Exercise 10C
Q1 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1
Find the value of each variable in the parallelograms. Give brief reasons.
Sol :
(a)
∵AD||BC
AB||DC
∠A=50°=∠C
∴x=50°
a=AB=DC=8
b=AD=BC=5
(b)
Diagonals bisect each other at the point 'O'
∴y=OC=AO=3
∴x=BO=OD=2
(c)
PQ||SR
PS||RQ
C=∠D=∠B=95°
∵a=b
∴b=180°-95°=85°
∴a=b=85°
(d)
∠A=∠C=2x
x=y
2x+x=180°
$x=\frac{180}{3}$=60°
∴∠A=2×60=120°
∠B=y=60°
∠C=2×60=120°
∠D=x=60°
(e)
AB||DC
AD||BC
∠B=∠D=120°
∠A=∠C=180°-120°=60°
∴3m=120°
m=40°
(f)
∠A=∠C=45°=5y
∴5y=45°
$y=\frac{45}{5}$=9°
x=AB=DC=6
(g)
AB=AC and BCD is a straight line
Sol :
AB=BC
x=∠DEF=∠ACD
∠A=36°
Again, ∠ABC=∠ACB
∴∠BAC+∠ABC=180°-∠ACB
36°+∠ABC+∠ACB=180°
∠ABC+∠ABC=180°-36° [∵∠ABC=∠ACB]
2∠ABC=144°
∠ABC$=\frac{144}{2}$=72°
∴∠ABC=∠ACB=72°
∴∠ACD=180°-∠ACB=180°-72°=108°
∴x=∠DEF=∠ACD=108°
(h)
Sol :
⇒∠BAC+∠ABC=∠ACO
50°+65°=∠ACO
∴∠ACO=115°=∠ADE
∴∠ADE=∠AFE=x=115°
Q2 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 2
In the figure, AO = OD = OC and reflex ∠AOC = 230°. Find the value of x.
Sol :
AB||DC AO=OD=OC
AD||BC
∠AOC=230°
∠COD+∠DOA=360°-230°=130°=y
In ΔODC
∠OAD=∠ODA (isosceles triangle)
In ΔODA
∠OCD=∠ODC (isosceles triangle)
In quadrilateral ADCO ,
y+∠OCD+∠OAD+∠ADC=360°
130°+∠ODC+∠ADO+∠ADO+ODC=360°
2∠ODC+2∠ADO=360°-130°
2(∠ODC+∠ADO)=230°
∠ADC$=\dfrac{230^{\circ}}{2}$
∠ADC=115°
∠ADC=∠ABC=x=115° (Opposite angles of parallelogram is equal)
Q3 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 3
The measure of two consecutive angles of a parallelogram are in the ratio 4 : 5. Find the measure of each angle of the parallelogram.
Sol :
Let, common angles is x
∴ratio of two
adjacent angles= 4 : 5
∴∠A=4x
∠B=5x
We know that :
4x+5x=180°
9x=180°
x=20°
∴∠A=4×20=60°
∠B=5×20=100°
Q4 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 4
Two opposite angles of a parallelogram are (5x – 21)° and (x + 75)° . Find the measure of each of the angles of the parallelogram.
Sol :
∠ABC=5x-21°
∠ADC=x+75°
Both are opposite angle
AB||DC
AD||BC
We know that
∠ABC=∠ADC
5x-21°=x+75°
5x-x=75°+21°=96°
4x=96°
$x=\dfrac{96}{4}$=24°
∠ABC=(524°)-21°=120°-21°=99°
∠ADC=(24°+75°)=99°
∠BAD=(180°-99°)=81°
∠DCB=180°-99°=81°
Q5 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 5
The ratio of two adjacent sides of a parallelogram is 7:9 and its perimeter is 96 cm. Find the sides of the parallelogram.
Sol :
Let , common factor=x
two adjacent side ratio=7 : 9
∴1st adjacent side=AD=7x
2nd adjacent side=AB=9x
∴Perimeter=(7x+9x)=2×16x=32x
According to question
32x=96°
$x=\frac{96}{32}=3$
∴AD=BC=7×3=21 unit
AB=DC=9×3=27 unit
Q6 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 6
State whether the following quadrilaterals are parallelograms or not. Give reasons.
Sol :
(a) AO=OC=3cm
BO=DO=4cm
∴AC and DB bisect each other
∴So , this is parallelogram
(b) AD=BC=5cm
AB=DC=6cm
∴ABCD is a parallelogram
(c) ∠ABC=∠ADC=70°
∠BAD=∠DCB=110°
∴opposite angle are equal
∴ADCB is parallelogram
(d) AB||BC
But angle are not equal. It is not parallelogram
(e) AC and BC are not bisect each other.
∴It is not parallelogram
(f) Angles are not equal
∴It is not parallelogram
Q7 | Ex-10C | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 7
Find the value of a and b that would make the quadrilateral, a parallelogram.
Sol :
(a) ABCD parallelogram
AD=BC
AB=CD
∴AD=BC
5a-13=3a-5
5a-3a=13-5=8
2a=8
a=4
AB=CD
8b=10b-3
10b-8b=3
$b=\frac{3}{2}$
(b) ∠DAB=∠DCB
5b+6=8a-10
8a-5b=10+6=16
8a-5b=16..(i)
∴again,
(4a-8)+(8a-10)=180
4a-8+8a-10=180
12a-18=180
12a=180+18=198
a=16.5
∴Putting value of a in (i) equation
8a-5b=16
(8×16.5)-5b=16
132-5b=16
-5b=16-132=-116
$b=\frac{116}{2}$=23.2
(c) 5b-7=3b+6
5b-3b=7+6
2b=13
$b=\frac{13}{2}$=6.5
Again,
2a=3b-5
2a=(3×6.5)-5
2a=19.5-5=14.5
$a=\frac{14.5}{2}$=7.25
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