Question

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​

Answer 1

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