Math, asked by ItxAttitude, 2 months ago

Class 10
Abcd is a trapezium AB||CD
Area related to circles​

Answers

Answered by EmperorSoul
14

\large\underline{\sf{Solution-}}

Given that,

A trapezium ABCD in which AB || CD such that

AO = x - 1

CO = BO = x + 1

OD = x + 4

Now,

\rm :\longmapsto\:In \: \triangle  \: AOB  \: and \: \triangle \:  COD

\rm :\longmapsto \: \angle  \: AOB  \: and \: \angle \:  COD \:  \:  \{vertically \: opposite \: angles \}

\rm :\longmapsto \: \angle  \:ABO  \: and \: \angle \:  CDO \:  \:  \{alternate \: interior \: angles \}

\bf\implies  \: \triangle  \: AOB \:  \sim \:  \triangle  \: COD \:  \:  \:  \{AA \: similarity \}

\bf\implies \:\dfrac{AO}{CO}  = \dfrac{BO}{DO}

\rm :\longmapsto\:\dfrac{x - 1}{x + 1}  = \dfrac{x + 1}{x + 4}

\rm :\longmapsto\:(x - 1)(x + 4) =  {(x + 1)}^{2}

\rm :\longmapsto\: {x}^{2} - x + 4x - 4 =  {x}^{2} + 1 + 2x

\rm :\longmapsto\: 3x - 4 =  1 + 2x

\rm :\longmapsto\: 3x  -  2x=  1 +4

\bf\implies \:x = 5

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SHORT CUT TRICK

The diagonals of the trapezium divides each other proportionally. i.e. If ABCD is a trapezium and diagonals AC and BD intersects at O, then

 \red{\bf\implies \:\boxed{\sf{ \dfrac{AO}{CO}  = \dfrac{BO}{DO} }}} \\

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LEARN MORE

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem :-

This theorem states that :- If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answered by itzblackhole
0

Step-by-step explanation:

AB=7cm,DE=4cm,BF=3.5cm

BC=DE+EC=(4+7)cm=11cm

Area pf trapezium ABCD=

2

1

×(Sum of parallelogram)×(Distance between them)

=

2

1

×(11+7)×3.5

=

2

1

×18×3.5

=31.5cm

2

Area of the sector BGEC=

360

30

×

9

22

×7×7

=

12

1

×22×7

=12.83cm

2

Area of shaded region = Area of trapezium - Arc of sector

(31.5−12.88)cm

2

=18.67cm

2

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