NCERT Solutions for Class 8 Math Chapter 3 - Understanding Quadrilaterals

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$5)
= 180$°$ + 180$°$ = 360$°$ [By angle sum property of a triangle]

Answer:

We know that, the sum of the angles $\angle$1, $\angle$2, $\angle$3 and $\angle$4 is 360$°$.
The sum of the measure of the four angles of a quadrilateral is 360$°$.
[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$3, etc.]
The sum of the measures of the four angles of a quadrilateral is 360$°$.

Answer:

Given that
x = 180$°$ – m$\angle$2 – m$\angle$3 ...(i)
y = 180$°$ – m$\angle$4 – m$\angle$5 ...(ii)
z = 180$°$ – m$\angle$6 – m$\angle$7 ...(iii)
w = 180$°$ – m$\angle$8 – m$\angle$1 ...(iv)
Adding equations (i), (ii), (iii) and (iv), we get
x + y + z + w = 720$°$ – ($\angle$1 + $\angle$2 + $\angle$3 + $\angle$4 + $\angle$5 + $\angle$6 + $\angle$7 + $\angle$8)
$\Rightarrow$ 360$°$ = 720$°$ – ($\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$8 = 720$°$ – 360$°$ = 360$°$
$\therefore$ $\angle$A + $\angle$B + $\angle$C + $\angle$D = 360$°$
$\therefore$ The sum of the measures of the four angles of a quadrilateral is 360$°$.

Answer:

The quadrilateral is concave.
Divide the quadrilateral ABCD into two triangles ABD and CBD.
In $∆$ABD, $\angle$A + $\angle$ABD + $\angle$ADB = 180$°$ ...(i)
and in $∆$BDC, $\angle$C + $\angle$CDB + $\angle$DBC = 180$°$ ...(ii)
Adding quations (i) and (ii), we get
$\angle$A + $\angle$ADB + $\angle$ABD + $\angle$C + $\angle$CDB + $\angle$DBC =
180$°$ + 180$°$ = 360$°$
$\Rightarrow$ $\angle$A + $\angle$ADB + $\angle$CDB + $\angle$C + ($\angle$CBD +
$\angle$DBA) = 360$°$
$\Rightarrow$ $\angle$A + $\angle$B + $\angle$C + $\angle$D = 360$°$

Answer:

(a) (1), (2), (5), (6), (7)
(b) (1), (2), (5), (6), (7)
(c) (1), (2), (4)
(d) (2)
(e) (1), (4)

Answer:

(a) A convex quadrilateral has 2 diagonals.
(b) A regular hexagon has 9 diagonals.
(c) A triangle has no diagonal.

Answer:

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

Answer:

(a) The angle sum of a convex polygon with 7 sides is given by:
(7 – 2) × 180$°$ = 5 × 180$°$ = 900$°$
(b) The angle sum of a convex polygon with 8 sides is given by:
(8 – 2) × 180$°$ = 6 × 180$°$ = 1080$°$
(c) The angle sum of a convex polygon with 10 sides is given by:
(10 – 2) × 180$°$ = 8 × 180$°$ = 1440$°$
(d) The angle sum of a convex polygon with n sides is given by:
(n – 2) × 180$°$.

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.

Answer:

(a) 50$°$ + 130$°$ + 120$°$ + x = 360$°$
300$°$ + x = 360$°$
$\Rightarrow$ x = 360$°$ – 300$°$ = 60$°$
(b) 90$°$ + 60$°$ + 70$°$ + x = 360$°$
$\Rightarrow$ 220$°$ + x = 360$°$
$\Rightarrow$ x = 360$°$ – 220$°$ = 140$°$
(c) Figure (c) has five sides:
$\therefore$ its angle sum = (5 – 2) × 180$°$
= 3 × 180$°$ = 540$°$
Also, exterior angles 70$°$ and 60$°$ are
given.
$\therefore$ Corresponding interior angles are (180$°$
– 70$°$) = 110$°$ and (180$°$ – 60$°$) = 120$°$ respectively.
$\therefore$ 110$°$ + 120$°$ + x + 30$°$ + x = 540$°$
260$°$ + 2x = 540$°$
$\Rightarrow$ 2x = 540$°$ – 260$°$
$\Rightarrow$ 2x = 280$°$

Figure (d) is a regular pentagon.
$\therefore$ Its angle sum = (5 – 2) × 180$°$
= 3 × 180$°$ = 540$°$
$\therefore$ x + x + x + x + x = 540$°$
$\Rightarrow$ 5x = 540$°$

Answer:

In the given figure:
Exterior angle x = (180$°$ – 90$°$) = 90$°$
Exterior angle z = (180$°$ – 30$°$) = 150$°$
As sum of interior angles of a triangle is 180$°$.
$\therefore$ 90$°$ + 30$°$ + p = 180$°$
120$°$ + p = 180$°$
$\Rightarrow$ p = 180$°$ – 120$°$ = 60$°$
Exterior angle y = (180$°$ – 60$°$) = 120$°$
$\therefore$ x + y + z = 90$°$ + 150$°$ + 120$°$
= 360$°$.

Answer:

In the given figure: Exterior angle x = (180$°$ – 120$°$) = 60$°$
Exterior angle y = (180$°$ – 80$°$) = 100$°$
Exterior angle z = (180$°$ – 60$°$) = 120$°$
As sum of interior angles of a quadrilateral is
360$°$.
$\therefore$ 120$°$ + 80$°$ + 60$°$ + q = 360$°$
$\therefore$ 260$°$ + q = 360$°$
$\Rightarrow$ q = 360$°$ – 260$°$
$\Rightarrow$ q = 100$°$
$\therefore$ Exterior angle w = (180$°$ – 100$°$)
= 80$°$
$\therefore$ x + y + z + w = 60$°$ + 100$°$ + 120$°$ + 80$°$
= 360$°$

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

Answer:

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

Answer:

The measure of each exterior angle is given
by x + x + x + x + x + x = 360$°$
$\Rightarrow$ 6x = 360$°$
$\Rightarrow$ x = 60$°$

Answer:

Measure of each interior angle
= 180$°$– 60$°$ = 120$°$

Answer:

The octagon being regular having 8 sides.
$\therefore$ All the exterior angles have equal measure,
say x.
$\therefore$ 8x = 360$°$

$\therefore$ Measure of each exterior angle = 45$°$
Measure of each interior angle = 180$°$ – 45$°$
= 135$°$

Answer:

The polygon being regular having 20 sides.
$\therefore$ All the exterior angles have equal measure, say x.
$\therefore$ 20 x = 360$°$

$\therefore$ Measure of each exterior angle = 180$°$
Measure of each interior angle = 180$°$ – 18$°$
= 162$°$

Answer:

As the sum of the measures of the external angles
of any polygon is 360$°$.
$\therefore$ 125$°$ + 125$°$ + x = 360$°$
250$°$ + x = 360$°$
$\Rightarrow$ x = 360$°$ – 250$°$ = 110$°$

Answer:

As the sum of the measures of the external angles
of any polygon is 360$°$.
$\therefore$ x + 90$°$ + 60$°$ + 90$°$ + 70$°$ = 360$°$
$\Rightarrow$ x + 310$°$ = 360$°$
$\Rightarrow$ x = 360$°$ – 310$°$
$\Rightarrow$ x = 50$°$

Answer:

The polygon being regular having 9 sides.
$\therefore$ All the exterior angles have equal measure,
say x.
$\therefore$ 9x = 360$°$

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

Answer:

The polygon being regular having 15 sides.
$\therefore$ All the exterior angles have equal measure, say x.
$\therefore$ 15x = 360$°$

Answer:

Total measure of all exterior angles = 360$°$
Measure of each exterior angle = 24$°$

$\therefore$ The polygon has 15 sides.

Answer:

Measure of each interior angle = 165$°$
$\therefore$ Measure of each exterior angle
= 180$°$ – 165$°$
= 15$°$
Total measure of all exterior angles = 360$°$

$\therefore$ The polygon has 24 sides.

Answer:

No; since 22 is not a divisor of 360.

Answer:

No; because each exterior angle is (180$°$ – 22$°$)
= 158$°$, which is not a divisor of 360$°$.

Answer:

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

Answer:

The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle is equal to 60$°$.
The greatest exterior angle of an equilateral triangle can be (180$°$ – 60$°$) = 120$°$.

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.

Answer:

Answer:

Answer:

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.

Answer:

Answer:

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$°$ and m$\angle$B = x$°$

Answer:

Answer:

Answer:

It is given that RISK and CLUE are parallelograms.
$\therefore$ In parallelogram RISK;
$\angle$RKS + $\angle$ISK = 180$°$ [Because, sum of any two
adjacent angles of a parallelogram is 180$°$.]
$\therefore$ 120$°$ + $\angle$ISK = 180$°$
$\Rightarrow$ $\angle$ISK = 180$°$ – 120$°$ = 60$°$
Also, in parallelogram CLUE,
$\angle$CLU = $\angle$CEU = 70$°$
[Because, opposite angles of a parallelogram are equal]
$\therefore$ In $∆$OES; sum of three angles is equal to 180$°$.
$\therefore$ $\angle$OES + $\angle$ESO + x = 180$°$
[$\therefore$ $\angle$OES = $\angle$CEU and $\angle$ESO = $\angle$ISK]
70$°$ + 60$°$ + x$°$ = 180$°$
130$°$ + x = 180$°$
$\Rightarrow$ x = 180$°$ – 130$°$ = 50$°$

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$°$.

Answer:

It is given that, ABCD is a trapezium having
$\angle$B = 120$°$ and two of its sides AB and CD are
parallel.
$\therefore$ $\angle$B + $\angle$C = 180$°$
$\Rightarrow$ 120$°$ + $\angle$C = 180$°$
$\Rightarrow$ $\angle$C = 180$°$ – 120$°$ = 60$°$

Answer:

It is given that, PQRS is a trapezium having $\angle$Q =
130$°$ and two of its sides PS and RQ are parallel.
$\therefore$ $\angle$P + $\angle$Q = 180$°$
$\Rightarrow$ $\angle$P + 130$°$ = 180$°$
$\Rightarrow$ $\angle$P = 50$°$
Also, $\angle$R and $\angle$S each have measure 90$°$.
We may find $\angle$P by one more method.
i.e., Sum of all the interior angles of a quadrilateral is 360$°$.
$\therefore$ $\angle$P + $\angle$Q + $\angle$R + $\angle$S = 360$°$
$\Rightarrow$ $\angle$P + 130$°$ + 90$°$ + 90$°$ = 360$°$
$\Rightarrow$ $\angle$P + 310$°$ = 360$°$
$\Rightarrow$ $\angle$P = 360$°$ – 310$°$
$\Rightarrow$ $\angle$P = 50$°$

Answer:

Yes, their mid-point is the same and the angle at O is 90$°$.
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
$∆$POQ $\cong$ $∆$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.

Answer:

FALSE

Answer:

TRUE

Answer:

TRUE

Answer:

FALSE

Answer:

FALSE

Answer:

TRUE

Answer:

TRUE

Answer:

TRUE

Answer:

(a) Square and Rhombus
(b) Square and 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.

Answer:

(i) Parallelogram, Rhombus, Square, Rectangle
(ii) Rhombus, Square.
(iii) Rectangle, Square.

Answer:

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

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.