Question

PLEASE SOLVE THIS QUESTION ​

Answer 1

$\large\mathrm{Given:}$

$\mathtt{ seg. AB\:||\: seg.DC}$

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$\large\mathrm{To \:Find:}$

$\mathtt{ x = ?}$

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$\large\mathrm{Solution}$

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$\mathtt{ In \: \triangle{AOB} \:and\:\triangle{DOC}}$

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$\mathtt{\angle{AOB}\: = \:\angle{BOC}\:\:(Vertically\: Opposite\:Angles)}$

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$\mathtt{\angle{OAB}\: = \:\angle{OCD}\:\:(Alternate \:angels\:of\:||\:lines)}$

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$\mathtt{\angle{ABO}\: = \:\angle{CDO}\:\:(Alternate \:angels\:of\:||\:lines)}$

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${\therefore{\mathtt{\triangle{AOB}\:\sim\:\triangle{DOC}\:\:\:(AAA\:similarity\: criteria}}}$

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$\mathtt{hence ,\dfrac{OD}{OB} = \dfrac{OC}{OA}\:\:(Ratio\: of \:corres. \:sides \:of \:similar \triangle\: is \:equal)}$

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${\longmapsto{\mathtt{\dfrac{3}{x-3} = \dfrac{x-5}{3x-15}}}}$

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${\longmapsto{\mathtt{3(3x - 19) = (x - 5)(x - 3)}}}$

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${\longmapsto{\mathtt{9x - 57 = {x}^{2} - 3x -5x + 15 }}}$

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${\longmapsto{\mathtt{ {x}^{2} - 3x - 5x + 15 - 9x + 57 = 0 }}}$

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${\longmapsto{\mathtt{{x}^{2} - 17x + 72 = 0 }}}$

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${\longmapsto{\mathtt{ {x}^{2} - ( 8 + 9 )x + 72 = 0 }}}$

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${\longmapsto{\mathtt{ {x}^{2} - 8x - 9x + 72}}}$

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${\longmapsto{\mathtt{ x( x - 8 ) - 9( x - 8 ) }}}$

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${\longmapsto{\mathtt{ (x - 8 )( x - 9 ) }}}$

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${\star{\mathrm{ For ( x - 8 ) = 0 }}}$

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${\longmapsto{\mathtt{ ( x - 8 ) = 0 }}}$

${\longmapsto{\mathtt{ x = 8 }}}$

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${\star{\mathrm{ For ( x - 9 ) = 0 }}}$

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${\longmapsto{\mathtt{ ( x - 9 ) = 0 }}}$

${\longmapsto{\mathtt{ x = 9 }}}$

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${\blue{\underline{\green{\boxed{\mathrm{ Value\:of\:x\:is\:8 \:and\:9}}}}}}$

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$\mathrm{Some\: similarity\: Criteria}$

$\star$AAA similarity Criterion. If two triangles are equiangular, then they are similar.

$\star$Corollary(AA similarity). If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

$\star$SSS Similarity Criterion. If the corresponding sides of two triangles are proportional, then they are similar.

$\star$SAS Similarity Criterion. If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

Answer 2

ANSWER:-

Given:-

Seg.AB|| Seg.DC

TO FIND:-

X=?

SOLUTION:-

In △AOB and △DOC

∠AOB =∠BOC (VERTICALLY OPPSITE ANGLE)

∠OAB = ∠OCD (ALTERNATE ANGLE OF||LINES)

∠ABO =∠CDO (ALTERNATE ANGLE OF LINES)

∴△AOB ~ △DOC (AAA SIMILARITY CRITERIA)

Hence,OD/OB = OC/OA (RATIO OF CORRES.SIDES OF SIMILAR△ IS EQUAL)

➠ 3/X-3 = X-5/3x-15

➠3(3x-19)=(x-5)(x-3)

➠9x -57 =x^2 -3x -5x+15

➠x^2-3x-5x+15-9x+57 =0

➠x^2 -(8+9)x+72=0

➠x^2 -8x -9x +72

➠x(x-8)-9(x-8)

➠(x-8) (x-9)

For (x-8)=0

➠(x-8)=0

➠x=8

For (x-9)=0

➠(x-9)=0

➠x=9