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QUESTION: 1

The area of a triangle with base 8 cm and height 10 cm is

Solution:

QUESTION: 2

The sides of a triangle are in the ratio of 3 : 4 : 5. If its perimeter is 36 cm, then what is its area?

Solution:
We know that area of parallelogram = base Ã— height = 60 Ã— 24 cm =1440 cm^2

QUESTION: 3

An isosceles right triangle has area 8 cm^{2}. The length of its hypotenuse is

Solution:

QUESTION: 4

The cost of turfing a triangular field at the rate of Rs. 45 per 100 m^{2} is Rs. 900. If the double the base of the triangle is 5 times its height, then its height is

Solution:

Let the height of triangular field be h metres.

It is given that 2 x (base) = 5 × (Height)

QUESTION: 5

The area of the the triangle having sides 1 m, 2 m and 2 m is :

Solution:
*Through Heronâ€™s formula =**S=Perimeter/2**S=5/2**Let a=1m, b=2m ,c=2m*

*Formulae= Root(s(s-a)(s-b)(s-c))**=root(5/2(5/2 - 1)(5/2 - 2)(5/2 - 2))**=root(5/2*3/2*1/2*1/2)**=root(15/16)**=root(15)/4*

QUESTION: 6

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be

Solution:

QUESTION: 7

The area of a triangle whose sides are 12 cm, 16 cm and 20 cm is

Solution:
S = 12+16+20/2

= 24

Area of a triangle = âˆšs(s-a)(s-b)(s-c)

= âˆš24(24-12)(24-16)(24-20)

= âˆš24 x 12 x 8 x 4

=6 x 4 x 4

= 96 cm^2

= 24

Area of a triangle = âˆšs(s-a)(s-b)(s-c)

= âˆš24(24-12)(24-16)(24-20)

= âˆš24 x 12 x 8 x 4

=6 x 4 x 4

= 96 cm^2

QUESTION: 8

The difference of semi-perimeter and the sides of △ABC are 8, 7 and 5 cm respectively. Its semi-perimeter ‘s’ is

Solution:

QUESTION: 9

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

Solution:

QUESTION: 10

The sides of a triangular flower bed are 5 m, 8 m and 11 m. the area of the flower bed is

Solution:

QUESTION: 11

A triangle ABC in which AB = AC = 4 cm and ∠A = 90^{o}, has an area of

Solution:

QUESTION: 12

The perimeter and area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm respectively are

Solution:

QUESTION: 13

The perimeter of a rhombus is 20 cm. One of its diagonals is 8 cm. Then area of the rhombus is

Solution:

Given, Perimeter of a rhombus = 20 cm

Perimeter of a rhombus = 4*side

Hence, side = 20/4 = 5 cm.

Now, we know that the diagonals of a rhombus bisect each other at right angles (90 degree).

Hence 'a right angled triangle can be visualised with 'side' as the hypotenuse'.

diagonal length = 8 cm

Half the length (since diagonal bisects each other) = 8/2 = 4 cm

(d/2)^{2} + (d1/2)^2 = 5^{2}

4^2 + (d1/2)^{2} = 5^{2}

(d1/2)^{2} = 9

d1/2 = 3

d1 = 3*2 = 6 cm

Hence other diagonal = 6 cm.

Area = 1/2 * d1*d = 1/2 * 8 * 6 = 24 cm^{2}

QUESTION: 14

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

Solution:

QUESTION: 15

The length of the sides of a triangle are 5 cm, 7 cm and 8 cm. Area of the triangle is :

Solution:

QUESTION: 16

If the side of an equilateral triangle is 4 cm, then its area is

Solution:

QUESTION: 17

The area of a triangle whose sides are 15 cm, 8 cm and 19 cm is

Solution:

QUESTION: 18

If one side and one diagonal of a rhombus and 20 m and 24 m, then its area =

Solution:

The rhombus has a perimeter of 80m.

A rhombus has 4 equal sides, each would be 20m.

One of the diagonals is 24m...this is bisected by the other diagonal into 12m segs.

The diagonals of a rhombus intersect at right angles. Use the Pythagorean Theorem to find the length of the other diagonal...

12² + b² = 20²

144 + b² = 400

b² = 256

b = 16

This is of course the length of just one segment, the diagonal is 32m.

Now we can use our area formula with the diagonals.

A = ½d¹d²

A = ½(24)(32)

A = 384 m²

QUESTION: 19

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm^{2 }is

Solution:

QUESTION: 20

Area of an equilateral triangle of side 2 cm is :

Solution:

QUESTION: 21

If the perimeter of an equilateral triangle is 24 m, then its area is

Solution:

Given:

Area of equilateral ∆ = 16√3 m²

Let the side of equilateral ∆ = a m

Area of equilateral ∆ = (√3/4)a²

16√3 = (√3/4)a²

16√3 × (4/√3) = a²

64 = a²

a = √64

a =√ 8 × 8 = 8 m

Side of equilateral ∆ = 8 m

Perimeter of equilateral ∆ = 3 × side

Perimeter of equilateral ∆ = 3 × 8 = 24 m.

Hence, the perimeter of equilateral ∆ is 24 m.

QUESTION: 22

Each of the equal sides of an isosceles triangle is 2 cm greater than its height. If the base of the triangle is 12 cm, then its area is

Solution:

and, equating this with the basic area formula,

i.e. A = 1/2.h.b

QUESTION: 23

The product of difference of semi-perimeter and respective sides of △ABC are given as 13200 m^{3}. The area of △ABC, if its semi-perimeter is 132m, is given by

Solution:

QUESTION: 24

The area of equilateral triangle of side ‘a’ is 4√3 cm^{2}. Its height is given by

Solution:

QUESTION: 25

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be :

Solution:

Given,

base= 8cm

hypotenuse= 10cm

First, let us find the height of the triangle:

Let height = h

By applying pythagorean theorem

h 2 + 8 2 = 10 2

h 2 + 64 = 100

h 2 = 100 - 64

h 2 = 36

h = 6cm

Now let us find the area,

Area = 1/2 x b x h

= 1/2 x 8 x 6

= 1/2 x 48

= 24cm 2

Therefore, area of the right triangle = 24cm 2

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