Question

(c)
How many angles are formed in each of the figures? Name them and measure
6.
them.
C
D
(b)
T
is
U
LABC.
B
saolo
А
5
St
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Answer 1

Step-by-step explanation:

ABCD is a trapezium in which \tt A B \| C DAB∥CD .

Also E and F are respectively the mid-points of sides A D and B C .

Let us join B E and produce it to meet C D produced at G, also join B D which intersects E F at O.

Now,

\tt\triangle G E D \cong \triangle B E A \Rightarrow G E=B E \qquad\qquad [\text{By C.P.C.T}]△GED≅△BEA⇒GE=BE[By C.P.C.T]

Hence, E is also the mid point of G B .

Now, in \Delta G C BΔGCB ,E and F are respectively the mid-points of B G and B C .

\text{\( \tt \Rightarrow \quad E F \| G C \) \: \: \: \: \: \: \: \: \: \: \: \: [By mid-point theorem]}⇒EF∥GC [By mid-point theorem]

But

\text{\( \tt G C \| A B \) or \( \tt C D \| A B \) \: \: \: \: \: \: \: [Given]}GC∥AB or CD∥AB [Given]

\tt \Rightarrow \quad E F \| A B⇒EF∥AB

In \tt\triangle A D B, A B \| E O△ADB,AB∥EO and E is the mid-point of A D.

\text{\( \Rightarrow O \) is mid-point of \( \tt B D \) and \( \tt E O=\dfrac{1}{2} A B \) \( \ldots \) (1) [By converse of mid-point theorem]}⇒O is mid-point of BD and EO=

2

1

AB … (1) [By converse of mid-point theorem]

\text{In \( \tt \triangle B D C, O F \| C D \) and \( O \) is the mid-point of \( B D \).}In △BDC,OF∥CD and O is the mid-point of BD.

\text{say \( \tt O F=\dfrac{1}{2} C D \) \( \ldots \) (2) [By converse of mid-point theorem]}say OF=

2

1

CD … (2) [By converse of mid-point theorem]

On adding eqn (1) & (2) we get

\begin{gathered}\begin{aligned} \tt E O+O F & \tt=\frac{1}{2} A B+\frac{1}{2} C D \\ \\ \tt \Rightarrow \quad E F & \tt=\frac{1}{2}(A B+C D) \end{aligned} \end{gathered}

EO+OF

⇒EF

=

2

1

AB+

2

1

CD

=

2

1

(AB+CD)