$=\sqrt{4^2}\times\sqrt{2^2\cdot3}$
$=4\times2\sqrt3=8\sqrt3$
1. Find the area of parallelogram ABCD.
1 answer:
Answer:
1: C =69.3 m2
2: F = ll units
3. C: 3 in.
4. G =65 units
Step-by-step explanation:
1: C =69.3 m2
We see from the figure that height of the parallelogram ABCD is not known and the base is 10 m
Taking the right angled triangle the height is given by
Sine theta= height/hypotenuse
height = hypotenuse* sine theta
= 8* sine 60
= 8* 0.866= 6.928
Area of parallelogram= base * height
= 10 * 6.928
= 69.28
=69.3 m²
2: F = ll units
We see from the figure that height of the parallelogram DEFG is not known and the base is 13 cm
Area of parallelogram= base * height
143 = 13 * height
Height = 143/13= 11 units
3. C: 3 in.
The area of the triangle = 1/2 * base * height
Let the height be x then the base is 3x
54= 1/2*x*3x
54= 3x²/2
27= 3x²
x²=27/3
x²= 9
x= 3
The height is 3 inches
4. G =65 units
Area of the quadrilateral = 1/2 *PR*QO + 1/2 *PR*OS
= 1/2*13*4+ 1/2*13*6
=26+ 39
= 65 units
These measurements are given in the diagram .
You might be interested in
Step-by-step explanation:
coordinate point of B(2,5)
and coordinate point ofC(3,1)
Distance of BC