Need Full Solution for both Questions(21 and 22).
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Answers
Step-by-step explanation:
Q.21.
Given:
ABCD is a parallelogram and AE and CF bisects
angle A and angle C.
To prove that-
AE || FC
we have,
Solution:
In angle A
angle EAF = angle DAE …given
Also, in angle C
angle ECF = angle BCF
So, we get
L EAF = L DAE = L ECF =L BCF
because, opposite sides of a parallelogram are equal.
In AECF
L EAF = L ECF
Since, DC || AB …opposite sides of parallelogram are equal
then, EC || AF also
We get AECF as a parallelogram.
Hence, AE || FC
________________________________
Q.22.
Given:
AB || DC , AD || BC,
AB = DC, AD = BC
therefore, ABCD is a parallelogram.
To find:
x, y, and z
we have
Solution:
AB = DC
since, alternate angles are equal
we have,
L DCA = L CAB
z = L D
by angle sum property of triangle
we have,
L D =
i.e. z = 106°
Hence, x = 11
y = 15
z = 106°
Answer:
In angle A
angle EAF = angle DAE …given
Also, in angle C
angle ECF = angle BCF
So, we get
L EAF = L DAE = L ECF =L BCF
because, opposite sides of a parallelogram are equal.
In AECF
L EAF = L ECF
Since, DC || AB …opposite sides of parallelogram are equal
then, EC || AF also
We get AECF as a parallelogram.
Hence, AE || FC
Q.22 mn
AB = DC
\begin{gathered}2x + 25 = 3x + 14 \\ 3x - 2x = 25 - 14 \\ x = 11\end{gathered}
2x+25=3x+14
3x−2x=25−14
x=11
since, alternate angles are equal
we have,
L DCA = L CAB
\begin{gathered} {(y + 9)}^{o} = {24}^{o} \\ y + 9 = 24 \\ y = 24 - 9 \\ y = 15\end{gathered}
(y+9)
o
=24
o
y+9=24
y=24−9
y=15
z = L D
by angle sum property of triangle
we have,
L D =
\begin{gathered} {180}^{o} - ({(3y + 5)}^{o} + {24}^{o} ) \\ = {180}^{o} - {(3y + 5)}^{o} - {24}^{o} \\ = {156}^{o} - {(3 \times 15 + 5)}^{o} \\ = {156}^{o} - {50}^{o} \\ = {106}^{o} \end{gathered}
180
o
−((3y+5)
o
+24
o
)
=180
o
−(3y+5)
o
−24
o
=156
o
−(3×15+5)
o
=156
o
−50
o
=106
o
i.e. z = 106°
Hence, x = 11
y = 15
z = 106°