find the perimeter and all the angles in a rhombus,if
√B=70°,PR=9 CM, OS = 8 CM.
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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°
yasummu:
I hope that this is right if yes please mark as the brainliest
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