Question

Xy parallel to AC and xy divides triangle ABC into two reasons equal in area so that a X upon ab is equal to 2 minus root 2upon 2

Answer 1

Step-by-step explanation:

answr

search

What would you like to ask?

MATHS

In given figure, XY||AC and XY divides triangular region ABC into two parts equal in area. Determine

AB

AX

.

1009457

Share

Study later

ANSWER

We have XY||AC

and, Area(△BXY)=Area(quadXYCA)

⇒ Area(△ABC)=2Area(△BCY)........(i)

Now, XY∣∣AC and BA is a transversal.

⇒ ∠BXY=∠BAC.......(ii)

Thus, in △

′

sBAC and BXY, we have

∠XBY=∠ABC [Common]

∠BXY=∠BAC [From(ii)]

Therefore, AA-criterion of similarity, we have

△BAC∼△BXY

⇒

Area(△BXY)

Area(△BAC)

=

BX

2

BA

2

⇒2=

BX

2

BA

2

[Using (i)]

⇒ BA=

2

BX⇒BA=

2

(BA−AX)

⇒ (

2

−1)BA=

2

AX

⇒

AB

AX

=

2

2

−1

Answer 2

Step-by-step explanation:

MATHS

In given figure, XY||AC and XY divides triangular region ABC into two parts equal in area. Determine

AB

AX

.

1009457

Share

Study later

ANSWER

We have XY||AC

and, Area(△BXY)=Area(quadXYCA)

⇒ Area(△ABC)=2Area(△BCY)........(i)

Now, XY∣∣AC and BA is a transversal.

⇒ ∠BXY=∠BAC.......(ii)

Thus, in △

′

sBAC and BXY, we have

∠XBY=∠ABC [Common]

∠BXY=∠BAC [From(ii)]

Therefore, AA-criterion of similarity, we have

△BAC∼△BXY

⇒

Area(△BXY)

Area(△BAC)

=

BX

2

BA

2

⇒2=

BX

2

BA

2

[Using (i)]

⇒ BA=

2

BX⇒BA=

2

(BA−AX)

⇒ (

2

−1)BA=

2

AX

⇒

AB

AX

=

2

2

−1