in angle Lmn, angle m=90°seg mp perpendicular seg Ln,Lp= 16 cm, pn=25 cm find pm
Answers
Answer:
1.The perimeter of trapezium MNQP=30 cm
2.The area of trapezium MNQP=40.5 square cm
Step-by-step explanation:
In right triangle LMN,
\angle M=90^{\circ}∠M=90
∘
LM=9 cm
MN=12 cm
LN=15 cm
P is the mid point of LM
Therefore, LP=PM=\frac{1}{2}LM=\frac{9}{2}=4.5
2
1
LM=
2
9
=4.5 cm
LQ=QN=\frac{1}{2}LN=\frac{1}{2}\times 15
2
1
LN=
2
1
×15 =7.5 cm
By mid- segment theorem
PQ=\frac{1}{2}MNPQ=
2
1
MN ,PQ\parallel MNPQ∥MN
PQ=\frac{1}{2}(12cm)=6 cmPQ=
2
1
(12cm)=6cm
1.Perimeter of trapezium=PM+MN+QN+PQ=4.5+12+7.5+6=30 cm
2.Area of trapezium PMNQ=\frac{1}{2}(sum\;of\;parallel\;sides)\times h
2
1
(sumofparallelsides)×h
Area of trapezium PMNQ=\frac{1}{2}(6+12)\times 4.5
2
1
(6+12)×4.5
Area of trapezium PMNQ=9\times 4.5=40.5cm^29×4.5=40.5cm
2
Answer:
PM =20 cm
Step-by-step explanation:
