An architect must lay out a parallelogram-shaped hall ABCD with AB = 9 m, BC = 7 m and the longer diagonal AC = 14 m. Using the parallelogram law, the diagonals satisfy AC^2 + BD^2 = 2(AB^2 + BC^2). Find the shorter diagonal BD (in metres).

Question

Accepted Answer

8. Parallelogram law: AC^2 + BD^2 = 2(AB^2 + BC^2) = 2(81 + 49) = 260. So BD^2 = 260 - 14^2 = 260 - 196 = 64, giving BD = 8 m.

An architect must lay out a parallelogram-shaped hall ABCD with $A B = 9$ m, $BC = 7$ m and the longer diagonal $A C = 14$ m.
Using the parallelogram law, the diagonals satisfy $A C^{2} + B D^{2} = 2 (A B^{2} + B C^{2})$ .
Find the shorter diagonal $B D$ (in metres).

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More Class 8 Constructing Quadrilaterals practice

An architect must lay out a parallelogram-shaped hall ABCD with AB=9 m, BC=7 m and the longer diagonal AC=14 m.Using the parallelogram law, the diagonals satisfy AC2+BD2=2(AB2+BC2).Find the shorter diagonal BD (in metres).

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 8 Constructing Quadrilaterals practice

An architect must lay out a parallelogram-shaped hall ABCD with $A B = 9$ m, $BC = 7$ m and the longer diagonal $A C = 14$ m.
Using the parallelogram law, the diagonals satisfy $A C^{2} + B D^{2} = 2 (A B^{2} + B C^{2})$ .
Find the shorter diagonal $B D$ (in metres).