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Area of a Quadrilateral

Question 5 iI...

Question

Question 5 (i)
Inthe given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)

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Solution

(i) It can be observed that parallelogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB.
$∴$ Area (PQRS) = Area (ABRS) ... (1)

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Q. Question 5 (i)
Inthe given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)

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)

P Q R S

A B R S

are parallelogram and

X

is any point on side

B R

. Show that
(i)

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

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

Q. 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)

P Q R S

A B R S

are parallelograms and

X

is any point on the side

B R

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

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Area of a Quadrilateral

Standard IX Mathematics

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