Question

3
The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 metres respectively, and
the angle contained by the first two sides is a right angle. Find its area.​

Answer 1

Answer:

Step-by-step explanation:

Here, AB = 5 m, BC = 12 m, CD = 14 m and DA = 15 m

 Join the diagonal AC.

Now, area of △ABC = 1/2 x AB x BC  = 1/2 x 5 x 12 = 30  

Area of △ABC is 30 m2  In △ABC, (right triangle).

 From Pythagoras theorem,  AC2 = AB2 + BC2  AC2 = 52 + 122  AC2 = 25 + 144 = 169  or AC = 13  

Now in △ADC,  All sides are know, Apply Heron’s Formula: Where a, b and c are sides of a triangle.

Perimeter of △ADC = 2s = AD + DC + AC  2s = 15 m + 14 m + 13 m s = 21 m

Area of △ADC = 84 m^2  

Area of quadrilateral ABCD = Area of △ABC + Area of △ADC  = (30 + 84) m2  = 114 m^2

Hope it helps!

Answer 2

Answer:

Area = 114m²

Step-by-step explanation:

Here, AB=5m, BC=12m, CD=14m and DA=15m

Join the diagonal AC

Now, area of ΔABC=1/2×AB×BC

→1/2×5×12

→30

Area of ΔABC is 30m²

In ΔABC, (right triangle)

From Pythagoras theorem,

⇒AC²=AB²+BC²

⇒AC²=5²+12²

⇒AC²=25+144

⇒AC²=169

⇒AC=√169

⇒AC=13

Now, in ΔABC,

All sides are know, Apply Heron's formula

Area of triangle:

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

Where a, b and c are sides of a triangle.

Perimeter of ΔADC=2s=AD+DC+AC

⇒2s=15m+14m+13m

⇒2s=42m

⇒s=42/2

⇒s=21m

Area of ΔADC

$\sqrt{21×(21-13)×(21-14)×(21-15)}$

$\sqrt{21×8×7×6}$

⇒84

Area of ΔADC=84m²

Area of quadrilateral ABCD=Area of ΔABC+Area of ΔADC

⇒(30+84)m²

⇒114m²

Therefore, Area = 114m²