**Q.1: The height of an equilateral triangle measures 9 cm. Find its area correct to 2 decimal places. (Take root 3 is 3–√=1.732)**

**Sol:**

Let, a be the side of an equilateral triangle.

Therefore, the height of an equilateral triangle =

Height of an equilateral triangle = 9cm (given)

Area of the equilateral triangle =

=

Area of the equilateral triangle = ^{2}

**Q.2: An umbrella is made by stitching 12 triangular pieces of cloth, each measuring 50cm×20cm×50cm. Find the area of the cloth used in it.**

** **

**Sol:**

Let a = 50 cm, b = 20 cm and c = 50 cm

Now,

S =

S =

Therefore, area of triangular piece of cloth =

=

=

=

=

=

=

Therefore, the area of one piece of cloth = 490

Now, the area of 12 pieces =

**Q.3: A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per cm ^{2}**

** **

**Sol:**

Let, a = 16 cm, b = 12 cm and c = 20 cm

Now,

S =

S =

Area of one triangular tile =

=

=

=

Area of one tile = 96 cm^{2}

Cost of polishing the tiles per sq. cm = Re. 1

Thus, the total cost of polishing all the tiles = Rs. (

**Q.4: Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12cm, ∠ACB=90∘ and AC = 15 cm.**

** **

**Sol:**

Consider the right triangle ABC.

Now, by Pythagoras theorem, we have

BC =

Therefore, BC =

Now, the Perimeter of quad. ABCD = 17 + 9 + 12 + 8 = 46 cm

Area of triangle

= 60

For area of triangle, ACD:

Let, a = 15 cm, b = 12 cm and c = 9 cm

Now,

S =

S =

Area of one triangular tile =

=

Thus the area of quadrilateral ABCD= area of

= 60 + 54 = 114 cm^{2}

**Q.5: Find the perimeter and area of the quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29cm, DA=34 cm and ∠CBD=90∘.**

**Sol:**

Perimeter of quad. ABCD = 34 + 29 + 21 + 42 = 126 cm

Area of triangle BCD =

Area of triangle BCD =

For area of triangle ABC,

Let, a = 42 cm, b = 20 cm and c = 34 cm

Now,

S =

S =

Area of triangle =

=

=

=

=

Area of quadrilateral ABCD = area of

Thus, the area of quadrilateral ABCD = 336 + 210 = 546 cm^{2}

**Q.6: Find the area of the quadrilateral ABCD in which AB = 24 cm, ∠BAC=90∘ and △BCD is an equilateral triangle having each side equals to 26cm. Also, find the perimeter of the quadrilateral. (Given 3–√=1.73 )**

** **

**Sol:**

Consider the right triangle ABD,

By pythagoras theorem, we have

Therefore, AB =

Area of the triangle ABD =

Area of equilateral triangle BCD =

^{2}

Area of quad. ABCD = area of

= 120 + 292.37 = 412.37 cm^{2}

**Q.7: Find the area of parallelogram ABCD in which AB = 28 cm, BC = 26cm and diagonal AC = 30 cm.**

** **

**Sol:**

Consider the triangle ABC,

Let, a = 26 cm, b = 30 cm and c = 28 cm

Now,

S =

S =

Area of triangle ABC =

=

=

=

=

=

In a parallelogram, diagonal divides the parallelogram in two equal area

Therefore area of quadrilateral ABCD= area of

= Area of area of

= 336 ^{2}

**Q.8: Find area of parallelogram ABCD in which AB = 14 cm, BC = 10cm and AC = 16 cm. (Given 3–√=1.73)**

** **

**Sol:**

Consider the triangle ABC,

Let, a = 10 cm, b = 16 cm and c = 14 cm

S =

S =

Area of triangle ABC =

=

=

=

=

=

In a parallelogram, diagonal divides the parallelogram in two equal areas

Therefore, the area of quadrilateral ABCD= area of

= Area of area of

=

= ^{2}

**Q.9: In the given figure ABCD is a quadrilateral in which diagonal BD =64 cm AL⊥BD and CM⊥BD such that AL= 16cm and CM =16.8 cm. Calculate the area of quadrilateral ABCD.**

** **

**Sol:**

Area of triangle ABD =

=

= ^{2}

Area of triangle BCD =

=

= ^{2}

Therefore, the area of quadrilateral ABCD= Area of ^{2}

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