Answer:
1)c)13
2)a)3/5
SECTION B
11)For the given A.P.,
First term(a)=3
Last term(t
n
)=253
Common Difference(d)=8−3=5
∵t
=a+(n−1)d
⇒253=3+(n−1)×5
⇒253−3=(n−1)×5
⇒250÷5=n−1
⇒50+1=n
∴n=51
In 51 terms, 20
th
term from the last term will be 51−20+1=32
∴t
32
=a+(32−1)d=3+31×5=3+155=158
12)In ΔABC
DE∣∣AC
Line drawn parallel to one side of triangle, insects the other two sides. It divides the other side in the same ratio.
EC
BE
=
DA
BD
__(i)
In ΔAEB
DF∣∣AE
Line drawn parallel to one side of triangle intersects the other sides. It divides the other sides in the same ratio.
FE
BF
__(ii)
From (i) & (ii)
∴ Hence, it proved.
13)Let point on X− axis be P(x,0) to be equidistant from A(2,−5) and B(−2,9)
∴PA=PB
Using the distance formula we get,
⇒
(x−2)
2
+(0+5)
(x+2)
+(0−9)
Squaring both sides
⇒ (x−2)
+25=(x+2)
+81
⇒ x
−4x+4−56=x
+4x+4
⇒ 8x= −56
It gives x=−7
So, point equidistant from (2,−5) and (−2,9) and lying on x axis is (−7,0).
14)The answer is the ratio in which the line is divided is 2:7
Let us consider a point P(-1, 6)(refer fig) that divides A(-3, 10) and B(6,-8) as per k:1 ratio, then we co-ordinates of P as (
k+1
6k−3
,
−8k+10
).
But we already have given co-ordinates of P,
∴
=−1 and rewriting we have
6k−3=−k−1 or 6k+k=3−1 ∴7k=2 or k=
7
Thus, we get the point P dividing AB in the ratio 2:7.
15)
According to the question:
Let cosA=
Hypotenuse
sideadjacentA
AB
AC
Similarly,
cosB=
sideadjacentB
BC
Given that
cosA=cosB
AC=BC
In triangle,
angles opposite equal sides are equal
∠B=∠A
please follow me also
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Verified answer
Answer:
1)c)13
2)a)3/5
SECTION B
11)For the given A.P.,
First term(a)=3
Last term(t
n
)=253
Common Difference(d)=8−3=5
∵t
n
=a+(n−1)d
⇒253=3+(n−1)×5
⇒253−3=(n−1)×5
⇒250÷5=n−1
⇒50+1=n
∴n=51
In 51 terms, 20
th
term from the last term will be 51−20+1=32
th
∴t
32
=a+(32−1)d=3+31×5=3+155=158
12)In ΔABC
DE∣∣AC
Line drawn parallel to one side of triangle, insects the other two sides. It divides the other side in the same ratio.
EC
BE
=
DA
BD
__(i)
In ΔAEB
DF∣∣AE
Line drawn parallel to one side of triangle intersects the other sides. It divides the other sides in the same ratio.
FE
BF
=
DA
BD
__(ii)
From (i) & (ii)
EC
BE
=
FE
BF
∴ Hence, it proved.
13)Let point on X− axis be P(x,0) to be equidistant from A(2,−5) and B(−2,9)
∴PA=PB
Using the distance formula we get,
⇒
(x−2)
2
+(0+5)
2
=
(x+2)
2
+(0−9)
2
Squaring both sides
⇒ (x−2)
2
+25=(x+2)
2
+81
⇒ x
2
−4x+4−56=x
2
+4x+4
⇒ 8x= −56
It gives x=−7
So, point equidistant from (2,−5) and (−2,9) and lying on x axis is (−7,0).
14)The answer is the ratio in which the line is divided is 2:7
Let us consider a point P(-1, 6)(refer fig) that divides A(-3, 10) and B(6,-8) as per k:1 ratio, then we co-ordinates of P as (
k+1
6k−3
,
k+1
−8k+10
).
But we already have given co-ordinates of P,
∴
k+1
6k−3
=−1 and rewriting we have
6k−3=−k−1 or 6k+k=3−1 ∴7k=2 or k=
7
2
Thus, we get the point P dividing AB in the ratio 2:7.
15)
According to the question:
Let cosA=
Hypotenuse
sideadjacentA
AB
AC
Similarly,
cosB=
Hypotenuse
sideadjacentB
=
AB
BC
Given that
cosA=cosB
AB
AC
=
AB
BC
AC=BC
In triangle,
angles opposite equal sides are equal
∠B=∠A
please follow me also