One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2, find the length of the shorter diagonal.
Answer: Area = 12 × d × 2d = d2 = 128, so d = √128 = 8√2 ≈ 11.31 cm.
Step-by-step solution
Given: Diagonals d and 2d; Area = 128 cm2
To find: Shorter diagonal d
To find: Shorter diagonal d
Idea: The diagonals of a rhombus cut each other at right angles, so its area is half the product of the diagonals.
- Let the shorter diagonal be d; the longer is 2d. Area = 12 × d × 2d = d2.1 mark
- d2 = 128 ⇒ d = √128 = √(64 × 2) = 8√2 ≈ 11.31 cm.1 mark
The shorter diagonal is 8√2 cm ≈ 11.31 cm.
Check: Diagonals 8√2 and 16√2: 12 × 8√2 × 16√2 = 12 × 256 = 128 ✓.
Answer to write in the exam
Area = 12 × d1 × d2 = 12 × d × 2d = d2
d2 = 128
∴ d = 8√2 cm ≈ 11.31 cm
Common mistakes that cost marks
- Writing area = d × 2d (forgetting the 12), giving d = 8.
- Giving the longer diagonal (16√2) instead of the shorter one.
How this can come in the exam
MCQ (1 mark)
The diagonals of a rhombus are 10 cm and 24 cm. Its area is
- 120 cm2
- 240 cm2
- 60 cm2
- 34 cm2
Show answer
(A) 120 cm2
12 × 10 × 24 = 120 cm2.
Try one yourself
One diagonal of a rhombus is three times the other, and its area is 150 cm2. Find the diagonals.
Show answer
12 × d × 3d = 150 ⇒ d2 = 100 ⇒ 10 cm and 30 cm.
More questions like this
- ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?
- O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
- If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered later, when quadrilaterals are studied.)
- In ∆ABC, the midpoint of BC is D (see the figure). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).
- Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (see the figure). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?