Question

Each side of a rhombus is 13cm and one diagonal is 10cm. What is the length of its other diagonal and the area of the rhombus? ​

Answer 1

Step-by-step explanation:

Let ABCD be the Rhombus in AB = 13cm, BC=13cm, CD=13 cm, DA=13 cm, AC=10 cm.

As we know, diagonals bisect each other at right angle

so, OA= 1/2 × AC = 1/2 × 10 = 5

In right angle ∆AOB, by Pythagoras theorm.

ab {}^{2} = oa {}^{2} + ob {}^{2}ab

2

=oa

2

+ob

2

13 {}^{2} = 5 {}^{2} + ob {}^{2}13

2

=5

2

+ob

2

169 = 25 + ob {}^{2}169=25+ob

2

ob {}^{2} = 169 - 25ob

2

=169−25

ob ^{2} = 144ob

2

=144

ob = \sqrt{144}ob=

144

Ob = 12

so, BD = 2 × OB

= 2 × 12

=24

Area of Rhombus

\frac{1}{2} \times d1 \times d2

2

1

×d1×d2

\frac{1}{2} \times ac \times bd

2

1

×ac×bd

\frac{1}{2} \times 10 \times 24

2

1

×10×24

=120 cm2

Answer 2

Answer:

12cm2

Step-by-step explanation:

Given side of rhombus =13 cm

Length of diagonal AC=10 cm

∴ OC=5 cm

Since, the diagonals of a rhombus bisect each other at right angles.

So, △BOC is right-angled.

Then, by Pythagoras Theorem we have

BC2=OC2+OB2

⇒13(up) 2=5(up) 2+OB (up)2

⇒OB(up)2 =169−25=144

⇒OB= 144  =12cm

Hence,

diagonal BD=2×OB=2×12=24 cm

Area of rhombus = 1/2 × d1 ×d2=1/2×10×24= 12cm2 (final answer)

Hope helps you