area of parallelogram is 30 cm^2. if the length of two adjacent sides is 6 cm and 10 cm. find diagonal
Answers
Step-by-step explanation:
The area of a parallelogram is 30cm^2. The length of its adjacent sides are 6 cm and 10 cm. What is the length of its diagonal side?
The area of the parallelogram = 30 sq cm.
The area of two congruent triangles forming the parallelogram = 30/2 = 15 sq cm.
The adjacent sides of the parallelogram as also the triangles = 6 cm and 10 cm.
Let the included angle between the two sides = x.
Area of the triangle = (ab/2)sin x, or
15 = (6x10/2) sin x, or
sin x = (15x2)/(6x10) = 0.5, or x = 30 deg.
Let the triangle be ABC, where AB = 6 cm, BC = 10 cm and <ABC = 30 deg.
AC^2 = AB^2+BC^2–2xABxBC cos 30
= 36+100 - 2x6x10x0.866025403
= 136 - 103.9230485
= 32.07695155
AC = 32.07695155^0.5 = 5.663651785 cm
In triangle ABD, AB = 6 cm, AD = 10 cm and <BAD = 150 deg.
BD^2 = AB^2+AD^2–2xABxAD cos 150
= 36+100 +2x6x10x0.866025403
= 136 + 103.9230485
= 230.9230485
BD = 230.9230485^0.5 = 15.48944959 cm.
Check: AB = 6, BC = 10 , <ABC = 30 deg. Area of ABC = (6x10/2)xsin 30 = 15. Correct
The diagonals are 5.663651785 cm and 15.48944959 cm.
*Question:-
The area of a parallelogram is 30cm^2. The length of its adjacent sides are 6 cm and 10 cm. What is the length of its diagonal side?
*Answer:-
I'd extend the long side a little (the base in my mind's eye) and drop an altitude to it to make a small right-angled triangle. The height of this triangle is 3 (becase base x height gives area of a parallelogram)
The third side of the triangle (Pythagoras) is sqrt(29)
Now you have height (3) and base (10 + sqrt29) of the big triangle, so you can get its hypotenuse, the long diagonal of the parallelogram.
On reflection (should've sketched it first) using 6 as the base would give a little right-angled triangle with hypotenuse 10 and height 5 — a 30–60–90 triangle.
The large angle of the parallelogram is 120
You could use the Cosine Rule in the half-parallelogram.
The area of the parallelogram = 30 sq cm.
The area of two congruent triangles forming the parallelogram = 30/2 = 15 sq cm.
The adjacent sides of the parallelogram as also the triangles = 6 cm and 10 cm.
Let the included angle between the two sides = x.
Area of the triangle = (ab/2)sin x, or
15 = (6x10/2) sin x, or
sin x = (15x2)/(6x10) = 0.5, or x = 30 deg.
Let the triangle be ABC, where AB = 6 cm, BC = 10 cm and <ABC = 30 deg.
AC^2 = AB^2+BC^2–2xABxBC cos 30
= 36+100 - 2x6x10x0.866025403
= 136 - 103.9230485
= 32.07695155
AC = 32.07695155^0.5 = 5.663651785 cm
In triangle ABD, AB = 6 cm, AD = 10 cm and <BAD = 150 deg.
BD^2 = AB^2+AD^2–2xABxAD cos 150
= 36+100 +2x6x10x0.866025403
= 136 + 103.9230485
= 230.9230485
BD = 230.9230485^0.5 = 15.48944959 cm.
Check:
AB = 6, BC = 10 , <ABC = 30 deg. Area of ABC = (6x10/2)xsin 30 = 15. Correct
The diagonals are
5.663651785 cm
and
15.48944959 cm.