In a triangle ABC, AB & BE are altitudes and EC = 8, CD = 6, AE = 3, Find BC.
Answers
Answer:
Let P & R are the mid-point of AD & BC respectively.
Now draw RQ perpendicular to AB.
Now, in △CDB & △RQB,
∠D=∠Q=90°
∠B=∠B
∴△CDB∼△RQB (by AA)
So, basic propertionality theorem, if R is mid point of BC.
∴Q is midpoint of BD.
We can say,
⟹AP+DQ=PD+QB
⟹2(AP+DQ)=PD+QB
A
P+DQ
⟹2(AP+DQ)=AB
⟹AP+DQ=
2
AB
⟹AP+DQ=
2
8
=4
Also,
QR=
2
1
CD=
2
1
×6=3
∴PR=
(QR)
2
+(PQ)
2
(By pythagoras theorem)
=
(3)
2
+(4)
2
=
25
=5
Answer:
Class 7
>>Maths
>>The Triangle and Its Properties
>>Recalling Basics of Triangles and Its Types
>>Δ ABC is an acute angled tr...
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ΔABC is an acute angled triangle CD be the altitude through C. If AB = 8, CD = 6. Find the distance between the midpoints of AD and BC.
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Solution
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Let P & R are the mid-point of AD & BC respectively.
Now draw RQ perpendicular to AB.
Now, in △CDB & △RQB,
∠D=∠Q=90°
∠B=∠B
∴△CDB∼△RQB (by AA)
So, basic propertionality theorem, if R is mid point of BC.
∴Q is midpoint of BD.
We can say,
⟹AP+DQ=PD+QB
⟹2(AP+DQ)=PD+QB
A
P+DQ
⟹2(AP+DQ)=AB
⟹AP+DQ=
2
AB
⟹AP+DQ=
2
8
=4
Also,
QR=
2
1
CD=
2
1
×6=3
∴PR=
(QR)
2
+(PQ)
2
(By pythagoras theorem)
=
(3)
2
+(4)
2
=
25
=5