The angles P- Q- R and S of a trapezium PQRS are in the ratio of 1-3-7-9- Then-A- - P -81- - Q -45- - R -126- and - S -162-B- - P -18- - Q -54- - R -126- and - S -162-C- - P -18- - Q -54- - R -216- and - S -62-D- - P -8- - Q -94- - R -26- and - S -62-

The correct option is B ∠ P = 18 ^∘, ∠ Q = 54 ^∘, ∠ R = 126 ^∘ and ∠ S = 162 ^∘As the angles are in the ratio 1:3:7:9 nbsp; We can take the angles as x, nb ...

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Byju's Answer

Standard IX

Mathematics

Trapezium and Its Property

The angles P,...

Question

The angles P, Q, R and S of a trapezium PQRS are in the ratio of 1:3:7:9. Then:

∠ P

18^{\circ}

∠ Q

54^{\circ}

∠ R

126^{\circ}

and

∠ S

162^{\circ}

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∠ P

81^{\circ}

∠ Q

45^{\circ}

∠ R

126^{\circ}

and

∠ S

162^{\circ}

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∠ P

18^{\circ}

∠ Q

54^{\circ}

∠ R

216^{\circ}

and

∠ S

62^{\circ}

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∠ P

8^{\circ}

∠ Q

94^{\circ}

∠ R

26^{\circ}

and

∠ S

62^{\circ}

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Solution

The correct option is A $∠ P$ = $18^{\circ}$ , $∠ Q$ = $54^{\circ}$ , $∠ R$ = $126^{\circ}$ and $∠ S$ = $162^{\circ}$
As the angles are in the ratio 1:3:7:9
We can take the angles as $x$ , $3 x$ , $7 x$ and $9 x$ .
$x + 3 x + 7 x + 9 x = 360^{\circ}$
$\Rightarrow 20 x = 360^{\circ}$
$\Rightarrow x = 18^{\circ}$
$∠ P = x = 18^{\circ}$ ,
$∠ Q = 3 x = 54^{\circ}$
$∠ R = 7 x = 126^{\circ}$
$∠ S = 9 x = 162^{\circ}$

Suggest Corrections

The angles P, Q, R and S of a trapezium PQRS are in the ratio of 1:3:7:9. Then:

The correct option is A ∠P = 18∘, ∠Q = 54∘, ∠R = 126∘ and ∠S = 162∘As the angles are in the ratio 1:3:7:9 We can take the angles as x, 3x, 7x and 9x. x+3x+7x+9x=360∘ ⇒20x=360∘ ⇒x=18∘ ∠P=x=18∘, ∠Q=3x=54∘ ∠R=7x=126∘ ∠S=9x=162∘