Math, asked by aryangiri5224, 7 months ago

ABCD is a rectangle. PQRS is a rhombus whose vertices are mid-points of the sides of the rectangle. AB = 16 cm and AD = 12 cm. Find the: i) Area of Rhombus PQRS ii) Area of triangle PBQ iii) length of PT where PT is perpendicular to SR.

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Answered by itzBrainlymaster
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Understanding Quadrilaterals

Some Special Parallelograms

ABCD is a rectangle and P, ...

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Asked on December 20, 2019 by

Dumpa Tirkey

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

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Here, we are joining A and C.

In ΔABC

P is the mid point of AB

Q is the mid point of BC

PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]

PQ=

2

1

AC

In ΔADC

R is mid point of CD

S is mid point of AD

RS∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]

RS=

2

1

AC

So, PQ∣∣RS and PQ=RS [one pair of opposite side is parallel and equal]

In ΔAPS & ΔBPQ

AP=BP [P is the mid point of AB)

∠PAS=∠PBQ(All the angles of rectangle are 90

o

)

AS=BQ

∴ΔAPS≅ΔBPQ(SAS congruency)

∴PS=PQ

BS=PQ & PQ=RS (opposite sides of parallelogram is equal)

∴ PQ=RS=PS=RQ[All sides are equal]

∴ PQRS is a parallelogram with all sides equal

∴ So PQRS is a rhombus.

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Answered by satishsoni11223
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