Exercise 10A
Find the value of unknown angle in each of the following figures.
Q1(a) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1
(a) 75°+125°+105°+x=360°
305°+x=360°
x=360°-305°=55°
Q1(a) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(b)
(b) 80°+100°+70°+y=360°
y+250°=360°
y=360°-250°=110°
Q1(c) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(c)
(c) 90°+140°+30°+a=360°
a+260°=360°
a=360°-260°=100°
Q1(d) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(d)
Sol :
(d) ΔABC
∠CAB=(20°+z)
∠BCA=(10°+y)
∠ABC=40°
∴∠ABC+∠CAB+∠BCA=180°
40°+(20°+z)+(10°+y)=180°
(z+y)=180°-70°=110°
Now ΔAOC
∠CAO+∠ACO+∠AOC=180°
z+y+∠AOC=180°
110°+∠AOC=180°
∠AOC=180°-110°=70°
x=360°-70°=290°
Q1(e) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(e)
(e) p+70°+90°+90°=360°
p+250°=360°
p=360°-250°=110°
Q1(f) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(f)
(f) 138°+90°+x+2x=360°
288°+2x+x=360°
3x=360°-288°
3x=72°
$x=\frac{72}{3}=24^{\circ}$
∴2x=2×24=48°
Q1(g) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(g)
(g) m+2m+m+2m=360°
6m=360°
$m=\frac{360}{6}$=60°
∴∠A=60° ,∠B=2×60°=120°
∠C=60° ,∠D=2×60°=120°
Q1(h) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 1(h)
Sol :
(h) ∠O=360°-240° =120°
∴ΔBOC⟶
∠BOC+∠OBC+∠OCB=180°
,120°+a+b=180°
a+b=180°-120°=60°
Now , ΔABC⟶
∠ABC+∠ACB+∠BAC=180°
(y+a)+(30°+b)+65°=180°
y+(a+b)+30°+65°=180°
y+60°+30°+65°=180°
y=180°-155°=25°
Find the value of x
Q2(a) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 2(a)
(a) ∠BCD=180°-60°=120°
∴120°+110°+50°+x=360°
280°+x=360°
x=360°-280°=80°
Q2(b) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 2(b)
Sol :
(b) ∠EDF=∠ADC=35° (Vertically Opposite Angles)
∴90°+136°+35°+x=360°
x=360°-261°=99°
Q2(c) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 2(c)
(c) 3x+4x+2x+x=360°
10x=360°
$x=\frac{360}{10}$=36°
∴3x=3×36°=108°
4x=4×36°=144°
2x=2×36°=72°
Q2(d) | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 2(d)
(d) ∠DAB=90°
∠ABC=180-100=80°
∠DCB=x
∠CDA=180-70=110°
x+110+80+90=360°
x+280=360°
x=360°-280°=80°
Q3 | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 3
Find the measures of the four angles of a quadrilateral if they are in the ratio 2 : 4 : 6 : 8.
Sol :
Angle ratio=2 : 4 : 6 : 8
∴common factor=x
2x+4x+6x+8x=360°
20x=360°
$x=\frac{360}{20}$=18°
2x=2×18°=36°
4x=4×18°=72°
6x=6×18°=108°
8x=8×18°=144°
Q4 | Ex-10A | Class 8 | SChand New learning Composite Maths | Quadrilaterals | myhelper
Question 4
Find the value of x if x, (2x+13°), (3x+10°) and (x–6°) are all the angles of a quadrilateral?
Sol :
x+(2x+13°)+(3x+10°)+(x-6°)=360°
x+2x+3x+x+13°+10°-6°=360°
7x+17=360
7x=360°-17°=343°
$x=\frac{343}{7}$=49°
2x+13°=2×49°+13°=111°
3x+10°=3×49°+10°=157°
x–6°=49°-6°=43°
Thanks for solution help
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