Question

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

Answer 1

375.85  m² is area of Quadrilateral

Step-by-step explanation:

the angle contained by last two side is a right angle

=> area of right angle triangle

= (1/2) * 7 * 24

= 84 m²

Hypotenuse of right angle triangle

= √7² + 24²

= √49 + 576

= √625

= 25 m

25 m will be one of the diagonal of Quadrilateral

Splitting Quadrilateral in two triangle

Right angle triangle and triangle with sides 25 , 26 , 27 m

Area of  triangle with sides 25 , 26 , 27 m

s = (25 + 26 + 27)/2 = 39

Area = √39(39 - 25)(39 - 26)(39 - 27)

= √39 * 14 * 13 * 12

= √13 * 3 * 2 * 7 * 13 * 2 * 2 * 3

= 13 * 3 * 2 * √14

= 78 √14

= 291.85 m²

Area of Quadrilateral = 84 +  291.85 = 375.85  m²

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Answer 2

Last two side containes a right angle and make a triangle

• Height of triangle = 7 m
• Base of triangle = 24 m

Area of triangle :

$\boxed{ \bf \blue{ Area = \frac{1}{2} \times b \times h}} \\ \\ \bf Area = \frac{1}{ \cancel2} \times \cancel{24} \times 7 \\ \\ \large \bf \green{Area = 84 {m}^{2} }$

Hypotenuse of triangle :

$\bf {h}^{2} = {b}^{2} + {p}^{2} \\ \\ {h}^{2} = {24}^{2} + {7}^{2} \\ \\ \bf {h}^{2} = 576 + 49 \\ \\ \bf h = \sqrt{625} \\ \\ \bf h = 25$

This hypotenuse is the third side of triangle which have side 25 , 26 , 27

Area of this triangle :

$s = \frac{a + b + c}{2} \\ \\ s = \frac{25 + 26 + 27}{2} \\ \\ s = 39 \\ \\ Area = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ Area = \sqrt{39(39 - 25)(39 - 26)(39 - 27)} \\ \\ Area = \sqrt{39 \times 14 \times 13 \times 12} \\ \\ Area = \sqrt{13 \times 3 \times 7 \times 2 \times 13\times 2 \times 2 \times 3} \\ \\ Area = 13 \times 3 \times 2 \sqrt{14} \\ \\ Area = 78 \sqrt{14} \\ \\ Area = 291.85 {m}^{2}$

Area of quadrilateral :

Area of 1st triangle + Area of second triangle

$84 + 291.85 = 375.85 \\ \\ \boxed{ \bf \orange{ area \: of \: quadrilateral =375.85 {m}^{2} }}$