Question

Construct a LMN in which <M = 90degree , LM= 5 cm and <50 degree ​

Answer 1

ANSWER

Since P and Q are mid-points of LM and LN.

LM

LP

=

LN

LQ

=

2

1

In △LPQ and △LMN

LM

LP

=

LN

LQ

∠PLQ=∠MLN(Common)

=>△LPQ∼△LMN(ByS.A.Ssimilaritycriterion)

∴

MN

PQ

=

LM

LP

=

2

1

(CPST)

or

MN

PQ

=

2

1

12

PQ

=

2

1

(∵MN=12cm)

PQ=

2

12

=6cm

PM=

2

1

LM=

2

9

=4.5cm [Since P and Q are mid-points of LM and LN]

$$QN=\dfrac { 1 }{ 2 } LN=\dfrac { 15 }{ 2 } =7.5cm\\ MN=12cm\\ $$

(i) Perimeter of trapezium MNQP

=MN+NQ+PQ+PM

=12+7.5+6+4.5

=30cm

(ii)Area of Trapezium MNQP

=Base×Height

=MN×PM

=12×4.5

=54cm 2

$ANSWER</h3><h3></h3><h3>Since P and Q are mid-points of LM and LN.</h3><h3></h3><h3>LMLP=LNLQ=21</h3><h3></h3><h3>In △LPQ and △LMN</h3><h3></h3><h3>LMLP=LNLQ∠PLQ=∠MLN(Common)=>△LPQ∼△LMN(ByS.A.Ssimilaritycriterion)∴MNPQ=LMLP=21(CPST)orMNPQ=2112PQ=21(∵MN=12cm)PQ=212=6cm</h3><h3></h3><h3>PM=21LM=29=4.5cm [Since P and Q are mid-points of LM and LN]</h3><h3></h3><h3> QN=\dfrac { 1 }{ 2 } LN=\dfrac { 15 }{ 2 } =7.5cm\\ MN=12cm\\ </h3><h3></h3><h3>(i) Perimeter of trapezium MNQP</h3><h3></h3><h3>=MN+NQ+PQ+PM=12+7.5+6+4.5=30cm</h3><h3></h3><h3>(ii)Area of Trapezium MNQP</h3><h3></h3><h3>               =Base×Height=MN×PM=12×4.5=54cm2</h3><h3></h3><h3></h3><h3>$

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