Answer:
17ans :In ΔADE,
∠EAD+∠BAF=90
o
and
⇒∠EDA=∠BAF
⇒
AF
DE
=
BF
AE
3
5
....(1)
similarly ΔCFB:ΔDEC
CE
CF
7
...(2)
adding (1) and (2)
AF+CF
21
50
AC
10
⇒BF=4.2
18 ans : sorry don't know
19ans:Correct option is A)
oin BD, BE
From isosceles △BCD we get
∠CBD=∠CDB=
2
1
(180
∘
−110
)=
×70
=35
From the cyclic quadrilateral ABDE
∠BDE=180
−∠BAE=180
−120
=60
From the cyclic quadrilateral BCDE
∠BED=180
=70
ArcAB=ArcBC=∠AEB=∠BDC=35
Sum of all the interior angles of a Pentagon =(2×5−4)rt.∠s=6×90
=540
(i) ∠ABC=540
−(110
+35
+60
+70
+120
)=540
−430
=110
(ii) ∠CDE=∠CDB+∠BDE=35
=95
(iii) ∠AED=∠AEB+∠BED=35°
+70°
=105°
(iv) ∠EAD=∠EBD=180°
−(70°
+60°
)=180°
−130°
=50°
Given, ∠A=90
By Pythagoras theorem,
BC
=AB
+AC
=5
+12
BC=13 cm
Area of triangle =
×base×height
Thus,
AB×AC=
AD×BC
5×12=13×AD
AD=
13
60
cm.
20 ans : Given:- perimeter of rectangle = area of rectangle
2(l+b)=lb..(1)
if b=2×
4
11
cm
l=?
2(l+
)=l(
)
2l−(
11l
)=−
3l
⇒l=
22
=7
∴ if breadth =2
cm than length = 7
21 ans: Given 5th term and 10th term of an AP are 26 and 51 respectively.
a+4d=26
a+9d=51
Solving the above equations, we get
−25=−5d
⟹d=5
Plugging in d = 5 in either of above two equations, we get
26=a+4(5)
⟹26=a+20
⟹a=6
To find the 15th term,
t
15
=a+(n−1)d
⟹t
=6+5(15−1)
=76
The 15th term is 76.
Step-by-step explanation:
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Answers & Comments
Answer:
17ans :In ΔADE,
∠EAD+∠BAF=90
o
and
∠EAD+∠BAF=90
o
⇒∠EDA=∠BAF
⇒
AF
DE
=
BF
AE
⇒
BF
AF
=
AE
DE
=
3
5
....(1)
similarly ΔCFB:ΔDEC
⇒
BF
CE
=
CF
DE
⇒
BF
CF
=
CE
DE
=
7
5
...(2)
adding (1) and (2)
BF
AF+CF
=
21
50
⇒
BF
AC
=
21
50
⇒
BF
10
=
21
50
⇒BF=4.2
18 ans : sorry don't know
19ans:Correct option is A)
oin BD, BE
From isosceles △BCD we get
∠CBD=∠CDB=
2
1
(180
∘
−110
∘
)=
2
1
×70
∘
=35
∘
From the cyclic quadrilateral ABDE
∠BDE=180
∘
−∠BAE=180
∘
−120
∘
=60
∘
From the cyclic quadrilateral BCDE
∠BED=180
∘
−110
∘
=70
∘
ArcAB=ArcBC=∠AEB=∠BDC=35
∘
Sum of all the interior angles of a Pentagon =(2×5−4)rt.∠s=6×90
∘
=540
∘
(i) ∠ABC=540
∘
−(110
∘
+35
∘
+60
∘
+70
∘
+35
∘
+120
∘
)=540
∘
−430
∘
=110
∘
(ii) ∠CDE=∠CDB+∠BDE=35
∘
+60
∘
=95
∘
(iii) ∠AED=∠AEB+∠BED=35°
+70°
=105°
(iv) ∠EAD=∠EBD=180°
−(70°
+60°
)=180°
−130°
=50°
Given, ∠A=90
∘
By Pythagoras theorem,
BC
2
=AB
2
+AC
2
BC
2
=5
2
+12
2
BC=13 cm
Area of triangle =
2
1
×base×height
Thus,
2
1
AB×AC=
2
1
AD×BC
5×12=13×AD
AD=
13
60
cm.
20 ans : Given:- perimeter of rectangle = area of rectangle
2(l+b)=lb..(1)
if b=2×
4
3
=
4
11
cm
l=?
2(l+
4
11
)=l(
4
11
)
2l−(
4
11l
)=−
2
11
⇒
4
3l
=
2
11
⇒l=
3
22
=7
3
1
cm
∴ if breadth =2
4
3
cm than length = 7
3
1
cm
21 ans: Given 5th term and 10th term of an AP are 26 and 51 respectively.
a+4d=26
a+9d=51
Solving the above equations, we get
−25=−5d
⟹d=5
Plugging in d = 5 in either of above two equations, we get
26=a+4(5)
⟹26=a+20
⟹a=6
To find the 15th term,
t
15
=a+(n−1)d
⟹t
15
=6+5(15−1)
⟹t
15
=76
The 15th term is 76.
Step-by-step explanation:
17ans :In ΔADE,
∠EAD+∠BAF=90
o
and
∠EAD+∠BAF=90
o
⇒∠EDA=∠BAF
⇒
AF
DE
=
BF
AE
⇒
BF
AF
=
AE
DE
=
3
5
....(1)
similarly ΔCFB:ΔDEC
⇒
BF
CE
=
CF
DE
⇒
BF
CF
=
CE
DE
=
7
5
...(2)
adding (1) and (2)
BF
AF+CF
=
21
50
⇒
BF
AC
=
21
50
⇒
BF
10
=
21
50
⇒BF=4.2
18 ans : sorry don't know
19ans:Correct option is A)
oin BD, BE
From isosceles △BCD we get
∠CBD=∠CDB=
2
1
(180
∘
−110
∘
)=
2
1
×70
∘
=35
∘
From the cyclic quadrilateral ABDE
∠BDE=180
∘
−∠BAE=180
∘
−120
∘
=60
∘
From the cyclic quadrilateral BCDE
∠BED=180
∘
−110
∘
=70
∘
ArcAB=ArcBC=∠AEB=∠BDC=35
∘
Sum of all the interior angles of a Pentagon =(2×5−4)rt.∠s=6×90
∘
=540
∘
(i) ∠ABC=540
∘
−(110
∘
+35
∘
+60
∘
+70
∘
+35
∘
+120
∘
)=540
∘
−430
∘
=110
∘
(ii) ∠CDE=∠CDB+∠BDE=35
∘
+60
∘
=95
∘
(iii) ∠AED=∠AEB+∠BED=35°
+70°
=105°
(iv) ∠EAD=∠EBD=180°
−(70°
+60°
)=180°
−130°
=50°
Given, ∠A=90
∘
By Pythagoras theorem,
BC
2
=AB
2
+AC
2
BC
2
=5
2
+12
2
BC=13 cm
Area of triangle =
2
1
×base×height
Thus,
2
1
AB×AC=
2
1
AD×BC
5×12=13×AD
AD=
13
60
cm.
20 ans : Given:- perimeter of rectangle = area of rectangle
2(l+b)=lb..(1)
if b=2×
4
3
=
4
11
cm
l=?
2(l+
4
11
)=l(
4
11
)
2l−(
4
11l
)=−
2
11
⇒
4
3l
=
2
11
⇒l=
3
22
=7
3
1
cm
∴ if breadth =2
4
3
cm than length = 7
3
1
cm
21 ans: Given 5th term and 10th term of an AP are 26 and 51 respectively.
a+4d=26
a+9d=51
Solving the above equations, we get
−25=−5d
⟹d=5
Plugging in d = 5 in either of above two equations, we get
26=a+4(5)
⟹26=a+20
⟹a=6
To find the 15th term,
t
15
=a+(n−1)d
⟹t
15
=6+5(15−1)
⟹t
15
=76
The 15th term is 76.