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Q4- area of llgm = b×h
If we take AB as base
so, area = AB×DE
=4×6 ( opp. side of a parallelogram are equal so, AB =DC=6)
area = 24cm^2 _(1)
and if we take AD as base
so, area = AD×BF
= 3×BF _(2)
now from (1) and (2)
24 = 3×BF
24/3= BF
8= BF
=> [BF=8cm]
Q5- as ABCD is a llgm
so, AB=DC
If we take DC as base
•°•, area of ABCD= AM×DC
= 16×6
= 96cm^2 _(1)
now if we take AD as base
•°•, area = AD×NC
= AD× 8 _(2)
From (1) and (2)
96 = AD× 8
96/8 = AD
AD= 12
=> [AD=12cm]
Q7- In ∆ ADL
( By using Pythagoras Theorem)
H^2 = P^2+B^2
(5)^2 = (4)^2+B^2
25 = 16 + B^2
25-16 = B^2
B = √9= 3cm
DM= 3cm
as DM= MC =3cm
so, DC= DM+MC+LM
DC = 3+3+9= 15cm
area of trapeziumABCD
= 1/2 [a+b] h
•°•, = 1/2 [ 15+9]×4
= 1/2 × 24×4
= 24×2
= 48cm^2
If we take AB as base
so, area = AB×DE
=4×6 ( opp. side of a parallelogram are equal so, AB =DC=6)
area = 24cm^2 _(1)
and if we take AD as base
so, area = AD×BF
= 3×BF _(2)
now from (1) and (2)
24 = 3×BF
24/3= BF
8= BF
=> [BF=8cm]
Q5- as ABCD is a llgm
so, AB=DC
If we take DC as base
•°•, area of ABCD= AM×DC
= 16×6
= 96cm^2 _(1)
now if we take AD as base
•°•, area = AD×NC
= AD× 8 _(2)
From (1) and (2)
96 = AD× 8
96/8 = AD
AD= 12
=> [AD=12cm]
Q7- In ∆ ADL
( By using Pythagoras Theorem)
H^2 = P^2+B^2
(5)^2 = (4)^2+B^2
25 = 16 + B^2
25-16 = B^2
B = √9= 3cm
DM= 3cm
as DM= MC =3cm
so, DC= DM+MC+LM
DC = 3+3+9= 15cm
area of trapeziumABCD
= 1/2 [a+b] h
•°•, = 1/2 [ 15+9]×4
= 1/2 × 24×4
= 24×2
= 48cm^2
soyabbhatta87:
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