Question

HEYA GUYS ...
Question is in the attachment. .
answer both 5 and 6

Answer 1

5] Let the diagonals AC and BD of rhombus ABCD intersect at Point M

Diagonals of rhombus bisects each other

AM = $\dfrac{1}{2}$ × AC

$\dfrac{1}{2}$ × 16

= 8 cm

BM = $\dfrac{1}{2}$ × BD

$\dfrac{1}{2}$ × 12

= 6cm

i) In right angled ΔAMB,

(AB)² = (AM)² + (BM)² [Pythagoras theorem]

= (8)² + (6)²

= 64 + 36

(AB)² = 100

AB = 10 [By taking the square root]

Sides of rhombus are congruent.

$therefore{}$ AB = BC = CD = AD = 10 cm

ii) Perimeter = 4 × side

= 4 × 10

= 40 cm

_______________________


6] Let Square ABCD is a square

AB = BC = CD = AD = 8cm [Sides of square]

In right angle ΔABC,

(AC)² = (AB)² + (BC)²

= (8)² + (8)²

= 64 + 64

= 128

(AC)² = 64 × 2

AC = 8$\sqrt{2}$ [Taking square root]

$\therefore{}$ Length of diagonal = 8$\sqrt{2}$

Answer 2

5] Let the diagonals AC and BD of rhombus ABCD intersect at Point M

Diagonals of rhombus bisects each other

AM = × AC

× 16

= 8 cm

BM = × BD

× 12

= 6cm

i) In right angled ΔAMB,

(AB)² = (AM)² + (BM)² [Pythagoras theorem]

= (8)² + (6)²

= 64 + 36

(AB)² = 100

AB = 10 [By taking the square root]

Sides of rhombus are congruent.

AB = BC = CD = AD = 10 cm

ii) Perimeter = 4 × side

= 4 × 10

= 40 cm

_______________________

6] Let Square ABCD is a square

AB = BC = CD = AD = 8cm [Sides of square]

In right angle ΔABC,

(AC)² = (AB)² + (BC)²

= (8)² + (8)²

= 64 + 64

= 128

(AC)² = 64 × 2

AC = 8 [Taking square root]

Length of diagonal = 8.

Help u