Question

In a triangle ABC, AB & BE are altitudes and EC = 8, CD = 6, AE = 3, Find BC.​

Answer 1

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 2

Answer:

Class 7

>>Maths

>>The Triangle and Its Properties

>>Recalling Basics of Triangles and Its Types

>>Δ ABC is an acute angled tr...

Question

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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.

Medium

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