Question

A field in the form of a parallelogram has an area of 432m find the length of its diagonal on which the perpendicular drawn from two vertices on either side of it are 12 m long.​

Answer 1

Answer:

Given: A field in the form of a parallelogram has an area of 432 m².

We need to find the length of diagonal. Let length of diagonal be x m.

As we know a diagonal divides parallelogram into two triangles.

Formula: Area of triangle = 1/2 * Base * Height

Calculation:

:Area of triangle 1 + Area of triangle 2 = Area of parallelogram

1/2 * x * h1 + 1/2 * x * h2 = 432

1/2 * x * 12 + 1/2 * x * 12 = 432

6x + 6x = 432

12x = 432

x = 36m .

So,, The length of diagonal is 36 m .

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Answer 2

• Given area = 432 m²✓

• And the perpendicular drawn from the either sides are of length 12 cm ✓

So it is common thing that the perpendicular forms 90° with the horizontal so the due to the intersections of diagonals the parallelogram is divided into two triangls(∆)

So in the above attachment notice that the parallelogram is divided into two traingle ∆PQR and ∆PSR respectively

So

$\red{ \bold{area \: of \: triangle \: pqr \: + area \: of \: \:triangle = area \: of \: parallelogram\: }}$

Therefore :

_______________

,

$= > \frac{1}{2} \times (base1) \times( height 1)+ \frac{1}{2} \times( base \: 2)\times( height \: 2) =432 \\ \\ = > \bold{ \: let \: bas2e \: be \: x} \\ \\ =>\frac{1}{2}× x ×12+\frac{1}{2}× x ×12=432\\ \\</p><p>= > x \times 6 + x \times 6 =432 \\ \\ = > 12x = 432 \\ \\ = > 24x = 432\\ \\ = > 12x = 432 \\ \\ = > x = \frac{432}{12} \\ \\ = > \red{ x =36} \\ \\ = > therefore \: the \: base \: of \: one \: triangle \: is= to \: the \: diagonal \: of \: the \: paralleogram$

Therefore diagonal is 36 m✓