagonals.

Answer:

True.
In triangle,
Sum of interior angle = 180º
Sum of exterior angle = 360º
Hence, triangle is a polygon whose sum of exterior angle is double the sum of interior angles.

Answer:

False.

It is not a polygon, because it is not a simple closed curve as it intersects with itself more than once.

Answer:

False.
A kite is a convex quadrilateral.

Answer:

True.
Sum of interior and exterior angles of quadilateral, both are equal to 360º.

Answer:

True.
Since, the sum of exterior angle of a hexagon is 360º and the sum of interior angles of a hexagon is 720º. i.e., double sum of exterior angles.

Answer:

False.
Because, a polygon is regular if all sides and all angles are equal.

Answer:

False.
Because, in rectangle all angle are equal to 90º but all sides are not equal.

Answer:

True.
If diagonals of quadrilateral are equal, it must be a rectangle.

Answer:

True.
If opposite angles are equal, it must be a parellogram.

Answer:

False. 

Let the angles be x, 2x and 3x (∵ Ratio = 1 : 2 : 3)
∴ x + 2x + 3x = 180º
⇒ 6x =  180º 
⇒ x = 30º

Thus, angles are 30º, 60º, 90º.

Let exterior angles be  $\alpha , \beta \mathrm{and} \gamma$.

Now $\alpha$ = 180º – 30º
            = 150º

Also, $\beta$ = 180º – 60º
             = 120º

And $\gamma$ = 180º – 120º
           = 60º
Now,

 $\alpha : \beta : \gamma$ = 150º : 120º : 60º = 5 : 4 : 3

Answer:

False.
Pentagon has five sides, but here we have 6 sides.

Hence, it is a concave hexagon.

Answer:

False.
Because diagonal of rhombus are perpendicular but are not equal to each other.

Answer:

True.
Diagonal of a rectangle are equal.

Answer:

False.
Diagonals of a rectangle do not bisect each other at right angles.

Answer:

False.
Kite is not a parallelogram, as its opposite sides are not equal and parallel.

Answer:

False.
Because in trapezium only one pair of sides is parallel.

Answer:

False.
Every rectangle is a parellogram but the inverse is not true.

Answer:

False.

Since in a rectangle, opposite sides are equal and parallel but in a trapezium it is not.

So, every rectangle is a trapezium but every trapezium is not a rectangle.

Answer:

True.
Every rectangle is a trapezium as a rectangle satisfy all the properties of trapezium.

Answer:

True.
Because a square possess all the properties of a rhombus but vice-versa is not true.

Answer:

True.
Because a square possess all the properties of a parallelogram.

Answer:

True.
Because a square possess all the properties of a trapezium.

Answer:

True.
Because a rhombus possess all the properties of a trapezium but vice-versa is not true.

Answer:

False.
Because we require atleast five measurement to determine a quadilateral uniquely.

Answer:

False.
Because, if all four angles are obtuse, then the sum will exceed 360º.

Answer:

True
A quadrilateral can be drawn if four sides and one diagonal is known.

Answer:

False
We cannot draw a unique-quadrilateral, if four angles and one side is known.

Answer:

True
A quadrilateral can be drawn, if all four sides and one angle is known.

Answer:

True
A quadrilateral can be drawn, if three sides and two diagonals are given.

Answer:

True
If diagonals of a quadrilateral bisect each other, it must be a parallelogram. It is the property of a parallelogram.

Answer:

True
We can construct a unique quadrilateral can be constructed uniquely if three angles and any two sides are given.

Answer:

True
We can construct a unique parallelogram, if both diagonals and the angle between them is given.

Answer:

True
A rhombus can be constructed uniquely, if both diagonals are given.

Answer:

Let ABCD be a rhombus.

Given: AC = 15 cm and BD = 8 cm

Since the diagonals of a rhombus bisect each other at 90º.

Therefore, in âˆ†AOB,

AB² = OA² + OB²

$$\begin{gathered} \Rightarrow {\mathrm{AB}}^2={\left(\frac{15}{2}\right)}^2+{\left(\frac{8}{2}\right)}^2 \\ \Rightarrow {\mathrm{AB}}^2=72.25 \\ \Rightarrow \mathrm{AB}=8.5 \end{gathered}$$

Since it is a rhombus all sides are equal.

Hence, length of each side is 8.5 cm.
 

Answer:

Let the adjacent angles of a parallelogram be x and 3x.
 
Since adjacent angles of parallelogram are supplementary

∴ x + 3x = 180°
⇒ 4x = 180°
⇒ x = 45°
Thus, the angles are 45°, 135°.

Opposite angles in a parallelogram are equal.

Hence, the angles are 45°, 135°, 45°, 135°. 

Answer:

In square, rectangle and rhombus, opposite sides are parallel and equal.
Also, opposite angles are equal, i.e. they all are parallelograms.
But in trapezium, there is only one pair of parallel sides, i.e. it is not a parallelogram.
Hence, trapezium has different design.

Answer:

We have, AB = 25 cm and BC = 15 cm

Let AO be x, then

OB = AB − x = 25 − x

In rectangle ABCD,
CO is the bisector of ∠C and divides AB. So,
∠OCB = ∠OCD = 45º
In âˆ†OCB,
∠CBO + ∠OCB + ∠COB = 180º
⇒ ∠COB = 180º − 90º − 45º 
⇒ ∠COB = 45º 

In âˆ†OCB,
∠OCB = ∠COB
As the angle opposite to equal sides are equal.

∴ OB = BC = 15 cm

⇒ 25 − x = 15

⇒ x = 10 cm

Hence, AO : OB = x : 25 − x = 10 : 15 = 2 : 3.

Answer:

Given, ST ⊥ PR and ST divides ∠S in the ratio 2 : 3.

So,
∠TSP = $\frac{2}{5}\times 90°=36°$

∠TSR = $\frac{3}{5}\times 90°=54°$

Also, by angle sum property of a triangle,
∠TPS = 180º − (∠STP + ∠TSP)
          = 180º − (90º + 36º)
          = 54º 

Now, ∠SPQ = 90º
⇒ ∠TPS + ∠TPQ = 90º
⇒ ∠TPQ = 36º

Answer:

No, the photo frame cannot be a rectangle as, in a rectangle both the diagonals are of equal lengths.

Answer:

(2x – 4)° + (3x – 1)° = 180°  (The adjacent angles of a parallelogram are supplementary)

$$\begin{gathered} \Rightarrow 5x-5°=180° \\ \Rightarrow 5x=185° \\ \Rightarrow x=37° \end{gathered}$$

Thus, the adjacent angles are:
$$\begin{gathered} 2\left(37°\right)-4=70° \\ 3\left(37°\right)-1=110° \end{gathered}$$

Hence, all the angles are 70°, 110°, 70° and 110°.

Answer:

No, it can never be a parallelogram, as the diagonals of a parallelogram intersect each other in the ratio 1 : 1.

Answer:

Let the exterior and interior angle be x and 5x respectively.

Since exterior angle and corresponding interior angles are supplementary.

Then, 
x + 5x = 180º
⇒ 6x = 180º
⇒ x = 30º

$\begin{array}{rcl}\therefore \mathrm{Number} \mathrm{of} \mathrm{sides}& =& \frac{360°}{\mathrm{Exterior} \mathrm{Angle}}\\ & =& \frac{360°}{30°}\\ & =& 12\end{array}$

Hence, there are 12 sides in the polygon.

Answer:

Sticks can be taken as the diagonals of a quadrilateral.
Now, since they are bisecting each other, therefore the shape formed by joining their end points will be a parallelogram.
Hence, it may be a rectangle or a square depending on the angle between the sticks.

Answer:

Sticks can be treated as the diagonals of a quadrilateral.
Since the diagonals are equal and bisecting each other at right angles, therefore the shape formed by joining their end points will be a square.

Answer:

Let the length of other non-consecutive sides be x metres.
Then, perimeter of the playground = 23 + 23 + x + x
⇒ 106 = 2(23 + x)
⇒ 46 + 2x = 106 
⇒ 2x = 106 – 46
⇒ 2x = 60
⇒ x = 30 m

Since a kite has two pairs of equal consecutive sides.

Hence, the lengths of other three sides are 23 m, 30 m and 30 m.

Answer:

∠EOA = 180º − 60º = 120º            (Linear pair angles)
Now, in âˆ†EOA, 
∠OEA = ∠OAE 
⇒ OE = OA                                    (In triangle sides opposite to equal angles are equal)
So,
∠EOA + ∠OEA + ∠OAE = 180º   (Angle sum property of a triangle)
⇒ ∠OAE = ∠EAR = 30º

∠RAD = 90º − 30º = 60º
∠ROD = ∠EOA = 120º                  (Vertically opposite angles)

Hence, ∠EAR = 30º, ∠RAD = 60º and ∠ROD = 120º.

Answer:

Given: ∠RAI = 35º
So, ∠PRA = 35º              (Alternate interior angles)


In â–³ARI, using Angle sum property, we get

∠ARI + ∠RAI + ∠RIA = 180º
⇒ ∠ARI + 90º + 35º = 180º
⇒ ∠ARI = 55º

Now, AM = MI                  (Diagonals of a rectangle bisect each other)
So, ∠RAI = ∠PIA = 35º     (Angle opposite to equal sides of a triangle are equal)

In âˆ†AMI, using Angle sum property, we get
∠MAI + ∠MIA + ∠AMI = 180º
⇒ 35º + 35º + ∠AMI = 180º
⇒ ∠AMI = 110º

Now, 
∠AMI + ∠RMI = 180º         (Linear pair of angles)

⇒ 110º + ∠RMI = 180º

⇒ ∠RMI = 70º

And, ∠PMA = ∠RMI = 70º  (Vertically opposite angles)

Hence, ∠ARI = 55º, ∠RMI = 70º and ∠PMA = 70º.

Answer:

∠C = ∠A = 80º     (Opposite angles in a parallelogram are equal)
Also,
∠A + ∠D = 180º   (Adjacent angles in a parallelogram are supplementary)
⇒ 80º + ∠D = 180º
⇒ ∠D = 100º 
And,
∠D = ∠B = 100º

Hence, ∠B = 100º, ∠C = 80º and ∠D = 100º.

Answer:

∠PQR + ∠RQY = 180º       (Linear pair of angles)
⇒ ∠PQR + 60º = 180º
⇒ ∠PQR = 120º

Now, 
∠PQR = ∠S = 120º             (Opposite angles in a parallelogram are equal)

Also, 
∠RQY = ∠R = 60º               (SR || PQY, RQ is a transversal and the angles are alternate interior angles)
And,
∠P = ∠R = 60º                     (Opposite angles in a parallelogram are equal)

Now, PQ = SR = 15 cm        (Opposite sides of a parallelogram are equal)
And, PS = QR = 11 cm    (Opposite sides of a parallelogram are equal)

Diagonal PR = 2 × PO     (Diagonals of a parallelogram bisect each other)
⇒ PR = 12 cm

Hence, ∠S = 120º, ∠R = 60º, PQ = 15 cm, QR = 11 cm and diagonal PR = 12 cm.

Answer:

Since, the diagonals of a rhombus bisect each other at 90º, ∠AOM = 90º.
So, in âˆ†AMO,
∠AOM + ∠OMA + ∠MAO = 180º
⇒ 90º + ∠OMA + 70º = 180º
⇒ ∠OMA = 20º
⇒ ∠AME = 20º                     (∵ ∠OMA = ∠AME)

Also, AM = AE so,

Since angles opposite to equal sides are equal.

∴ ∠OMA = ∠AEM = 20º

Answer:

In parallelogram FIST, FI || TS and TI is a transversal. So,
∠FIT = ∠STI = ∠STO = 35º
∠SIT = ∠OTF = 25º

And, 
∠FOT + ∠SOT = 180º    (Linear pair)
⇒ 110º + ∠SOT = 180º 
⇒ ∠SOT = 70º

In âˆ†SOT,
∠SOT + ∠STO + ∠OST = 180º   (Angle sum property)
⇒ 70º + 35º + ∠OST = 180º
⇒ ∠OST = 75º

In âˆ†FOT,
∠FOT + ∠OTF + ∠TFO = 180º   (Angle sum property)
⇒ 110º + 25º + ∠TFO = 180º
⇒ ∠TFO = 45º = ∠SFT

Hence, ∠SFT = 45º, ∠OST = 75º and ∠STO = 35º.

Answer:

∠RUO = ∠RYO = 120°  (Opposite angles of a quadrilateral are equal)
And, 
∠RYO + ∠RYS = 180°   (Linear pair)
⇒ 120° + ∠RYS = 180°
⇒ ∠RYS = 180° − 120° = 60°

In âˆ†RSY,
∠YSR + ∠SRY + ∠RYS = 180°
⇒ ∠YSR + 50° + 60° = 180°
⇒ ∠YSR = 70°

Answer:

In âˆ†WEA,
∠WEA + ∠EAW + ∠AWE = 180º
⇒ 70º + 2∠EAW = 180º                    (∠EAW = ∠AWE as angles opposite to equal sides are equal)
⇒ 2∠EAW = 110º
⇒ ∠EAW = 55º = ∠AWE

Similarly in âˆ†WRA,
∠WRA + ∠RAW + ∠AWR = 180º
⇒ 80º + 2∠RAW = 180º                    (∠RAW = ∠AWR as angles opposite to equal sides are equal)
⇒ 2∠RAW = 100º
⇒ ∠RAW = 50º = ∠AWR

So, ∠RWE = ∠AWR + ∠AWE = 50º + 55º = 105º
And, ∠RAE = ∠RAW + ∠EAW = 50º + 55º = 105º
 

Answer:

(i) Yes, RE = OM as opposite sides of a rectangle are equal.

(ii) Yes, ∠MYO = ∠RXE as RX and MY are perpendicular to OE.

(iii) Yes, ∠MOY = ∠REX as RE || OM, OE is a transversal and the angles are alternate interior angles.

(iv) Yes, âˆ†MYO ≅ ∆RXE as RE = OM, ∠MYO = ∠RXE and ∠MOY = ∠REX so ASA criteria applies.

(v) Yes, MY = RX by CPCT as âˆ†MYO ≅ ∆RXE. 

Answer:

In âˆ†STM,
∠STM + ∠SMT + ∠TSM = 180º   (Angle sum property)
⇒ ∠STM + 90º + 40º = 180º
⇒ ∠STM = 50º

Now, 
∠STM = ∠SON = 50º                    (Opposite angles in a parallelogram are equal)

And, in âˆ†SON,
∠SON + ∠SNO + ∠NSO = 180º   (Angle sum property)
⇒ 50º + 90º + ∠NSO = 180º
⇒ ∠NSO = 40º

Also, 
∠SON + ∠TSO = 180º
⇒ ∠SON + ∠TSM + ∠NSO + ∠NSM = 180º
⇒ 50º + 40º + 40º + ∠NSM = 180º
⇒ ∠NSM = 50º

Hence, ∠STM = 50º, ∠SON = 50º and ∠NSM = 50º.

Answer:

Since, EP and RP are angle bisectors of ∠E and ∠R therefore,
∠E = 2 × 25º = 50º
∠R = 2 × 30º = 60​º

Also, ER || AH and EH is a transversal so,
∠E + ∠H = 180º              (Interior angles on the same side of the transversal sum up to 180º)
⇒ 50º + ∠H = 180º
⇒ ∠H = 130º = ∠EHA

Similarly,
∠R + ∠A = 180º
⇒ 60º + ∠A = 180º
⇒ ∠A = 120º = ∠HAR

Hence, ∠HAR = 120º and ∠HAR = 120º.

Answer:

Since, MODE is a parallelogram,
∠EMO + ∠DOM = 180º                (Adjacent angles are supplementary)
$\Rightarrow \frac{1}{2}\angle \mathrm{EMO}+\frac{1}{2}\angle \mathrm{DOM}=90°$
⇒ ∠OMQ + ∠MOQ = 90º

Now, in âˆ†OMQ,
∠OMQ + ∠MOQ + ∠MQO = 180º
⇒ 90º + ∠MQO = 180º
⇒ ∠MQO = 90º 

Answer:

In âˆ†EFB,
∠EFB + ∠FBE + ∠BEF = 180º    (Angle sum property)
⇒ 2∠FBE + 90º  = 180º                (∠FBE = ∠BFE as angles opposite to equal sides are equal)
⇒ 2∠FBE = 90º
⇒ ∠FBE = 45º = ∠BFE

Since, every angle in a rectangle is 90º therefore,
∠BFE + y = 90º
⇒ 45º + y = 90º
⇒ y = 45º

And, 
∠FBE + x = 180º                 (Linear pair)
⇒ 45º + x = 180º
⇒ x = 135º

Hence, x = 135º and y = 45º.
 

Answer:

FE || GC and FG is a transversal so,
∠EFG = ∠CGH = 30º            (Corresponding angles)

In parallelogram ABDH,
∠DBA = ∠DHA = 130º          (Opposite angles are equal)

Now, ∠DHA + ∠DHG = 180º 
⇒ 130º + ∠DHG = 180º
⇒ ∠DHG = 50º

By Angle sum property, we have

∠DHG + ∠CGH + x = 180º
⇒ 50º + 30º + x = 180º
⇒ x = 100º
 

Answer:

In âˆ†AOB, 
(AB)² = (OA)² + (OB)² 
⇒ (AB)² = 4² + 3²
⇒ (AB)² = 25
⇒ AB = 5 units

AB is the diameter of the semicircle.
So, radius = $\frac{5}{2}=2.5$ units.

Answer:

AM is perpendicular to DC.
∴ ∠AMC = 90º

Answer:

Here, EG = 2 × JF    (Diagonals of a rectangle are equal and bisect each other)
$$\begin{gathered} \Rightarrow 24x-8=2\left(8x+4\right) \\ \Rightarrow 24x-8=16x+8 \\ \Rightarrow 8x=16 \\ \Rightarrow x=2 \end{gathered}$$

Answer:

6y = 120º    (Opposite angles of a parallelogram are equal)
⇒ y = 20º

Also,
5x + 10º + 6y = 180º   (Adjacent angles of a parallelogram are supplementary)
⇒ 5x + 10º + 120º = 180º
⇒ 5x = 50º
⇒ x = 10º

Answer:

y = 110º                                   (One pair of angles of a kite are equal)
Also,
x + y + 110º + 60º = 360º        (Angle sum property of quadrilaterals)
⇒ x + 110º + 110º + 60º = 360º
⇒ x + 280º = 360º
⇒ x = 80º

Hence, x = 80º and y = 110º.

Answer:

Since the adjacent angles of a trapezium are supplementary,
$$\begin{gathered} \Rightarrow x-20°+x+40°=180° \\ \Rightarrow 2x+20°=180° \\ \Rightarrow 2x=160° \\ \Rightarrow x=80° \end{gathered}$$

Answer:

Let the equal angles be x.
Then,
$$\begin{gathered} 75°+75°+x+x=360° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow 2x+150°=360° \\ \Rightarrow 2x=210° \\ \Rightarrow x=105° \end{gathered}$$

The possible figure is a parallelogram since opposite angles are equal.

Answer:

∠P + ∠Q + ∠R + ∠S = 360º     (Angle sum property)
⇒ 50° + 50° + 60° + ∠S = 360º
⇒ ∠S = 200°

Since, one angle is obtuse, the quadrilateral is concave.
 

Answer:

Let the angles of the quadrilateral be ∠A, ∠B, ∠C, ∠D.
Then, ∠A = ∠C and ∠B = ∠D.
And,
∠A + ∠C = 180º
⇒ ∠A + ∠A = 180º
⇒ 2∠A = 180º
⇒ ∠A = 90º = ∠C

⇒ ∠A = ∠C = 90º

Similarly, ∠B = ∠D = 90º.

Answer:

Sum of interior angles of a regular polygon = (n − 2) × 180º
So, sum of interior angles of a regular octagon = (8 − 2) × 180º     (Number of sides of an octagon = 8)
                                                                           = 6 × 180º 
                                                                           = 1080º

So, measure of each angle = $\frac{1080º}{8}=135°$

Answer:

Exterior angle of a regular polygon = $\frac{360°}{n}$  (Where n is the number of sides of the polygon)
So,
Exterior angle of a regular pentagon = $\frac{360°}{5}=72°$   (Number of sides of a pentagon = 5)

Exterior angle of a regular decagon = $\frac{360°}{10}=36°$     (Number of sides of a decagon = 10)

∴ Required ratio = $\frac{72°}{36°}=2:1$

Answer:

Sum of exterior angles of a regular polygon = 360º

$$\begin{gathered} \Rightarrow 85°+20°+92°+89°+x°=360° \\ \Rightarrow x°=74° \end{gathered}$$

Hence, the value of x is 74.

Answer:

Let the three equal angles be x each.
Then,
$$\begin{gathered} x+x+x+120°=360° \\ \Rightarrow 3x=240° \\ \Rightarrow x=80° \end{gathered}$$

Hence, the measure of each equal angle is 80°.

Answer:

**Disclaimer: The data given in this question is insufficient to solve the problem.

Answer:

x + 100º = 180º   (Linear pair of angles)
⇒ x = 80º

Also, AD || BC and BD is a transversal. So,
z = 30º     (Alternate interior angles) 

Now, in âˆ†BOC, 
∠BOC + ∠COB + ∠BCO = 180º
⇒ 80º + 30º + y = 180º
⇒ y = 70º

Hence, x = 80º, y = 70º and z = 30º.

 

Answer:

Diagonals of a trapezium can also be perpendicular to each other. Hence, the quadrilateral does not necessarily need to be a rhombus.

Answer:

Let the angles be ∠A = 2x, ∠D = x, ∠B = 7y and ∠C = 5y.
Then,
∠A + ∠D = 180º    (Adjacent angles are supplementary)
⇒ 2x + x = 180º
⇒ 3x = 180º
⇒ x = 60º

Also,
∠B + ∠C = 180º    (Adjacent angles are supplementary)
⇒ 7y + 5y = 180º
⇒ 12y = 180º
⇒ y = 15º

So, angles of the trapezium are 
∠A = 2 × 60º = 120º
∠B = 7 × 15º = 105º
∠C = 5 × 15º = 75º
∠D = 1 × 60º = 60º
 

Answer:

 

Since, l || m and p is a transversal,
∠DXY = ∠AYX       (Alternate interior angles)
⇒ $\frac{1}{2}$∠DXY = $\frac{1}{2}$∠AYX
⇒ ∠PXY = ∠QYX                ( XP and YQ are angle bisectors)

So, XP || YQ as alternate interior angles are equal.
Similarly, XQ || YP.

Also, 
∠CXY + ∠DXY = 180º    (Linear pair)
⇒ $\frac{1}{2}$∠CXY + $\frac{1}{2}$∠DXY = 90º
⇒ ∠QYX + ∠PXY = 90º
⇒ ∠QYP = 90º

Since opposite sides are parallel and one angle is 90º, PXQY forms a rectangle.


 

Answer:

AY || XC     (Since, AB || CD)
Also,
∠A = ∠C
⇒ $\frac{1}{2}$∠A = $\frac{1}{2}$∠C
⇒ ∠XAY = ∠YCX

Also, ∠YCX = ∠CYB    (Alternate interior angles)
⇒ ∠CYB = ∠XAY

So, XA || CY   (Corresponding angles are equal)  

Hence, AXCY is a parallelogram.



 

Answer:

Given: Parallelogram ABCD bisects ∠A
To prove: Parallelogram ABCD bisects ∠C
Proof: ∠1 = ∠4 and ∠2 = ∠3    (AB || CD and the angles are alternate interior angles)
But, ∠1 = ∠2   (Since, AC bisects ∠A)

So, ∠3 = ∠4
Hence, proved.

Answer:

∠BED + ∠BEC = 180º    (Linear pair)
⇒ 90º + ∠BEC = 180º
⇒ ∠BEC = 90º

In âˆ†BEC,
∠BEC + ∠ECB + ∠CBE = 180º     (Angle sum property)
⇒ 90º + x + ∠CBE = 180º
⇒ ∠CBE = 90º − x

Similarly in âˆ†BAF,
∠BAF + ∠AFB + ∠FBA = 180º     (Angle sum property)
⇒ 90º + x + ∠FBA = 180º
⇒ ∠FBA = 90º − x

Now,
∠A + ∠B = 180º       (Adjacent angles of a parallelogram are supplementary)
⇒ x + ∠FBA + ∠FBE + ∠CBE = 180º
⇒ x + 90º − x + 45º + 90º − x = 180º
⇒ x = 45º

So, 
∠A = 45º
∠C = 45º

∠B = ∠FBA + ∠FBE + ∠CBE 
      = 90º − x + 45º + 90º − x
      = 90º − 45º + 45º + 90º − 45º
      = 135º 

∠D = ∠B = 135º

Answer:

Let ABCD be a rhombus and let DE be the perpendicular bisector of AB.
Now in âˆ†ADE and âˆ†BDE,
AE = BE                 (DE is the perpendicular bisector)
∠AED = ∠BED      (Both 90º)
DE = DE                 (Common)

⇒ âˆ†ADE â‰Œ ∆BDE by SAS congruency rule.
⇒ AD = DB = AB    (CPCT)

So, âˆ†ABD is an equilateral triangle.
⇒ ∠DAB = ∠ABD = ∠BDA = 60º
⇒ ∠DCB = ∠DAB = 60º           (Opposite angle of a rhombus are equal)

Now, ∠DAB + ∠ABC = 180º     (Adjacent angles are supplementary)
⇒ 60º + ∠ABD + ∠DBC = 180º
⇒ 60º + 60º + ∠DBC = 180º
⇒ ∠DBC = 60º

So, ∠ABC = 120º = ∠ADC

Answer:

∠C = 45° = ∠A   (Opposite angles in a parallelogram are equal)

Then, 
∠R = ∠A = 45°   (Opposite angles in a parallelogram are equal)

Answer:

Let ABCD be a parallelogram and the angle bisector of ∠A meet BC at F.
Construction: Join FE || AB.

Then, ABFE is a parallelogram by construction.
So, ∠1 = ∠6     (Alternate interior angles)
But, ∠1 = ∠2    (AF bisect ∠A)

So, ∠2 = ∠6.
⇒ AB = BF       (Sides opposite to equal angles are equal)
⇒ ABFE is a rhombus.     (Parallelogram with equal sides is a rhombus)

Now, in âˆ†ABO and âˆ†FBO,
OA = OF         (Diagonals of a rhombus bisect each other)
AB = BF          (Sides of a rhombus)
OB = OB         (Common)

So, âˆ†ABO â‰Œ âˆ†FBO by SSS criteria.
⇒ ∠3 = ∠4

Also, BF = $\frac{1}{2}$BC    (Given)

⇒ BF = $\frac{1}{2}$AD         (BC = AD)

⇒ EA = $\frac{1}{2}$AD         (BF = EA)

Hence, the angle bisector of B also bisects AD.

Answer:

Construction: Join FC 

Now, 
Measure of each interior angle of a regular polygon = $\frac{\left(n-2\right)}{n}\times 180°$

Measure of each interior angle of a regular pentagon = $\frac{\left(5-2\right)}{5}\times 180°=108°$

So, ∠CBA = 108º

Now, 
∠CBA + ∠ABF + ∠FBC = 360º
⇒ 108º + 90º + ∠FBC = 360º
⇒ ∠FBC = 162º

∆FBC is an isosceles triangle as FB = BC.
So, ∠BCF = ∠BFC

In âˆ†FBC,
∠BCF + ∠BFC + ∠FBC = 180º
⇒ ∠BCF + ∠BCF + 162º = 180º
⇒ 2∠BCF = 18º
⇒ ∠BCF = 9º

Answer:

If an angle is acute, then the corresponding exterior angle is greater than 90°.
Now, suppose a convex polygon has four or more acute angles.
Since, the polygon is convex, all the exterior angles are positive. So, sum of the exterior angle is at least the sum of the interior angles.
Now, supplementary of the four acute angles, which is greater than 4 × 90° = 360°
However, this is impossible.
Since, the sum of exterior angle of a polygon must equal to 360° and cannot be greater than it.
It follows that the maximum number of acute angle in convex polygon is 3.

Answer:

Produce DF to meet AB at G.


In âˆ†ABC,
∠ABC + ∠BCA + ∠CAB = 180º
⇒ 64º + ∠BCA + 52º = 180º
⇒ ∠BCA = 64º

Now BC || FG so,
∠BCA = ∠GFA = 64º  (Corresponding angles)

And, 
∠GFA + ∠AFD = 180º   (Linear pair)
⇒ ∠AFD = 116º       

Since opposite angles of a parallelogram are equal.

∴ x = 116º
 

Answer:

Construction: Draw a line through D parallel to BC that intersects AB at E.


Then, DEBC is a parallelogram. So,
CD = EB = 20 cm
DE = BC = AD = 10 cm

In âˆ†ADE, AD = DE so,
∠DAE = ∠DEA = 60º

So, âˆ†ADE is an equilateral triangle by angle sum property.
⇒ AD = AE = 10 cm

Hence, x = AE + EB = 10 + 20 = 30 cm
 

Answer:

∠A + ∠D = 180°    (Adjacent angles are supplementary)
⇒ ∠D = 75°

Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: Draw AX such that ∠BAX = 105°
Step 3: Cut an arc on AX at 3 cm that cuts AX at D.
Step 4: Draw DY such that ∠ADY = 75°
Step 5: Cut an arc on DY at 8 cm that cuts DY at C.
Step 6: Join BC

ABCD is the required trapezium.

Answer:

Since opposite sides of a parallelogram are equal,
AB = DC = 4 cm
BC = AD = 5 cm
Also, since adjacent angles of a parallelogram are supplementary,
∠B + ∠A = 180°
⇒ ∠A = 120°

Steps of construction:
Step 1: Draw AB = 4 cm.
Step 2: Draw BX such that ∠ABX = 60º.
Step 3: Cut an arc at 5 cm on BX and mark it C.
Step 4: Draw AY such that ∠BAY = 120º.
Step 5: Cut an arc at 5 cm on AY and mark it D.
Step 6: Join CD.

 
ABCD is the required parallelogram.

Answer:

If one angle is 60°, the adjacent angle will be 120°.   (Adjacent angles are supplementary)

Steps of construction:
Step 1: Draw AB = 5 cm
Step 2: Draw AY such that ∠BAY = 120º
Step 3: Cut an arc at 5 cm on AY and mark it D
Step 4: Draw BX such that ∠ABX = 60º
Step 5: Cut an arc at 5 cm on BX and mark it C
Step 6: Join CD.

  
ABCD is the required rhombus.

Answer:

Steps of construction: 
Step 1: Draw AB = 3 cm
Step 2: Draw BX such that ∠ABX = 90º
Step 3: Draw an arc on BX such that it is 5 cm away from point A. Mark the intersection as C.
Step 4: Draw an arc of radius 5 cm with B as centre and with C as centre draw another arc of radius 3 cm. 
Step 5: Mark the intersection D.
Step 6: Join AD and CD.

  

ABCD is the required rectangle.
 

Answer:

Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: Draw BX such that ∠ABX = 90º
Step 3: With B as the centre and radius = 4 cm, cut an arc at BX and mark the intersection C.
Step 4: With C as the centre and radius = 4 cm, draw an arc.
Step 5: With A as the centre and radius = 4 cm, draw another arc cutting the previous arc.
Step 6: Mark the intersection as D. Join AD and CD.



ABCD is the required square.
 

Answer:

Steps of construction:
Step 1: Draw LC = 7.5 cm
Step 2: With C as the centre and radius = 7.5 cm, draw an arc.
Step 3: With L as the centre and radius = 6 cm, draw an arc intersecting the previous arc. Mark the point E.
Step 4: Join CE
Step 5: With L as the centre and radius = 7.5 cm, draw an arc.
Step 6: With E as the centre and radius = 7.5 cm, draw an arc intersecting the previous arc. Mark the point U.
Step 7: Join LU and UE.



CLUE is the required rhombus.

 

Answer:

Steps of construction:
Step 1: Draw BE = 6 cm
Step 2: With B as centre and radius = 5 cm, draw an arc.
Step 3: With E as centre and radius = 5 cm, draw another arc intersecting the previous arc at R.
Step 4: Join BR
Step 5: With E as centre and radius = 7 cm, draw an arc.
Step 3: With B as centre and radius = 9 cm, draw another arc intersecting the previous arc at A.
Step 4: Join EA and RA.



Now when we measure the fourth side i.e., EA by the scale, the length of the EA will be 4 cm.

 

Answer:

Steps of construction:

Step 1: Draw PO = 5.5 cm
Step 2: Construct ∠POX = 70º
Step 3: With O as centre and radius = 7.2 cm, draw an arc cutting OX at U.
Step 4: With U as centre and radius = 5.5 cm, draw an arc.
Step 5: With P as centre and radius = 7.2 cm, draw another arc cutting the previous arc. Mark it R.
Step 6: Join PR and UR


POUR is the required parallelogram.

Answer:

Steps of construction:
Step 1: Draw a circle of radius = OC = 3 cm
Step 2: Draw the diameter AC.
Step 3: Draw the perpendicular bisector of AC and let it cut the circle at points B and D.



Then, the quadrilateral is a cyclic quadrilateral since all its vertices are lying on the circle.

Answer:

Steps of construction: 
Step 1: Draw HO = 6 cm
Step 2: With H as centre and radius = 4 cm, draw an arc.
Step 3: With O as centre and radius = 3 cm, draw another arc intersecting the previous arc at E.
Step 4: Join HE
Step 5: With E as centre and radius = 6 cm, draw an arc.
Step 6: With O as centre and radius = 4 cm, draw another arc intersecting the previous arc at M.
Step 4: Join OM and EM.



HOME is the required parallelogram.

Answer:

No, it is not possible to construct such a quadrilateral.
Given measures are AB = 3 cm, BC = 4 cm, CD = 5.4 cm, DA = 59 cm and AC = 8cm.
Now, AB + BC = 3 + 4 = 7 cm and AC = 8 cm
i.e. the sum of two sides of a triangle is less than the third side, which is absurd.

Answer:

No, it is not possible to construct such a quadrilateral.
Given measures are RO = 4 cm, OA = 5 cm, ∠O = 120°, ∠R = 105° and ∠A = 135°.
Now,  ∠O + ∠R + ∠A = 360°.
i.e. the sum of three angles of a quadrilateral is 360°, which is absurd.

Answer:

Steps of construction:
Step 1: Draw AC = 5 cm
Step 2: With A and C as centres respectively, and radius more than their halves, cut two arcs each intersecting each other.
Step 3: Join the intersection points and let the line cut AC at O.
Step 4: With O as centre, cut off OB = OD = 2.5 cm along the perpendicular bisector.
Step 5: Join AB, BC, CD and DA.



ABCD is the required square.

Answer:

Steps of construction:
Step 1: Draw NE = 7 cm
Step 2: Make ∠NEX = 110º
Step 3: With E as a centre and radius 6 cm, draw an arc cutting EX at W.
Step 4: Make ∠EWY = 105º
Step 5: Make ∠ENZ = 60º such that NZ and WY intersect each other at S.


​Then, NEWS is the required quadrilateral.
 

Answer:

Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: With A as centre and radius = 2.8 cm, draw an arc.
Step 3: With B as centre and radius = 3.5 cm, draw an arc. Mark the intersection O.
Step 4: Join OA and OB
Step 5: Produce AO to C such that AO = OC and produce BO to D such that OB = OD.
Step 6: Join AD, BC and DC.


ABCD is the required parallelogram.

Answer:

Sum of interior angles of an n-polygon = (n − 2) × 180º
Since, n = 20, sum of the interior angles = (20 − 2) × 180º
                                                                 = 3240º

So, measure of each angle = $\frac{3240}{20}=162°$

Answer:

Steps of construction:
Step 1: Draw RI = 7 cm
Step 2: Make ∠RIX = 60º
Step 3: With I as centre and radius = 5 cm, draw an arc on IX and mark the intersection S.
Step 4: Make ∠ISY = 120º
Step 5: With R as centre and radius = 6.5 cm, draw an arc on SY and mark the intersection K.
Step 6: Join RK.



RISK is the required parallelogram.

Answer:

Since, CD − AB = 3.2 cm, âˆ†BEC is an equilateral triangle with each angle = 60º.

Steps of construction:
Step 1: Draw DC = 9.6 cm
Step 2: With D as the centre, draw an angle of 60º. Draw an arc of radius 3.2 cm and mark the intersection A.
Step 3: Draw AB || CD and AB = 6.4 cm
Step 4: Join CB.
Step 5: Make ∠CBE = 60º.




ABCD is the required trapezium with ∠A = 120º and ∠B = 60º + 60º = 120º.