Question

1. Find the area of a triangle having sides of length 15 cm, 28 cm and 4lom
Also find the length of cel altitude drown from opposite Wester to
the side of length 28 cm​

Answer 1

Step-by-step explanation:

Perimeter of triangle is 40 (cm). Area of triangle is 60(cm^{2})(cm

2

) . Altitude on side 17 cm is 7.059 (cm).

Step-by-step explanation:

1. Side of triangle is 8 cm, 15 cm and 17 cm.

So

Perimeter of triangle = Sum of side of triangle

Perimeter of triangle= 8+15+ 17=40 (cm)

2. If we see side of triangle

8^{2}+15^{2}=17^{2}8

2

+15

2

=17

2

64+225=17^{2}64+225=17

2

289=17^{2}289=17

2

17^{2}=17^{2}17

2

=17

2

Means it satisfied Pythagoras rule

It must be a right angle triangle, and right angle opposite to side longest side 17 (cm).

3. Base =15 cm

Perpendicular= 8 cm

Hypotenuse =17 cm

4. So area of triangle =\frac{1}{2}\times base\times perpendicular=\frac{1}{2}\times 15\times 8=60(cm^{2})=

2

1

×base×perpendicular=

2

1

×15×8=60(cm

2

)

5.

Area of triangle=\frac{1}{2}\times base\times Perpendicular=\frac{1}{2}\times Hypotenuse\times altitude=

2

1

×base×Perpendicular=

2

1

×Hypotenuse×altitude

We can also write

\frac{1}{2}\times Hypotenuse\times altitude=

2

1

×Hypotenuse×altitude= Area of triangle

\frac{1}{2}\times 17\times altitude= 60

2

1

×17×altitude=60

On solving

Altitude =7.059 (cm)

Answer 2

Answer:

Here,

a=5,b=12,c=13

s=

2

1

(a+b+c)=15

Area of the triangle, A =

s(s−a)(s−b)(s−c)

A=

15(15−5)(15−12)(15−13)

A=30 cm

2

Let p be the length of perpendicular. Then,

A=

2

1

×13×p

Therefore,

2

1

×13×p=30

p=

13

60

cm