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Standard VIII

Mathematics

Construction of a Quadrilateral When Other Special Properties Are Known

In the given ...

Question

In the given figure, ∠MNP = 90°, seg NQ ⊥seg MP, MQ = 9, QP = 4, find NQ.

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Solution

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

$∴ {QN}^{2} = MQ \times QP = 9 \times 4 = 36 QN = \sqrt{36} = 6$

Hence, NQ = 6.

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In the given figure, ∠MNP = 90°, seg NQ ⊥seg MP, MQ = 9, QP = 4, find NQ.

We know that, In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided. ∴ QN2=MQ×QP =9×4 =36QN=36 =6 Hence, NQ = 6.

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

$∴ {QN}^{2} = MQ \times QP = 9 \times 4 = 36 QN = \sqrt{36} = 6$

Hence, NQ = 6.