Question

please answer this questionanswer 10th and 11 question

Answer 1

I hope this helps u

Answer 2

ANS.11(Rhombus diagram is above in picture)

let a rhombus be ABCD
in above picture ABCD is a rhombus in which
AB=BC=CD=DC=20 cm
with centre O where diagonals bisect each other at 90 degree.

Area of rhombus=1/2×diagonal1×diagonal2

sides of rhombus=20cm
Diagonal1=24 cm

So,
by Pythagoras theoram in triangle OAB

${ab}^{2} = {ob}^{2} + {ao}^{2}$
IF DIAGONAL1 BD=24 cm
So,its half OB will be 24=2=12cm
${20}^{2} = {12}^{2} + {ao}^{2}$
$400 = 144 + {ao}^{2} \\ {ao}^{2} = 400 - 144 \\ {ao}^{2} = 256 \\ ao = \sqrt{256 } \\ ao = \sqrt{16 \times 16 } \\ ao = 16cm$
so we get AO=16cm
rhombus diagonal bisect each other,then
AO+OC=Diagonal2
AO+AO=diagonal2 (AO=OC)
2×AO=Diagonal2
2×16=diagonal2
diagonal2=32cm

Area of rhombus=1/2×diagonal1×diagonal2
=1/2×24×32
=12×32
area of rhombus=384 sq.cm

HOPE IT HELPS YOU!