If each diagonal of a rhombus is 12cm find the area of a rhombus.

Question

If each diagonal of a rhombus is 12cm find the area of a rhombus.

Answer 1

We know that the diagonals of a rhombus bisect each other and are perpendicular. Go ahead and draw a picture to model this situation. Call the vertices of the rhombus A, B, C, and D, and call the point where the diagonals intersect P.

 

Since diagonals AC and BD are perpendicular, all four angles with P as their vertex are right angles. That means we have four right triangles, each with right angle P: ΔAPB, ΔBPC, ΔCPD, and ΔDPA.

 

Since diagonals AC and BD bisect each other, segments AP and PC are congruent and segments BP and PD are congruent. AC is 12 cm, so AP and PC are each 6 cm. BD is also 12 cm, so BP and PD are each 6 cm as well. That means those four segments are all congruent: AP ≅ BP ≅ CP ≅ DP = 6 cm.

 

By Side-Angle-Side congruency, all four triangles from above are congruent, since each has two legs of length 6 cm and an included 90° angle. So: ΔAPB ≅ ΔBPC ≅ ΔCPD ≅ ΔDPA. Since these four triangles are congruent and together make up the area of the rhombus, we need only calculate the area of one of them and then multiply that by four to find the area of the rhombus.

 

The area of a triangle is (1/2)bh, where b = base and h = height meet at a 90° angle. For ΔAPB, its base is AP = 6 and its height is BP = 6. So, it's area is (1/2)(6)(6) = (1/2)(36) = 18. That means the rhombus' area is (18)(4) = 72 cm².

 

 

 

By the way, if you have the formula for the area of a rhombus, this problem becomes much simpler:

 

Area = (length of first diagonal×length of second diagonal)/2

A = (12×12)/2

A = 144/2

A = 72