a plot is in the shape of a quadrilateral abcd in which angle c is 90 ,ab is 7 m , bc is 4 m, cd is 3m and ad is 6m. how much area does the plot occupy
Answers
Answer:
32.4m^2
Step-by-step explanation:
In quadrilateral ABCD, join BD,
Now we have two triangles, ∆ABD and ∆BCD
In ∆BCD, right angle at C, so it is a right angle triangle
by Pythagoras Theorem.
(BD)^2= (BC)^2 + (DC)^2
(BD)^2= (4)^2 + (3)^2
(BD)^2= 16+9
(BD)^2=25
BD=√25
BD= 5m
Area of ∆BCD=1/2*b*h
=1/2*3*4
=6m^2
In ∆ABD,
Semi perimeter of ∆ABD (s)= (a+b+c)/2
s= 7+5+6/2
s=20/2
s=10
using herons formula
perimeter of∆ABD=√s(s-a) (s-b) (s-c)
=√10(10-7) (10-5) (10-6)
=√10*3*5*6
=√600
=24.49(approx)
area of quadrilateral ABCD =area of ∆ABD+area of∆BCD
=6+24.49
=30.49m^2
hence, the are occupied by the plot is 30.49m^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