Question

find the area of a quadrilateral one of whose diagonals is 40 cm and the length of the perpendiculars drawn from the opposite vertices on the diagonals are 16 cm and 12 cm​

Answer 1

Given:

For a quadrilateral,

Length of a diagonal=40CM

Perpendiculars dropped onto the diagonal are

16cm and 12cm

To find:

Area of quadrilateral

Explanation:

As per the given Information,

$\sf Length \:of\: the\: diagonal ,d=40cm$

$\sf h_1=16cm$

$\sf h_2=12cm$

we know that

$\boxed{\sf area \: of \: quadrilateral=\dfrac{d}{2}(h_1+h_2)}$

Substituting values

$\longrightarrow \sf \bigtriangleup = \dfrac{40}{2} (12 + 16) \\ \\ \longrightarrow \sf \bigtriangleup = 20(28) \\ \\ \longrightarrow \sf \bigtriangleup = 560cm{} ^{ 2}$

Therefore,

Area of the given quadrilateral=560cm²

Answer 2

SOLUTION:-

Given:

A quadrilateral one of whose diagonal is 40cm & the length of the perpendicular drawn from the opposite vertices on the diagonals are 16cm & 12cm.

To find:

The area of quadrilateral.

Explanation:

We have,

• One diagonal of quadrilateral= 40cm
• Opposite vertices diagonals= 16cm & 12cm
• Height (H1)= 16cm
• Height (H2)= 12cm

We know that, area of quadrilateral:

Area of ∆ABC + Area of ∆BCD

$Area \: of \: quadrilateral = \frac{1}{2} \times BC \times H1 + \frac{1}{2} \times BC \times H2 \\ \\ Area \: of \: quadrilateral = \frac{1}{2} \times BC \times (H1 + H2) \\ \\ Area \: of \: quadrilateral = \frac{1}{2} \times 40(16 + 12) \\ \\ Area \: of \:quadrilateral = 20(28) {cm}^{2} \\ \\ Area \: of \: quadrilateral = 560 {cm}^{2}$

Thus,

The area of quadrilateral is 560cm².