Question

D
6. P(x) is a second degree polynomial with P(2) = 0 P(-5) = 0
(a
Find two first degree factors of P(x)
(b) Find the polynomial P(x)​

Answer 1

Answer:

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Step-by-step explanation:

Given :

Sides of triangle :

8 , 10 and 14 cm

To find :

Radius of inscribed circle.

Solution :

The sides of triangle are

a = 8 cm

b = 10 cm

c = 14 cm

The semi perimeter of triangle s :

\begin{gathered} \bf \: s = \frac{a + b + c}{2} \\ s = \frac{8 + 10 + 14}{2} = \frac{32}{2} \end{gathered}s=2a+b+cs=28+10+14=232

So, Semi perimeter = 16 cm

According to Heron's formula,

The area of triangle :

\bf \: A = \sqrt{s(s - a)(s - b)(s - c)}A=s(s−a)(s−b)(s−c)

So, putting the values of sides a, b, c and Semi perimeter s

Area of triangle :

\begin{gathered} \bf \: A = \sqrt{16(16 - 8)(16 - 10)(16 - 14)} \\ \bf \: A = \sqrt{16 \times 8 \times 6 \times 2} \\ \bf \: A = \sqrt{1536} = 39.19 \: {cm}^{2} \end{gathered}A=16(16−8)(16−10)(16−14)A=16×8×6×2A=1536=39.19cm2

Area of triangle with inscribed circle is:

A \: = r \times sA=r×s

where

r = radius of inscribed circle.

s = semi perimeter

So

Putting the values of A and s in above equation

we get

\begin{gathered}39.19 \: = r \times 16 \\ so \\ r = \frac{39.19}{16} = 2.45 cm\end{gathered}39.19=r×16sor=1639.19=2.45cm

So, the radius of inscribed circle r = 2.45 cm