if ab and cd are perpendicular find perimeter of triangle abc
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Answered by
0
Answer:
Consider △ACD&△ABC
∠CAD=∠CAB
∠CDA=∠ACB
∴△ACD∼△ABC,{ By AA similarty criterion}
⟹
AB
AC
=
AC
AD
∴AC
2
=AB×AD−(1)
Similarily,△BCD∼△BAC{ by AA similarity criterion}
⟹
BA
BC
=
BC
BD
∴BC
2
=BA×BD−(2)
∴
AC
2
BC
2
=
AB×AD
AB×BD
∴BC
2
×AD=AC
2
×BD
Answered by
0
Answer:
(i) From ΔABC and ΔDEC,
∠ABC=∠DEC=90
∘
(Given)
and ∠ACB=∠DCE= Common
∴ ΔABC∼ΔDEC (By A−A similarity)
(ii) In ΔABC and ΔDEC,
ΔABC∼ΔDEC (proved in (i) part)
∴
DE
AB
=
CD
AC
Given: AB=6 cm,DE=4 cm,AC=15 cm,
∴
4
6
=
CD
15
⇒6×CD=15×4
⇒CD=
6
60
⇒CD=10 cm.
(iii)
Area of ΔDEC
Area of ΔABC
=
DE
2
AB
2
(∵ΔABC∼ΔDEC)
=
(4)
2
(6)
2
=
16
36
=
4
9
∴ Area of ΔABC: Area of ΔDEC=9:4
Step-by-step explanation:
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