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Given ∠P = 70°

Diagonals PR = 9 cm and QS = 8cm

We know that the diagonals are perpendicular to each other.

So we have right angle triangles

Let the diagonals meet each other at O.

Then PO = 1/2 PR = 9/2 = 4.5 cm

And QO = 1/2 QS = 8/2 = 4 cm

We have ΔPOQ, in ΔPOQ

PQ is the hypotenuse

According to the pythagoras theorem

PQ² = PO² + QO²

⇒PQ² = 4.5² + 4²

⇒PQ² = 20.25 + 16

⇒PQ² = 36.25

⇒PQ = √36.25

∴ PQ = 6.02 cm

Side of the rhombus = 6.02 cm

Perimeter = 4(6.02) = 24.08 cm

Opposite angles are equal in a rhombus

∴∠P = ∠R

∴∠R = 70°

And ∠Q = ∠S ----------(1)

In a rhombus adjacent angles are supplementary

∴∠P + ∠Q = 180°

70° + ∠Q = 180°

∠Q = 180° - 70°

∴∠Q = 110°

From (1)

∠Q =∠S =110°

∴The angles are 70°, 110°, 70°, 110°

Diagonals PR = 9 cm and QS = 8cm

We know that the diagonals are perpendicular to each other.

So we have right angle triangles

Let the diagonals meet each other at O.

Then PO = 1/2 PR = 9/2 = 4.5 cm

And QO = 1/2 QS = 8/2 = 4 cm

We have ΔPOQ, in ΔPOQ

PQ is the hypotenuse

According to the pythagoras theorem

PQ² = PO² + QO²

⇒PQ² = 4.5² + 4²

⇒PQ² = 20.25 + 16

⇒PQ² = 36.25

⇒PQ = √36.25

∴ PQ = 6.02 cm

Side of the rhombus = 6.02 cm

Perimeter = 4(6.02) = 24.08 cm

Opposite angles are equal in a rhombus

∴∠P = ∠R

∴∠R = 70°

And ∠Q = ∠S ----------(1)

In a rhombus adjacent angles are supplementary

∴∠P + ∠Q = 180°

70° + ∠Q = 180°

∠Q = 180° - 70°

∴∠Q = 110°

From (1)

∠Q =∠S =110°

∴The angles are 70°, 110°, 70°, 110°