27. Construct a quadrilateral ABCD in whic
B = 120° and 2C = 90°.
quadrilateral ABCD in which AB = 5 cm, BC = 6 cm, CD = 8 cm,
Answers
Answer:
Step-by-step explanation:
1. Construct the line AB, BC and CD.
2. To get a 120 angle you need to first construct a 90 degrees which is very simple where after you should construct a 60 and bisect it to get the 30.
The 90 and 30 will give you the 120 degrees and there you construct the final line of the quadrilateral.
PS: I TRIED MY BEST, DON'T KNOW IF THIS WILL HELP.
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