Question

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

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : value \: ofx \\ \\ prove \: that \\ △AOB \: and \: △DOC \: are \: similar \\ by \: AA \: test \: of \: similarity \\ then \: by \: c.s.s.t \\ we \: get \\ \\ \frac{AO}{OC} = \frac{BO}{DO} \\ \\ \frac{x - 1}{x + 1} = \frac{x + 1}{x + 4} \\ \\ applying \: componendo \: dividendo \\ on \: both \: sides \\ we \: get \\ \\ \frac{x - 1 + x + 1}{x - 1 - (x + 1)} = \frac{x + 1 + x +4 }{x + 1 - (x + 4)} \\ \\ \frac{2x}{ - 2} = \frac{2x + 5}{1 - 4} \\ \\ \frac{x}{ - 1} = \frac{2x + 5}{ - 3} \\ \\ \frac{x}{1} = \frac{2x + 5}{3} \\ \\ 3x = 2x + 5 \\ \\ x = 5$

Answer 2

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

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.