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Standard VIII

Mathematics

Height & Base of Parallelogram

Let P Q R S b...

Question

Let PQRS be a kite such that PQ > PS. Prove that ∠PQR > ∠PSR. (Hint: Join QS).

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Solution

In any triangle, the angle opposite to a longer side is greater.

Here, PS = SR and PQ = QR

It is given that PQ > PS

∴In triangles SPQ and QRS:

Adding both inequalities, we get:

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Let PQRS be a kite such that PQ > PS. Prove that ∠PQR > ∠PSR. (Hint: Join QS).

In any triangle, the angle opposite to a longer side is greater. Here, PS = SR and PQ = QR It is given that PQ > PS ∴In triangles SPQ and QRS: Adding both inequalities, we get:

In any triangle, the angle opposite to a longer side is greater.

Here, PS = SR and PQ = QR

It is given that PQ > PS

∴In triangles SPQ and QRS:

Adding both inequalities, we get: