Question

Find the area of the following shaded regions:-

Don't give wrong answer then i will report and your will be hack​

Answer 1

$\huge\mathfrak\red{Answer}$

LOOK AT THE FIGURE CAREFULLY.

BP = NM = 4cm [Given]

AN = 3cm

AM = 5cm

Area of ∆ANM

$\frac{ h_{b}b}{2} \\ = \frac{3 \times 4}{2} \\ = \frac{12}{2} \\ = 6cm^{2}$

In ∆MPC

BN = PM = 3cm

PC = 4cm

CM = 5cm

Area of ∆MPC

$\frac{ h_{b}b}{2} \\ = \frac{3 \times 4}{2} \\ = \frac{12}{2} \\ = 6 {cm}^{2}$

Therefore, Area of Shaded portion = 6cm² + 6cm² = 12cm².

Answer 2

Answer:

BP = NM = 4cm [Given]

AN = 3cm

AM = 5cm

Area of ∆ANM

\begin{gathered} \frac{ h_{b}b}{2} \\ = \frac{3 \times 4}{2} \\ = \frac{12}{2} \\ = 6cm^{2} \end{gathered}

2

h

b

b

=

2

3×4

=

2

12

=6cm

2

In ∆MPC

BN = PM = 3cm

PC = 4cm

CM = 5cm

Area of ∆MPC

\begin{gathered} \frac{ h_{b}b}{2} \\ = \frac{3 \times 4}{2} \\ = \frac{12}{2} \\ = 6 {cm}^{2} \end{gathered}

2

h

b

b

=

2

3×4

=

2

12

=6cm

2

Therefore, Area of Shaded portion = 6cm² + 6cm² = 12cm².

Step-by-step explanation:

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