In a △ A B C , A D is a median and A L ⊥ B C . Prove that A C 2 = A D 2 + B C · D L + B C 2 2

Given: In ∆ ABC , AD is a median and AL ⊥ BC In right ∆ ACL , AC 2 = AL 2 + LC 2 ---------- (i) In right △ ALD , AL 2 = AD 2 - DL 2 ----------(ii) ∴ AC 2 = AD 2 - DL 2 + LC 2 [ from (i) and (ii) ] = AD 2 - DL 2 + ( DL + DC ) 2 [ Since , LC = DL + DC ] = AD 2 - DL 2 + DL + 1 2 BC 2 [ Since , DC = 1 2 BC ] = AD 2 - DL 2 + DL 2 + 1 4 BC 2 + 2 DL . 1 2 BC = A D 2 + B C · D L + 1 4 B C 2 = A D 2 + B C · D L + B C 2 2 Hence, proved.

In a △ A B C , A D is a median and A L ⊥ B C . Prove that A B 2 = A D 2 - B C · D L + B C 2 2

In right △ A B L and △ A L D , we get, Δ A B L : A B 2 = A L 2 + L B 2 … . (i) Δ A L D : A L 2 = A D 2 - D L 2 . … . (ii) ∴ A B 2 = A D 2 - D L 2 + L B 2 [ from (i) and (ii) ] = A D 2 - D L 2 + ( B D - D L ) 2 = A D 2 - D L 2 + 1 2 B C - D L 2 = A D 2 - B C · D L + B C 2 2 Hence, proved.

In a △ A B C , A D is a median and A L ⊥ B C . Prove that A C 2 + A B 2 = 2 . A D 2 + 1 2 B C 2

In right ∆ A C L , A C 2 = A L 2 + L C 2 ---------- (i) In right △ A L D , A L 2 = A D 2 - D L 2 ----------(ii) ∴ A C 2 = A D 2 - D L 2 + L C 2 [ from (i) and (ii) ] = A D 2 - D L 2 + ( D L + D C ) 2 = A D 2 - D L 2 + D L + 1 2 B C 2 = A D 2 + B C · D L + 1 4 B C 2 = A D 2 + B C · D L + B C 2 2 - - - - - - ( A ) In right △ A B L and △ A L D , we get, Δ A B L : A B 2 = A L 2 + L B 2 … . (iii) Δ A L D : A L 2 = A D 2 - D L 2 . … . (iv) ∴ A B 2 = A D 2 - D L 2 + L B 2 [ from (iii) and (iv) ] = A D 2 - D L 2 + ( B D - D L ) 2 = A D 2 - D L 2 + 1 2 B C - D L 2 = A D 2 - B C · D L + B C 2 2 - - - ( B ) Adding ( A ) and ( B ) , we get, = A C 2 + A B 2 = A D 2 + B C · D L + B C 2 2 + A D 2 - B C · D L + B C 2 2 = 2 A D 2 + B C 2 4 + B C 2 4 = 2 A D 2 + B C 2 4 + B C 2 4 = 2 A D 2 + 1 2 B C 2 . Hence, proved.

E M B I B E

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 19

MEDIUM

10th CBSE

IMPORTANT

Earn 100

$A B C$ is an isosceles triangle, right-angled at $B$ . Similar triangles $A C D$ and $A B E$ are constructed on sides $A C$ and $A B$ . Find the ratio between the areas of $Δ A B E$ and $Δ A C D$ .

Important Questions on Triangles

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 20

An aeroplane leaves an airport and flies due north at a speed of

1000 km per hour

. At the same time, another aeroplane leaves the same airport and flies due west at a speed of

1200 km per hour

. How far apart will be the two planes after

1 \frac{1}{2} hours

?

HARD

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 21

In a

△ A B C

,

A D

is a median and

A L ⊥ B C

.

Prove that

A C^{2} = A D^{2} + B C \cdot D L + {(\frac{B C}{2})}^{2}

HARD

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 21

In a

△ A B C

,

A D

is a median and

A L ⊥ B C

.

Prove that

A B^{2} = A D^{2} - B C \cdot D L + {(\frac{B C}{2})}^{2}

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 21

In a

△ A B C

,

A D

is a median and

A L ⊥ B C

.

Prove that

A C^{2} + A B^{2} = 2 . A D^{2} + \frac{1}{2} B C^{2}

HARD

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 7 - Triangles>Exercise 7D>Q 22

Naman is doing fly-fishing in a stream. The tip of his fishing rod is $1.8 m$ above the surface of the water and the fly at the end of the string rests on the water $3.6 m$ away from him and $2.4 m$ from the point directly under the tip of the rod. Assuming that the string (from the top of his rod to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of $5 cm$ per second, what will be the horizontal distance of the fly from him after $12$ seconds.