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**Find the area in square centimetres of a triangle whose base and altitude are as under: (i) base = 18 cm, altitude = 3.5 cm (ii) base = 8 dm, altitude = 15 cm**

We know that the area of a triangle = 1/2Ã—BaseÃ—Height

(i) Here, base = 18 cm and height = 3.5 cm

âˆ´ Area of the triangle = (1/2Ã—18Ã—3.5)=31.5 cm^{2}

(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]

and height = 3.5 cm

âˆ´ Area of the triangle = (1/2Ã—80Ã—15)=600 cm^{2}

**Find the altitude of a triangle whose area is 42 cm ^{2} and base is 12 cm.**

We have,

Altitude of a triangle =

Here, base = 12 cm and area = 42 cm^{2}

**The area of a triangle is 50 cm ^{2}. If the altitude is 8 cm, what is its base?**

We have,

Base of a triangle =

Here, altitude = 8 cm and area = 50 cm^{2}

**Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.**

In a right-angled triangle, the sides containing the right angles are of lengths 20.8 m and 14.7 m.

Let the base be 20.8 m and the height be 14.7 m.

Then,

Area of a triangle

**The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.**

For the first triangle, we have,

Base = 15 cm and altitude = 7 cm

Thus, area of a triangle =

It is given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 cm^{2}

One side of the second triangle = 10.5 cm

Therefore,

The other side of the second triangle =

Hence, the other side of the second triangle will be 10 cm.

**A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?**

We have,

Length of the rectangular field = 48 m

Breadth of the rectangular field = 20 m

Area of the rectangular field = Length x Breadth = 48 m x 20 m = 960 m^{2}

Area of one right triangular flower bed = 1/2Ã—12 mÃ—5 m = 30 m^{2}

Therefore,

Required number of right triangular flower beds

**In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL âŠ¥ AC, BM âŠ¥ AC, DL = 16.5 cm and BM= 12 cm. Find the area of quadrilateral ABCD.**

Hence,

Area of quadrilateral *ABCD* = Area of Î”* ADC* + Area of Î”* ABC*

= (693 + 504) cm^{2}^{ }= 1197 cm

**Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.**

We have,

Diagonal *AC* = 48 cm and diagonal *BD* = 32 m

âˆ´ Area of a quadrilateral = 1/2 x Product of diagonals

**In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF âŠ¥ AD and EF= 14 cm. Calculate the area of the shaded region.**

We have,

Area of the rectangle = *AB* x* BC*

= 32 m x 18 m

= 576 m^{2}

Area of the triangle =

= 1/2x BC x FE [Since AD = BC]

= 1/2x 18 m x 14 m

= 9 m x 14 m = 126 m^{2}

âˆ´ Area of the shaded region = Area of the rectangle âˆ’ Area of the triangle

=(576 âˆ’ 126) m^{2}^{ }=^{ }450^{ }m

In Fig. 32, *ABCD* is a rectangle of length *AB* = 40 cm and breadth *BC* = 25 cm. If *P*, *Q*, *R*, *S* be the mid-points of the sides *AB*, *BC*, *CD* and *DA* respectively, find the area of the shaded region.

We have,

Join points *PR *and *SQ*.

These two lines bisect each other at point* O.*

Here, *AB* = *DC* = *SQ* = 40 cm and *AD = BC =RP* = 25 cm

From the figure we observed that,

Area of Î” *SPQ* = Area of Î” *SRQ*

Hence, area of the shaded region 2 x (Area of Î” *SPQ*)