##### Question 1:

Take any quadrilateral, say ABCD shown in figure. Divide it into two triangle, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6.

Use the angle sum property of a triangle and argue how the sum of the measures of $\angle $A, $\angle $B, $\angle $C and $\angle $D amounts to 180$\xb0$ + 180$\xb0$ = 360$\xb0$.

##### Answer:

$\angle $A + $\angle $B + $\angle $C + $\angle $D

= ($\angle $1 + $\angle $4) + $\angle $6 + ($\angle $5 + $\angle $2) + $\angle $3

= ($\angle $3 + $\angle $1 + $\angle $2) + ($\angle $4 + $\angle $6 + $\angle $5)

= 180$\xb0$ + 180$\xb0$ = 360$\xb0$ [By angle sum property of a triangle]

##### Question 2:

Take four congruent card-board copies of any quadrilateral ABCD, with angles as shown fig (i). Arrange the copies as shown in the figure, where angles $\angle $1 + $\angle $2 + $\angle $3 + $\angle $4 meet at a point as shown in fig. (ii)

What can you say about the sum of the angles $\angle $1, $\angle $2, $\angle $3 and $\angle $4?

##### Answer:

We know that, the sum of the angles $\angle $1, $\angle $2, $\angle $3 and $\angle $4 is 360$\xb0$.

The sum of the measure of the four angles of a quadrilateral is 360$\xb0$.

[Note: We denote the angles by $\angle $1, $\angle $2, $\angle $3, etc., and their respective measures by m$\angle $1, m$\angle $2, m$\angle $3, etc.]

The sum of the measures of the four angles of a quadrilateral is 360$\xb0$.

##### Question 3:

Take any quadrilateral ABCD shown in figure. Let P be any point in its interior. Join P to vertices A, B, C and D. In the figure, consider $\u2206$PAB. From this we see x = 180$\xb0$ – m$\angle $2 – m$\angle $3; similarly from $\u2206$PBC, y = 180$\xb0$ – m$\angle $4 – m$\angle $5, from $\u2206$PCD, z = 180$\xb0$ – m$\angle $6 – m$\angle $7 and from $\u2206$PDA, w = 180$\xb0$ – m$\angle $8 – m$\angle $1.

Use this to find the total measure m$\angle $1 + m$\angle $2 +… + m$\angle $8, does it help to arrive at the result?

Remember: $\angle $x + $\angle $y + $\angle $z + $\angle $w = 360$\xb0$.

##### Answer:

Given that

x = 180$\xb0$ – m$\angle $2 – m$\angle $3 ...(i)

y = 180$\xb0$ – m$\angle $4 – m$\angle $5 ...(ii)

z = 180$\xb0$ – m$\angle $6 – m$\angle $7 ...(iii)

w = 180$\xb0$ – m$\angle $8 – m$\angle $1 ...(iv)

Adding equations (i), (ii), (iii) and (iv), we get

x + y + z + w = 720$\xb0$ – ($\angle $1 + $\angle $2 + $\angle $3 + $\angle $4 + $\angle $5 + $\angle $6 + $\angle $7 + $\angle $8)

$\Rightarrow $ 360$\xb0$ = 720$\xb0$ – ($\angle $1 + $\angle $2 + $\angle $3 + $\angle $4 + $\angle $5 + $\angle $6 + $\angle $7 + $\angle $8)

$\Rightarrow $ $\angle $1 + $\angle $2 + $\angle $3 + $\angle $4 + $\angle $5 + $\angle $6 + $\angle $7 + $\angle $8 = 720$\xb0$ – 360$\xb0$ = 360$\xb0$

$\therefore $ $\angle $A + $\angle $B + $\angle $C + $\angle $D = 360$\xb0$

$\therefore $ The sum of the measures of the four angles of a quadrilateral is 360$\xb0$.

##### Question 4:

Consider quadrilateral ABCD. Split it into two triangles and find the sum of the interior angles shown in figure.

##### Answer:

The quadrilateral is concave.

Divide the quadrilateral ABCD into two triangles ABD and CBD.

In $\u2206$ABD, $\angle $A + $\angle $ABD + $\angle $ADB = 180$\xb0$ ...(i)

and in $\u2206$BDC, $\angle $C + $\angle $CDB + $\angle $DBC = 180$\xb0$ ...(ii)

Adding quations (i) and (ii), we get

$\angle $A + $\angle $ADB + $\angle $ABD + $\angle $C + $\angle $CDB + $\angle $DBC =

180$\xb0$ + 180$\xb0$ = 360$\xb0$

$\Rightarrow $ $\angle $A + $\angle $ADB + $\angle $CDB + $\angle $C + ($\angle $CBD +

$\angle $DBA) = 360$\xb0$

$\Rightarrow $ $\angle $A + $\angle $B + $\angle $C + $\angle $D = 360$\xb0$

##### Question 5:

Given here are some figures:

Classify each of them on the basis of the

following:

(a) Simple curve

(b) Simple closed curve

(c) Polygon

(d) Convex polygon

(e) Concave polygon

##### Answer:

(a) (1), (2), (5), (6), (7)

(b) (1), (2), (5), (6), (7)

(c) (1), (2), (4)

(d) (2)

(e) (1), (4)

##### Question 6:

How many diagonals does each of the
following have:

(a) A convex quadrilateral

(b) A regular hexagon

(c) A triangle

##### Answer:

(a) A convex quadrilateral has 2 diagonals.

(b) A regular hexagon has 9 diagonals.

(c) A triangle has no diagonal.

##### Question 7:

What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex?

##### Answer:

360$\xb0$, Yes. (make a non-convex quadrilateral and try!)

##### Question 8:

Examine the table (Each figure is divided into triangles and the sum of the angles deduced from that?

What can you say about the angle sum of a convex polygon with number of sides?

(a) 7 (b) 8 (c) 10 (d) n

##### Answer:

(a) The angle sum of a convex polygon with 7 sides is given by:

(7 – 2) × 180$\xb0$ = 5 × 180$\xb0$ = 900$\xb0$

(b) The angle sum of a convex polygon with 8 sides is given by:

(8 – 2) × 180$\xb0$ = 6 × 180$\xb0$ = 1080$\xb0$

(c) The angle sum of a convex polygon with 10 sides is given by:

(10 – 2) × 180$\xb0$ = 8 × 180$\xb0$ = 1440$\xb0$

(d) The angle sum of a convex polygon with n sides is given by:

(n – 2) × 180$\xb0$.

##### Question 9:

What is a regular polygon?

State the name of a regular polygon of:

(i) 3 sides (ii) 4 sides (iii) 6 sides.

##### Answer:

Regular polygon: A regular polygon is both “equiangular” and “equilateral”.

(i) A regular polygon having 3 sides is called equilateral triangle.

(ii) A regular polygon having 4 sides is called square.

(iii) A regular polygon having 6 sides is called regular hexagon.

##### Question 10:

Find the angle measure x in the following figures:

##### Answer:

(a) 50$\xb0$ + 130$\xb0$ + 120$\xb0$ + x = 360$\xb0$

300$\xb0$ + x = 360$\xb0$

$\Rightarrow $ x = 360$\xb0$ – 300$\xb0$ = 60$\xb0$

(b) 90$\xb0$ + 60$\xb0$ + 70$\xb0$ + x = 360$\xb0$

$\Rightarrow $ 220$\xb0$ + x = 360$\xb0$

$\Rightarrow $ x = 360$\xb0$ – 220$\xb0$ = 140$\xb0$

(c) Figure (c) has five sides:

$\therefore $ its angle sum = (5 – 2) × 180$\xb0$

= 3 × 180$\xb0$ = 540$\xb0$

Also, exterior angles 70$\xb0$ and 60$\xb0$ are

given.

$\therefore $ Corresponding interior angles are (180$\xb0$

– 70$\xb0$) = 110$\xb0$ and (180$\xb0$ – 60$\xb0$) = 120$\xb0$ respectively.

$\therefore $ 110$\xb0$ + 120$\xb0$ + x + 30$\xb0$ + x = 540$\xb0$

260$\xb0$ + 2x = 540$\xb0$

$\Rightarrow $ 2x = 540$\xb0$ – 260$\xb0$

$\Rightarrow $ 2x = 280$\xb0$

Figure (d) is a regular pentagon.

$\therefore $ Its angle sum = (5 – 2) × 180$\xb0$

= 3 × 180$\xb0$ = 540$\xb0$

$\therefore $ x + x + x + x + x = 540$\xb0$

$\Rightarrow $ 5x = 540$\xb0$

##### Question 11:

##### Answer:

In the given figure:

Exterior angle x = (180$\xb0$ – 90$\xb0$) = 90$\xb0$

Exterior angle z = (180$\xb0$ – 30$\xb0$) = 150$\xb0$

As sum of interior angles of a triangle is 180$\xb0$.

$\therefore $ 90$\xb0$ + 30$\xb0$ + p = 180$\xb0$

120$\xb0$ + p = 180$\xb0$

$\Rightarrow $ p = 180$\xb0$ – 120$\xb0$ = 60$\xb0$

Exterior angle y = (180$\xb0$ – 60$\xb0$) = 120$\xb0$

$\therefore $ x + y + z = 90$\xb0$ + 150$\xb0$ + 120$\xb0$

= 360$\xb0$.

##### Question 12:

##### Answer:

In the given figure:
Exterior angle x = (180$\xb0$ – 120$\xb0$) = 60$\xb0$

Exterior angle y = (180$\xb0$ – 80$\xb0$) = 100$\xb0$

Exterior angle z = (180$\xb0$ – 60$\xb0$) = 120$\xb0$

As sum of interior angles of a quadrilateral is

360$\xb0$.

$\therefore $ 120$\xb0$ + 80$\xb0$ + 60$\xb0$ + q = 360$\xb0$

$\therefore $ 260$\xb0$ + q = 360$\xb0$

$\Rightarrow $ q = 360$\xb0$ – 260$\xb0$

$\Rightarrow $ q = 100$\xb0$

$\therefore $ Exterior angle w = (180$\xb0$ – 100$\xb0$)

= 80$\xb0$

$\therefore $ x + y + z + w = 60$\xb0$ + 100$\xb0$ + 120$\xb0$ + 80$\xb0$

= 360$\xb0$

##### Question 13:

Take a regular hexagon.

What is the sum of measure of its exterior

angles x, y, z, p, q, r?

##### Answer:

Let ABCDEF is a regular hexagon having each side equal.

As the sum of the measures of the external angles of any polygon is 360$\xb0$.

$\therefore $ x + y + z + p + q + r = 360$\xb0$

##### Question 14:

Take a regular hexagon.

Is x = y = z = p = q = r? Why?

##### Answer:

Yes, x = y = z = p = q = r, because the hexagon is regular.

##### Question 15:

Take a regular hexagon.

What is the measure of each exterior angle?

##### Answer:

The measure of each exterior angle is given

by x + x + x + x + x + x = 360$\xb0$

$\Rightarrow $ 6x = 360$\xb0$

$\Rightarrow $ x = 60$\xb0$

##### Question 16:

Take a regular hexagon.

What is the measure of each interior angle?

##### Answer:

Measure of each interior angle

= 180$\xb0$– 60$\xb0$ = 120$\xb0$

##### Question 17:

Take a regular octagon.

What is the measure of each exterior and interior angle?

##### Answer:

The octagon being regular having 8 sides.

$\therefore $ All the exterior angles have equal measure,

say x.

$\therefore $ 8x = 360$\xb0$

$\therefore $ Measure of each exterior angle = 45$\xb0$

Measure of each interior angle = 180$\xb0$ – 45$\xb0$

= 135$\xb0$

##### Question 18:

Take a regular 20-gon.

What is the measure of each exterior and interior angle?

##### Answer:

The polygon being regular having 20 sides.

$\therefore $ All the exterior angles have equal measure, say x.

$\therefore $ 20 x = 360$\xb0$

$\therefore $ Measure of each exterior angle = 180$\xb0$

Measure of each interior angle = 180$\xb0$ – 18$\xb0$

= 162$\xb0$

##### Question 19:

Find x in the following figures.

##### Answer:

As the sum of the measures of the external angles

of any polygon is 360$\xb0$.

$\therefore $ 125$\xb0$ + 125$\xb0$ + x = 360$\xb0$

250$\xb0$ + x = 360$\xb0$

$\Rightarrow $ x = 360$\xb0$ – 250$\xb0$ = 110$\xb0$

##### Question 20:

Find x in the following figures.

##### Answer:

As the sum of the measures of the external angles

of any polygon is 360$\xb0$.

$\therefore $ x + 90$\xb0$ + 60$\xb0$ + 90$\xb0$ + 70$\xb0$ = 360$\xb0$

$\Rightarrow $ x + 310$\xb0$ = 360$\xb0$

$\Rightarrow $ x = 360$\xb0$ – 310$\xb0$

$\Rightarrow $ x = 50$\xb0$

##### Question 21:

Find the measure of each exterior angle of a regular polygon of 9 sides.

##### Answer:

The polygon being regular having 9 sides.

$\therefore $ All the exterior angles have equal measure,

say x.

$\therefore $ 9x = 360$\xb0$

$\therefore $ Measure of each exterior angle = 40$\xb0$

##### Question 22:

Find the measure of each exterior angle of a regular polygon of 15 sides.

##### Answer:

The polygon being regular having 15 sides.

$\therefore $ All the exterior angles have equal measure,
say x.

$\therefore $ 15x = 360$\xb0$

##### Question 23:

How many sides does a regular polygon have if the measure of an exterior angle is 24$\xb0$?

##### Answer:

Total measure of all exterior angles = 360$\xb0$

Measure of each exterior angle = 24$\xb0$

$\therefore $ The polygon has 15 sides.

##### Question 24:

How many sides does a regular polygon have if each of its interior angles is 165$\xb0$?

##### Answer:

Measure of each interior angle = 165$\xb0$

$\therefore $ Measure of each exterior angle

= 180$\xb0$ – 165$\xb0$

= 15$\xb0$

Total measure of all exterior angles = 360$\xb0$

$\therefore $ The polygon has 24 sides.

##### Question 25:

Is it possible to have a regular polygon with measure of each exterior angle as 22$\xb0$?

##### Answer:

No; since 22 is not a divisor of 360.

##### Question 26:

Can 22$\xb0$ be an interior angle of a regular polygon? Why?

##### Answer:

No; because each exterior angle is (180$\xb0$ – 22$\xb0$)

= 158$\xb0$, which is not a divisor of 360$\xb0$.

##### Question 27:

What is the minimum interior angle possible for a regular polygon? Why?

##### Answer:

The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle is equal to 60$\xb0$.

##### Question 28:

What is the maximum exterior angle possible for a regular polygon?

##### Answer:

The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle is equal to 60$\xb0$.

The greatest exterior angle of an equilateral triangle can be (180$\xb0$ – 60$\xb0$) = 120$\xb0$.

##### Question 29:

Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, 5 cm. Arrange them as shown figure.

You get a trapezium. (Check it!) Which are the parallel sides here? Should the non-parallel sides be equal?

##### Answer:

Quadrilateral DCEA,

DC = AE = 5 cm

and AD = EC = 4 cm

DC || AE $\Rightarrow $ AD || CE

$\therefore $ DCEA is a parallelogram.

$\therefore $ ABCD is a trapezium. Its parallel sides are AB and DC. Non-parallel sides are AD and CB.

Two more examples of trapeziums using the same set of triangles.

##### Question 30:

In a parallelogram m$\angle $R = m$\angle $N = 70$\xb0$, find m$\angle $1 and m$\angle $G.

##### Answer:

##### Question 31:

##### Answer:

##### Question 32:

Consider the following parallelograms. Find the values of the unknown x, y, z.

##### Answer:

##### Question 33:

Can a quadrilateral ABCD be a parallelogramif

(i) $\angle $D + $\angle $B = 180$\xb0$

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?

(iii) $\angle $A = 70$\xb0$ and $\angle $C = 65$\xb0$?

##### Answer:

Fig. (i) is not a parallelogram, because, the
opposite angles i.e., $\angle $C and $\angle $A are not equal.
Fig. (ii) is not a parallelogram, because, the
opposite sides i.e, AB and CD and BC and DA
are not equal.

Fig. (iii) is also not a parallelogram, because,
the diagonals of a parallelogram bisect each
other and here it is not so and $\angle $A and $\angle $C are
not equal.

##### Question 34:

Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

##### Answer:

##### Question 35:

The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.

##### Answer:

##### Question 36:

Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

##### Answer:

It is given that, ABCD is a parallelogram in which two adjacent angles $\angle $A and $\angle $B have
equal measure, say x.

$\therefore $ m$\angle $A = x$\xb0$ and m$\angle $B = x$\xb0$

##### Question 37:

The adjacent figure HOPE is a parallelogram.

Find the angle measures x, y and z. State the
properties you use to find them.

##### Answer:

##### Question 38:

The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm.)

##### Answer:

##### Question 39:

##### Answer:

It is given that RISK and CLUE are parallelograms.

$\therefore $ In parallelogram RISK;

$\angle $RKS + $\angle $ISK = 180$\xb0$ [Because, sum of any two

adjacent angles of a parallelogram is 180$\xb0$.]

$\therefore $ 120$\xb0$ + $\angle $ISK = 180$\xb0$

$\Rightarrow $ $\angle $ISK = 180$\xb0$ – 120$\xb0$ = 60$\xb0$

Also, in parallelogram CLUE,

$\angle $CLU = $\angle $CEU = 70$\xb0$

[Because, opposite angles of a parallelogram are
equal]

$\therefore $ In $\u2206$OES; sum of three angles is equal to 180$\xb0$.

$\therefore $ $\angle $OES + $\angle $ESO + x = 180$\xb0$

[$\therefore $ $\angle $OES = $\angle $CEU and $\angle $ESO = $\angle $ISK]

70$\xb0$ + 60$\xb0$ + x$\xb0$ = 180$\xb0$

130$\xb0$ + x = 180$\xb0$

$\Rightarrow $ x = 180$\xb0$ – 130$\xb0$ = 50$\xb0$

##### Question 40:

Explain how this figure is a trapezium. Which of its two sides are parallel?

##### Answer:

The given figure KLMN is a trapezium, as its two sides KL and MN are parallel, because, sum of its adjacent angles $\angle $L and $\angle $M is 180$\xb0$.

##### Question 41:

Find m$\angle $C in the figure if AB || DC.

##### Answer:

It is given that, ABCD is a trapezium having

$\angle $B = 120$\xb0$ and two of its sides AB and CD are

parallel.

$\therefore $ $\angle $B + $\angle $C = 180$\xb0$

$\Rightarrow $ 120$\xb0$ + $\angle $C = 180$\xb0$

$\Rightarrow $ $\angle $C = 180$\xb0$ – 120$\xb0$ = 60$\xb0$

##### Question 42:

Find the measure of $\angle $P and $\angle $S if SP || RQ in Fig. (If you find m$\angle $R, is there more than one method to find m$\angle $P?).

##### Answer:

It is given that, PQRS is a trapezium having $\angle $Q =

130$\xb0$ and two of its sides PS and RQ are parallel.

$\therefore $ $\angle $P + $\angle $Q = 180$\xb0$

$\Rightarrow $ $\angle $P + 130$\xb0$ = 180$\xb0$

$\Rightarrow $ $\angle $P = 50$\xb0$

Also, $\angle $R and $\angle $S each have measure 90$\xb0$.

We may find $\angle $P by one more method.

i.e., Sum of all the interior angles of a quadrilateral
is 360$\xb0$.

$\therefore $ $\angle $P + $\angle $Q + $\angle $R + $\angle $S = 360$\xb0$

$\Rightarrow $ $\angle $P + 130$\xb0$ + 90$\xb0$ + 90$\xb0$ = 360$\xb0$

$\Rightarrow $ $\angle $P + 310$\xb0$ = 360$\xb0$

$\Rightarrow $ $\angle $P = 360$\xb0$ – 310$\xb0$

$\Rightarrow $ $\angle $P = 50$\xb0$

##### Question 43:

Take a square sheet, say PQRS (Fig (i)). Fold along both the diagonals. Are their mid-point the same? Check if the angle at O is 90$\xb0$ by using a set-square.

This verifies the property stated above.

##### Answer:

Yes, their mid-point is the same and the angle
at O is 90$\xb0$.

We know that PQSR is a square whose diagonals
meet at O.

PO = OR (Since, the square is a parallelogram)

By SSS congruency condition, we see that

$\u2206$POQ $\cong $ $\u2206$SOR

[$\therefore $ PQ = SR; PO = OR; SO

= OQ]

$\therefore $ m$\angle $POQ = m$\angle $SOR

These angles being a linear

pair, each is right angle.

##### Question 44:

All rectangles are squares.

##### Answer:

FALSE

##### Question 45:

All rhombuses are parallelograms.

##### Answer:

TRUE

##### Question 46:

All squares are rhombuses and also rectangles.

##### Answer:

TRUE

##### Question 47:

All squares are not parallelograms.

##### Answer:

FALSE

##### Question 48:

All kites are rhombuses.

##### Answer:

FALSE

##### Question 49:

All rhombuses are kites.

##### Answer:

TRUE

##### Question 50:

All parallelograms are trapeziums.

##### Answer:

TRUE

##### Question 51:

All squares are trapeziums.

##### Answer:

TRUE

##### Question 52:

Identify all the quadrilaterals that have:

(a) Four sides of equal length

(b) Four right angles

##### Answer:

(a) Square and Rhombus

(b) Square and Rectangle

##### Question 53:

Explain how a square is:

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle

##### Answer:

(i) Any four sided figure is called a quadrilateral and so is the square.

(ii) Opposite sides of a parallelogram are equal and parallel and so is in the square.

(iii) All the four sides of a rhombus are equal and so is in the square.

(iv) All the four angles of a rectangle are right angles and opposite sides are equal, same is the case with the square.

##### Question 54:

Name the quadrilaterals whose diagonals:

(i) bisect each other.

(ii) are perpendicular bisectors of each other.

(iii) are equal.

##### Answer:

(i) Parallelogram, Rhombus, Square, Rectangle

(ii) Rhombus, Square.

(iii) Rectangle, Square.

##### Question 55:

Explain why a rectangle is a convex quadrilateral.

##### Answer:

A rectangle is a convex quadrilateral, because no part of its diagonals lies in its exteriors.

##### Question 56:

ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

##### Answer:

As DABC is a right-angled triangle, right-angled
at B. And O is the mid-point of AC.

Complete the rectangle ABCD as shown in the
adjoining figure.

Since, the diagonals of a rectangle are equal.
And all rectangles are parallelograms.

Also, diagonals of a parallelogram bisect each
other.

$\therefore $ O is the mid-point of AC as well as BD.

$\therefore $ O is equidistant from A, B, C and D as well.