Measuring Space: Perimeter and Area | Exercise 6.2

Question 5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2128\text{ cm}^2, find the length of the shorter diagonal.

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Solution
Understand the Question
  • The area of a rhombus is given by the formula Area=12×d1×d2\text{Area} = \dfrac{1}{2} \times d_1 \times d_2, where d1d_1 and d2d_2 are the lengths of the diagonals.
  • Given that one diagonal is twice the length of the other, let the shorter diagonal be x cmx\text{ cm} and the longer diagonal be 2x cm2x\text{ cm}.
  • Substitute the given area (128 cm2128\text{ cm}^2) into the formula to solve for xx.

Step 1 · Set Up the Equation

Let the shorter diagonal be d1=x cmd_1 = x\text{ cm} and the longer diagonal be d2=2x cmd_2 = 2x\text{ cm}.Diagram 1

Using the area formula for a rhombus Area=12×d1×d2\text{Area} = \dfrac{1}{2} \times d_1 \times d_2

Given Area=128 cm2\text{Area} = 128\text{ cm}^2

128=12×x×2x128=x2\begin{aligned} 128 &= \dfrac{1}{2} \times x \times 2x \\[0.6em] 128 &= x^2 \end{aligned}

Step 2 · Calculate the Shorter Diagonal

Solve for xx by taking the square root

x=128=64×2=82\begin{aligned} x &= \sqrt{128} \\ &= \sqrt{64 \times 2} \\ &= 8\sqrt{2} \end{aligned}
Answer

82 cm8\sqrt{2}\text{ cm}

Common Mistakes
  • Forgetting the 12\dfrac{1}{2} Factor: Using Area=d1×d2\text{Area} = d_1 \times d_2 instead of Area=12×d1×d2\text{Area} = \dfrac{1}{2} \times d_1 \times d_2.
  • Incomplete Radical Simplification: Leaving the answer as 128\sqrt{128} instead of extracting the perfect square factor 64×2=82\sqrt{64 \times 2} = 8\sqrt{2}.

More questions in Exercise 6.2

Q1

Find the area of triangle ADE in Fig. 6.31.

Q2

The parallel sides of a trapezium are 40 cm40\text{ cm} and 20 cm20\text{ cm}. If its non-parallel sides are both equal, each being 26 cm26\text{ cm}, find the area of the trapezium.

Q3

Find the area of a triangle, given that its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm} long, and its perimeter is 32 cm32\text{ cm}.

Q4

The sides of a triangular plot are in the ratio 3:5:73 : 5 : 7; its perimeter is 300 m300\text{ m}. Find its area.

Q5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2128\text{ cm}^2, find the length of the shorter diagonal.

Q6

ABCDABCD is a parallelogram. PP and QQ are any two points on side ABAB. What can you say about the ratio area(ΔPCD):area(ΔQCD)\text{area}(\Delta \text{PCD}) : \text{area}(\Delta \text{QCD})?

Q7

OO is any point on the diagonal PRPR of a parallelogram PQRSPQRS. Prove that the areas of triangles PSOPSO and PQOPQO are equal.

Q8

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 in the chapter on quadrilaterals.)

Q9

In ΔABC\Delta\text{ABC}, the midpoint of BC\text{BC} is D\text{D} (Fig. 6.32). Median AD\text{AD} is drawn. P\text{P} is any point on AD\text{AD}. Show that area(ΔABP)=area(ΔACP)\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP}).

Q10

Given a square ABCDABCD, let PP be a point within it. Join PAPA, PBPB, PCPC, PDPD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB\Delta PAB and ΔPCD\Delta PCD) and the green region (ΔPBC\Delta PBC and ΔPDA\Delta PDA)?

Q11

In ΔABC\Delta\text{ABC}, DD is the midpoint of AB\text{AB}. PP is any point on BC\text{BC}, and QQ is a point on AB\text{AB} such that CQPD\text{CQ} \parallel \text{PD}. PQ\text{PQ} is joined (Fig. 6.34). Prove that

Area(ΔBPQ)=12Area(ΔABC)\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})

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