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**Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.**

Let *ABCD* be the rhombus where diagonals intersect at *O*.

Then* AB* = 15 cm and* AC* = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore, Î” *AOB* is a right-angled triangle, right angled at *O* such that

By Pythagoras theorem, we have,

(*AB*)^{2} = *(OA*)^{2} + (*OB*)^{2}

â‡’ (15)^{2} = (12)^{2} + (*OB*)^{2}

â‡’ (*OB*)^{2 }= (15)^{2 }âˆ’ (12)^{2}

â‡’ *(OB*)^{2 }= 225 âˆ’ 144 = 81

â‡’ (*OB*)^{2} = (9)^{2}

â‡’ *OB* = 9 cm

âˆ´ *BD* = 2 x *OB* = 2 x 9 cm = 18 cm

Hence,

Area of the rhombus* ABCD* =

**Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.**

Let* ABCD *be the rhombus whose diagonals intersect at *O*.

Then *AB* = 20 cm and *AC* = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore Î” *AOB* is a right-angled triangle, right angled at *O* such that*OA* = 1/2AC = 12 cm and *AB* = 20 cm

By Pythagoras theorem, we have,

(*AB*)^{2} = (*OA*)^{2} + (*OB*)^{2}

â‡’ (20)^{2} = (12)^{2} + (*OB*)^{2}

â‡’ (*OB*)^{2 }= (20)^{2 }âˆ’ (12)^{2}

â‡’ (*OB*)^{2 }= 400 âˆ’ 144 = 256

â‡’ (*OB*)^{2} = (16)^{2}

â‡’* OB* = 16 cm

âˆ´ *BD* = 2 x *OB* = 2 x 16 cm = 32 cm

Hence,

Area of the rhombus *ABCD* =

**The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?**

We have,

Side of a square = 4 m and one diagonal of a square = 2 m

Area of the rhombus = Area of the square of side 4 m

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

â‡’ AO=1/2AC = 8 m and BO =1/2BD=1 m

By Pythagoras theorem, we have:

*AO*^{2} + *BO*^{2} = *AB*^{2}

â‡’* **AB*^{2} = (8 m)^{2} + (1m)^{2} = 64 m^{2} + 1 m^{2} = 65 m^{2}

â‡’ Side of a rhombus = *AB* =

Let* DX *be the altitude.

Area of the rhombus = *AB* Ã— *DX*

16 m^{2} x Dx

Hence, the altitude of the rhombus will be

**Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.**

We have,*ABCD* is a parallelogram with longer side *AB *= 25 cm and altitude* AE* = 10 cm.

As ABCD is a parallelogram .hence *AB=CD* (opposite sides of parallelogram are equal)

The shorter side is* AD* = 20 cm and the corresponding altitude is* CF*.

Area of a parallelogram = Base Ã— Height

We have two altitudes and two corresponding bases.

So,

Hence, the altitude corresponding to the other pair of the side *AD* is 12.5 cm.

**The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.**

We have,*ABCD* is a parallelogram with side *AB* = *CD* = 10 cm (Opposite sides of parallelogram are equal) and corresponding altitude *AM* = 12 cm.

The other side is *AD* and the corresponding altitude is* CN* = 8 cm

Area of a parallelogram = Base Ã— Height

We have two altitudes and two corresponding bases.

So,

Hence, the length of the other pair of the parallel side = 15 cm.

**A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm ^{2}.**

We have,

Altitude of a tile = 3 cm

Base of a tile = 5 cm

Area of one tile = Altitude x Base = 5 cm x 3 cm = 15 cm^{2}

Area of 280 tiles = 280 x 15 cm^{2} = 4200 cm^{2}

Rate of polishing the tiles at 50 paise per cm^{2} = Rs. 0.5 per cm^{2}

Thus,

Total cost of polishing the design = Rs. (4200 x 0.5) = Rs. 2100