myhelper: S.chand books class 8 maths solution chapter 19 Special Types of Quadrilaterals Exercise 19D

EXERCISE 19 D

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Question 1

EFGH is a rhombus. Find a , b , c

Sol :

All sides of rhombus are equal .
Thus , GF=HG or c=10

Also,

Diagonals bisect each other .

So, OE=OG or a=8

In Right triangle EOF (Diagonals of rhombus are perpendicular bisectors)

Using Pythagoras theorem

Hypotenuse²=Height²+Base²
(EF)²=(OE)²+(OF)²

(10)²=(8)²+(OF)²
100=64+(OF)²
(OF)²=100-64

OF=√36=6 or b=6

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Question 2

PQRS is a rectangle whose diagonals intersect at point O. IF OP=4x-1 and OS=2x+7. Find the value of x.

Sol :

Diagonals of rectangle bisect each other.Also ,Diagonals of rectangle are of equal length.

Thus, PO=SO

4x-1=2x+7

4x-2x=7+1

2x=8

$x=\frac{8}{2}=4$

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Question 3

The diagonals of a rectangle ABCD intersect at O. If ∠AOB=114° , find ∠ACD and ∠ADB .

Sol :

Given : ∠AOB=114°

Diagonals of rectangle bisect each other. Also ,Diagonals of rectangle are of equal length.

Thus, OA=OB

So,ΔAOB is a isosceles triangle.

∴∠OAB=∠OBA=x

In ΔAOB

∠OAB+∠OBA+∠AOB=180° (Angle sum property of triangle)

x+x+114°=180°

2x=180°-114°=66

2x=66°

$x=\frac{66}{2}$

x=33°=∠OAB

Opposite sides of rectangle are parallel thus diagonals works as transversal.
∠ACD=∠AOB (alternate interior angle)
∴∠OAB=33°=∠ACD

∠ADB=∠DBC (alternate interior angle)
also, All corner angles of rectangle are of 90°
∴∠DBC=90°-∠ABO

∠DBC=90°-33°=57°

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Question 4

ABCD is a rhombus with ∠BDC=38°. Find ∠BCD

Sol :

∠DOC=90° ∵Rhombus diagonals bisect each other at 90°

In ΔDOC

∠CDO+∠DOC+∠OCD=180° (angle sum property of triangle)

38°+90°+∠OCD=180°

128°+∠OCD=180°

∠OCD=180°-128°=52°

Since, Rhombus diagonals bisect angles.

So,

∠BCD=∠BCO+∠OCD

∠BCD=∠OCD+∠OCD

∠BCD=2×52°=104°

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Question 5

PQRS is a rhombus with ∠SPR=44°. Find ∠PSQ and ∠PQR.

Sol :

∠POS=90° (Rhombus diagonals bisect each other at 90°)

In ΔPOS,

∠POS+∠OPS+∠OSP=180° (Angle sum property of triangle)

90°+44°+∠OSP=180°

∠OSP=180°-(99°+44°)

∠OSP=46° or ∠PSQ=46°

Also, ∠PSQ=∠SQR=46° (alternate interior angle)

∠PQR=∠PQS+∠SQR

Since, Rhombus diagonals bisect corner angles .

So, ∠PQS=∠SQR=46°

∠PQR=46°+46°=92°

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Question 6

(i) ABCD is a rhombus. ∠BAC=37°. Draw a sketch and find the four angles of the rhombus.

Sol :

Diagonals of rhombus are angle bisectors.

∴∠CAB=∠CAD

∠DAB=∠CAB+∠CAD

∠DAB=37°+37°=74°

In ΔAOB

∠AOB+∠OAB+∠OBA=180° (angle sum property of triangle)

90°+37°+∠OBA=180°

∠OBA=180°-(90°+37°)

∠OBA=180°-127°=53°

∠OBA=∠DBA=∠DBC=46° (Diagonals of rhombus are angle bisectors.)

∠ABC=DBA+DBC

∠ABC=53°+53°=106°

Opposite angles of rhombus are equal.

∴∠ABC=∠ADC=106°

∴∠DAB=∠BCD=74°

(ii) If an angle of a rhombus is 50°. Find the size of the angle of one of the triangles which are formed by the diagonals.

Sol :

Given : ∠A=50°

Diagonals of rhombus are angle bisector.
∴∠OAB$=\frac{50}{2}$=25°

Diagonals of rhombus are perpendicular bisector.

∴∠AOB=90°

In ΔAOB

∠OAB+∠AOB+∠OBA=180° (angle sum property of triangle)

25°+90°+∠OBA=180°

∠OBA=180°-(90°+25°)

∠OBA=180°-115°

∠OBA=65°

Three angles of triangle are 25°,65°,90°

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Question 7

If the base angles of an isosceles trapezium are 56° each, what is the measure of the other two angles ?

Sol :

In Isosceles trapezium, base angles are equal in measure.

∠D=∠C=56°

Also, Opposite angles are supplementary.

∠D+∠B=180°
56°+∠B =180°
∠B =180°-56°=124°
∠B=124°

∠B=∠A=124°

Measure of the other two angles is 124°

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Question 8

Calculate the angles marked with small letters in the following diagrams.

(i) Rectangle

Sol :

∠AOB=∠DOC=100° (Vertically Opposite Angle)

Diagonals of rectangle bisect each other.

∴DO=OC thus ΔDOC is isosceles therefore,their opposite angles are equal.

i.e. ∠ODC=∠OCD=a

In ΔDOC

∠DOC+∠ODC+∠OCD=180° (Angle sum property of triangle)

100°+a+a=180°

2a=180°-100°

2a=80°

$a=\frac{80}{2}$=40°

(ii) Rectangle

Sol :

∵Diagonals of rectangle are perpendicular bisector.

∠BOC=90°

Corner angles of rectangle is 90°

∴∠OBC=90°-23°=67°

In ΔBOC,

∠BOC+∠OBC+∠OCB=180° (Angle sum property of triangle)

∠90°+∠67°+∠OCB=180°

∠OCB=180°-(90°+67°)

∠OCB=180°-157°=23°

or ∠a=23°

∠OCD=90°-∠OCB

∠OCD=90°-23°=67°

or b=67°

(iii) Rhombus

Sol :

Diagonals of rhombus are perpendicular bisector and angle bisectors.

∠OFE=∠OHG=57.5°

In ΔHOG,

∠HOG+∠OHG+∠OGH=180° (Angle sum property of triangle)

90°+57.5°+x=180°

x=180°-147.5°=32.5°

(iv) Rhombus

∠z=∠PSQ=65° (alternate interior angle)

Diagonals of rhombus are perpendicular bisector.

∴∠x=90°

In ΔSLP,

∠PSL+∠SLP+∠SPL=180° (Angle sum property of triangle)
65°+90°+∠SPL=180°
155°+∠SPL=180°
∠SPL=180°-155°=25°
or y=25°

(v) Square

Sol :

Opposite sides are parallel in square.
∠DXA=∠CXY=84° (Vertically Opposite Angles)

Diagonals bisect interior angle. Also , all interior angle is 90°
∴∠DAX$=\frac{90}{2}$=45°

∠DAX=∠ACB=45° (alternate interior angle)

In ΔDXA,

∠DXA+∠ADX+∠DAX=180° (Angle sum property of triangle)
84°+∠ADX+45°=180°
∠ADX=180°-129°=51°
or ∠b=51°

∠ADX=∠XYC=51°
or ∠a=51°

(vi) Square

Sol :

Diagonals of square are angle bisector.

Also, Corner angles are 90°

∴∠MDN$=\frac{90}{2}$=45°

In ΔNMD

Exterior angle of a triangle is equal to sum of two interior angles

∴∠x=∠NMD+∠NDM

∠x=48°+45°=93°

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Question 9

ABCD is kite. Find the angles marked x and y in the given figures.
(i)

Sol :

Diagonals of kite are perpendicular bisector.

∴∠BOA=90°

In ΔBOA,

∠ABO+∠BOA+∠OAB=180°

18°+90°+∠OAB=180°

108°+∠OAB=180°

∠OAB=180°-108°=72°

In kite the longer diagonal bisects the pair of opposite angles.
∴∠OAB=OAD=72°
or x=72°

Also,

∠BOC=90°

∠BCO=29°

In ΔBOC,

∠BOC+∠BCO+∠OBC=180°

90°+29°+y=180°

119°+y=180°

y=180°-119°=61°

(ii)

Sol :

In kite,

It has one pair of opposite angles (obtuse) that are equal. Here, ∠B=∠D=y
or y=148°

In quadrilateral ABCD,
∠A+∠B+∠C+∠D=360° (Angle sum property of quadrilateral)
47°+148°+x+148°=360°
343°+x=360°

x=360°-343°=17°

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Question 10

ABCD is an isosceles trapezium and ABEF as a kite. ∠FAB= 43° and ∠AFE=137° . Find x and y .

Sol :

Kite has one pair of opposite angles (obtuse) that are equal.
Here, ∠F= ∠B

Also,

In quadrilateral ABEF
∠A+∠F+∠E+∠B=360° (Angle sum property of quadrilateral)
43°+137°+∠E+137°=360°

∠E=360°-317°=43°

∠CBE+∠ABE+∠ABC=360°

115°+137°+∠ABC=360°

∠ABC=360°-252°=108°

In isosceles trapezium sum of opposite angles is 180°

∠ABC+∠ADC=180°

108°+∠ADC=180°

∠ADC=180°-108°=72°

The base angles are the same in isosceles trapezium.

∠ADC=∠BCD=72°

or y=72°

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Question 11

If the diagonals of a rhombus are 12 cm and 16 cm, find the length of each side.

Sol :

Diagonals of rhombus bisect each other.

∴AO$=\frac{12}{2}$=6

∴BO$=\frac{16}{2}$=8

Rhombus diagonals are perpendicular bisector.

In Right angled triangle AOB

Using Pythagoras theorem

(Hypotenuse)²=(Height)²+(Base)²

(AB)²=(OA)²+(OB)²

(AB)²=(6)²+(8)²

(AB)²=36+64

AB=√100

AB=10

All sides of rhombus is equal so all other sides are 10

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Question 12

ABCD is a rhombus and its diagonals intersect at O.
(i) Is ΔAOB similarity ΔBOC? State the congruence condition used.
(ii) Also, state if ∠BCO=∠DCO .Deduce that each diagonal of a rhombus bisects the angles through which it passes.

Sol :

(i) Yes,

In ΔBOC and ΔAOB

AO=OC [In a rhombus diagonals bisect each other]

OB=OB (Common)
AB=BC [All sides of a rhombus are equal]

By using SSS Congruency,
ΔBOC≅ΔAOC

(ii) Yes,

∠BCO=∠DCO (Diagonals of rhombus are angle bisector)

S.chand books class 8 maths solution chapter 19 Special Types of Quadrilaterals Exercise 19D

EXERCISE 19 D

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

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