Question

answer the following question....plss​

Answer 1

Answer:

area of triangles = area of parallelogram

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Answer 2

Given:- It is a Parallelogram.

To Find:- Area of the Parallelogram.

Let's Start:- Area of ΔABD = $\frac{1}{2}$ × $Base$ × $Height$

                                           ⇒ $\frac{1}{2}$ × 5 cm × 2 cm

                                           ⇒ 5 cm²

Now, Area of ΔCBD = $\sqrt{S (S-a) (S-b) (S-c)}$    where, S = $\frac{a+b+c}{2}$

[ Note:- This formula is also known as Heron's formula ]

S = $\frac{5+3+4}{2}$  = $\frac{12}{2}$  = 6 cm         [ a + b + c = Perimeter of triangle ]

                                  ⇒ $\sqrt{6 (6 - 5) ( 6-4) (6-3)}$

                                  ⇒ $\sqrt{6*1*2*3}$

                                  ⇒ $\sqrt{36}$

                                  ⇒ 6 cm²

∴ Area of Parallelogram ABCD = Area of ΔABD + Area of ΔCBD

                                                   = 6 cm² + 6 cm²

                                                   = 12 cm²