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15. In the adjoining figure, ZABC = ZBCD = 90°,
ZDAB = 0, ZDBC = 0, AB = 14 cm,
BC = 12 cm and BD = 13 cm. Find the value of
[STANDARD
(i) sino
(ii) cote
(iii) sec
14 cm
D
13 cm
B 12 cm
Answers
Answered by
1
Answer:
We know Area of triangle=
2
1
×Base×Height.
i)In △ABC,Base=AB=3cm and Height=AC=4cm
So, Area of △ABC=
2
1
×3×4=6cm
2
ii)We apply Pythagoras theorem in △ABC and get
BC
2
=AB
2
+AC
2
, substitute all values we get
BC
2
=3
2
+4
2
=16
BC=5cm
iii)We apply Pythagoras theorem in △ABD and get
AB
2
=AD
2
+BD
2
AD
2
=AB
2
−BD
2
,substitute values we get
AD
2
=3
2
−BD
2
=9−BD
2
.......(1)
Apply Pythagoras theorem to △ACD and get
AC
2
=AD
2
+CD
2
,
AD
2
=AC
2
−CD
2
,
AD
2
=AC
2
−(BC−BD)
2
AD
2
=AC
2
−BC
2
−BD
2
+2BC.BD
AD
2
=3
2
−5
2
−BD
2
+2×5.BD
AD
2
=−16−BD
2
+10BD
9−BD
2
=−16−BD
2
+10BD
10BD=25
BD=2.5
Substitute the value in eqn(1), we get
AD
2
=9−(2.5)
2
=9−6.25=2.75
∴AD=1.65cm
∴ The length of altitude from A to BC is AD=1.65cm
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