Question

The sides of a quadrilateral field, taken in order are 26m 27m 7m 24m respectively. The angle contained by the last two sides is aright angle. Finds its area

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

REFER TO THE IMAGE FOR DIAGRAM

✤✤✤✤✤✤✤✤GIVEN✤✤✤✤✤✤✤

ABCD is a quadrilateral such that,

AB= 26cm ; BC= 27cm ; CD= 7cm ; AD= 24cm

and,

angle ADC = 90°

✤✤✤✤✤✤CONSTRUCTION ✤✤✤✤

join diagonal AC of quadrilateral ABCD

✤✤✤✤✤✤SOLUTION✤✤✤✤✤

In ∆ ADC by Pythagoras theorem

AC^2 = AD^2 + CD^2

AC^2 = (24) ^2 + (7) ^2

AC^2 = 576 + 49

AC^2 = 625

AC = 25 cm

Now, Consider ∆ ABC

let,

AB = 26 cm = a ; BC = 27 cm= b ; AC = 25 cm= c

semi- perimeter,

$s \: = \frac{a + b + c}{2}$

$s = \frac{26 + 27 + 25}{2}$

$s = \frac{78}{2} = 39$

HERON'S FORMULA

$ar(triangle) = \sqrt{(s(s - a)(s - b)(s - c))}$

PUTTING VALUES IN HERON'S FORMULA

Ar(∆ABC) =

$\sqrt{(39 \times (39 - 26)(39 - 27)(39 - 25))} \\ \\ = \sqrt{(39 \times 13 \times 12 \times 14)} \\ \\ = 291.8 \: {cm}^{2}$

Now consider ∆ ADC

ar (∆ADC) = 1/2 × AD × CD

$= \frac{1}{2} \times 24 \times 7 \\ \\ = 84 {cm}^{2}$

Ar( QUADRILATERAL ABCD )= AR( ABC) + AR(ADC)

ar(ABCD) = 291.8 cm^2 + 84 cm^2

ar(ABCD) = 375.8 cm^2

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