Question

The sides of a quadrilateral taken in order are 5,12,14 and 15 m respectively and the angle contained by first two sides is a right angle. Find its area.

Answer 1

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Answer 2

Answer:

The area of the Quadrilateral is 114 m² .

Step-by-step explanation:

Let the name of the Quadrilateral is ABCD.

The four sides of the Quadrilateral are 5m, 12m,14m and 15m.

The angle between the first two sides is right angle. So according to question ABC ∆ is a right angled triangle.

In right angled triangle AC is Hypotenuse ,BC is Base , AB is Altitude.

We can calculate the value of AC by using Hypotenuse formula.

Hypotenuse formula:

Hypotenuse formula:(Base)²+(altitude)²=(Hypotenuse)²

AB²+BC²=AC²

5²+12²=AC²

25+144=AC²

AC=√169

=13 m

The area of the ABC ∆

$= \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 12 \times 5$

$= 30 \: {m}^{2}$

ADC is a Scalene Triangle.

The three sides of the triangle are AC=13m DC=14m and AD=15m

The semiperemeter of the triangle is

S= AC+DC+AD/2

$\frac{13 + 14 + 15}{2}$

$= \frac{42}{2}$

$= 21 \: m$

Area of the ADC∆=

$\sqrt{s \times (s - a) \times (s - b) \times (s - c)}$

$= \sqrt{21 \times( 21 - 13)(21 - 14) \times (21 - 15)}$

Here, a=AC=13m, b=DC=14m, c=AD=15m

$= \sqrt{21 \times 8 \times 7 \times 6}$

$= \sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3}$

Factorise 21,8,7 and 6

$= 7 \times 3 \times 2 \times 2 \: {m}^{2}$

$= 84 \: {m}^{2}$

The area of the ABCD quadrilateral

= area of ABC∆ + area of ADC∆

$= 30 \: {m}^{2} + 84 \: {m}^{2}$

$= 114 \: {m}^{2}$