Question

Find the area of a quadrilateral whose diagonals measure 48 m and 32 m respectively and bisect each other at right angle
i want the answer with figure ​

Answer 1

$\green{\large\underline{\sf{Given- }}}$

A quadrilateral whose diagonals measure 48 m and 32 m respectively bisect each other at right angles.

$\purple{\large\underline{\sf{To\:Find - }}}$

Area of quadrilateral

$\red{\large\underline{\sf{Solution-}}}$

Let assume that ABCD be the required quadrilateral such that diagonals AC and BD measures 48 m and 32 m bisects each other at right angles.

Since, AC = 48 m and BD = 32 m

It means, diagonals are unequal.

So, the required quadrilateral ABCD is a rhombus as its diagonals are unequal and bisects each other at right angles.

We know,

$\pink{\boxed{\tt{ Area_{(rhombus)} = \frac{1}{2} \times product \: of \: diagonals}}}$

So,

$\rm :\longmapsto\:Area_{(rhombus)} = \dfrac{1}{2} \times 48 \times 32$

$\rm :\longmapsto\:Area_{(rhombus)} = 24 \times 32$

$\rm :\longmapsto\:Area_{(rhombus)} = 768 \: {m}^{2}$

Thus,

$\purple{\rm\implies \:\boxed{\tt{ Area_{(rhombus)} \: = \: 768 \: {m}^{2} \: }}}$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Question:-

Find the area of a quadrilateral whose diagonals measure 48 m and 32 m respectively and bisect each other at right angle.

Given:-

Diagonals of a quadrilateral are 48m and 32m respectively.

To Find:-

Area of a quadrilateral.

Solution:-

We know that, If diagonals of quadrilateral bisect each other at right angles, the area of the quadrilateral will be the half of the product of diagonals.

• ∠AOB = 90°,

• OB = OD = 16m,

• OA = OC = 24m.

$\sf \red { Area { \sf{{_{(quadrilateral)}}}} \sf = Ar( \triangle ABC) +Ar ( \triangle ACD)}$

$\sf \blue{ Area{\sf{{_{(traingle)}}}} = \frac{1}{2} = base \times height}$

$\sf \: Area = \frac{1}{2} \times AC × OB + \frac{1}{2} \times AC \times OD$

$\sf = \frac{1}{2} \times AC(OB×OD)$

$\sf = \frac{1}{ { \cancel{2}}} \times 48 \times {{ \cancel{32}}}$

$\sf = 48 \times 16$

$\sf = 768m {}^{2}$

Answer:-

$\sf \green{ \therefore { Area{\sf{{_{(traingle)}}}}} \sf = 786m {}^{2} .}$

Hope you have satisfied. ⚘