The adjacent sides of a parallelogram 14 cm and 21 cm and the measure of the smaller
angle is 58°. Find the length of the shorter diagonal round to the nearest centimeter..
(a) 11
(b) 13
(c) 16
(d) 18
Answers
Step-by-step explanation:
Given, AB=12 cm, BC=10 cm
BD is the diagonal of parallelogram ABCD.
Also BD=16 cm
We know, △ABD≅△DCB [S−S−S Congruence]
Also, area of parallelogram ABCD = Area of △ABD + Area of △DCB
= 2 × Area of △ABD
= 2
s(s−12)(s−10)(s−16)
s=
2
12+10+16
= 19
Therefore, area of parallelogram ABCD= 2
19(19−12)(19−10)(19−16)
= 2
3591
= 2(59.92)
= 119.8 cm
2
Also, h is the distance of the two shorter sides.
We know, area of parallelogram = base×height
= BC×h
Therefore, h=
BC
Area of parallelogram ABCD
⇒h=
10
119.8
⇒h=11.98 cm
solution