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"5. In Fig. 9.17 , PQRS and ABRS are parallelograms\nand \\( X \\) is any point on side \\( B R \\). Show that\n(i) ar (PQRS) = ar (ABRS)\n(ii) ar \\( ( A X S ) = \\frac { 1 } { 2 } \\) ar (PQRS)\n\\( \\begin{array} { l l } { \\text { (ii) } \\operatorname { ar } ( A X S ) = \\frac { 1 } { 2 } \\text { ar (PQRS) } } & { \\text { S } } \\end{array} \\)"

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Similar Questions

Q1

In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) area (PQRS) $=$ area (ABRS)

(ii) area (AXS) $= \frac{1}{2}$ area (PQRS)

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Q2

In figure, $$PQRS$$ and $$ABRS$$ are parallelograms and $$X$$ is any point on side $$BR$$. Show that
$$ar. (PQRS)=ar. ( \square ABRS)$$

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Q3

In fig,

P Q R S

and

A B R S

are parallelogram and

X

is any point on side

B R

. Show that

(i)

a r e a (P Q R S) = a r e a (A B R S)

(ii)

a r e a (A X S) = \frac{1}{2} a r e a (P Q R S)

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Q4

PQRS and ABRS are parallelogram and x is any point on the side BR. show that
(I) ar(PQRS) = ar(ABRS)
(ii) ar

(Δ A X S) = \frac{1}{2} a r (P Q R S)

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Q5

In Fig. $ PQRS$and $ ABRS$are parallelograms and $ X$ is any point on side $ BR.$ Show that

$ i) ar(PQRS)=ar(ABRS\left)\phantom{\rule{0ex}{0ex}}ii\right) ar\left(AXS\right)=\frac{1}{2}ar\left(PQRS\right)$

View Solution