Question

1. A park, in the shape of a quadrilateral ABCD, has Z C = 90°, AB = 9 m, BC = 120
CD = 5 m and AD = 8 m. How much area does it occupy?
by herons formulas ​

Answer 1

Given :

• ∠C = 90°
• AB = 9m
• BC = 12m
• CD = 5m
• AD = 8m

To Find :

• Total area that the park will occupy.

Solution :

In Δ BCD :

• Base = 5m.
• Height = 12m

By Pythagoras Theorem :

$\longmapsto\tt{{(BD)}^{2}={(BC)}^{2}+{(CD)}^{2}}$

$\longmapsto\tt{{(BD)}^{2}={(12)}^{2}+{(5)}^{2}}$

$\longmapsto\tt{{(BD)}^{2}=144+25}$

$\longmapsto\tt{{(BD)}^{2}=139}$

$\longmapsto\tt{BD=\sqrt{139}}$

$\longmapsto\tt\bold{BD=13cm}$

Now ,

• a = 5m
• b = 12m
• c = 13m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{s=\dfrac{5+12+13}{2}}$

$\longmapsto\tt{s=\cancel\dfrac{30}{2}}$

$\longmapsto\tt\bold{s=15m}$

$\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\longmapsto\tt{\sqrt{15(15-5)(15-12)(15-13)}}$

$\longmapsto\tt{\sqrt{15\times{10}\times{3}\times{2}}}$

$\longmapsto\tt{\sqrt{3\times{5}\times{2}\times{5}\times{3}\times{2}}}$

$\longmapsto\tt\bold{30{m}^{2}}$

In Δ ABD :

• a = 9m
• b = 8m
• c = 13m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{s=\dfrac{9+8+13}{2}}$

$\longmapsto\tt{s=\cancel\dfrac{30}{2}}$

$\longmapsto\tt\bold{s=15m}$

$\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\longmapsto\tt{\sqrt{15(15-8)(15-9)(15-13)}}$

$\longmapsto\tt{\sqrt{15\times{7}\times{6}\times{2}}}$

$\longmapsto\tt{\sqrt{3\times{5}\times{7}\times{2}\times{3}\times{2}}}$

$\longmapsto\tt{3\times{2}\sqrt{5\times{7}}}$

$\longmapsto\tt\bold{6\sqrt{35}{m}^{2}}$

Now ,

$\longmapsto\tt{Total\:Area=30+6\sqrt{35}}$

$\longmapsto\tt\bold{65.4{m}^{2}\:(Approx.)}$

Answer 2

Solution:

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.

In ΔBCD,

By applying Pythagoras Theorem

BD²=BC² +CD²

BD²= 12²+ 5²= 144+25

BD²= 169

BD = √169= 13m

∆BCD is a right angled triangle.

Area of ΔBCD = 1/2 ×base× height

=1/2× 5 × 12= 30 m²

For ∆ABD,

Let a= 9m, b= 8m, c=13m

Now,

Semi perimeter of ΔABD,(s) = (a+b+c) /2

s=(8 + 9 + 13)/2 m

= 30/2 m = 15 m

s = 15m

Using heron’s formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 9) (15 – 9) (15 – 13)

= √15 × 6 × 7× 2

=√5×3×3×2×7×2

=3×2√35

= 6√35= 6× 5.92

[ √6= 5.92..]

= 35.52m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD

= 30+ 35.5= 65.5 m²

Hence, area of the park is 65.5m²