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Question

The side of an equilateral triangle ABC is 28 cm. Find the side of another equilateral PQR whose area is 16 times the area of triangle ABC.

The correct answer is

112 cm

Understanding the Problem: Equilateral Triangle Area and Side

The question asks us to find the side length of an equilateral triangle PQR, given that its area is 16 times the area of another equilateral triangle ABC, which has a side length of 28 cm.

To solve this, we need to use the formula for the area of an equilateral triangle and the given relationship between the areas of the two triangles.

Key Concept: Area of an Equilateral Triangle

The area of an equilateral triangle with side length 's' is given by the formula:

\(\text{Area} = \frac{\sqrt{3}}{4} \times s^2\)

Step-by-Step Solution

Let \(s_{ABC}\) be the side length of triangle ABC, and \(s_{PQR}\) be the side length of triangle PQR.

We are given:

  • \(s_{ABC} = 28\) cm
  • Area of triangle PQR = 16 \(\times\) Area of triangle ABC

Using the area formula, we can write the areas of the two triangles:

  • Area of triangle ABC = \(\frac{\sqrt{3}}{4} \times (s_{ABC})^2 = \frac{\sqrt{3}}{4} \times (28)^2\)
  • Area of triangle PQR = \(\frac{\sqrt{3}}{4} \times (s_{PQR})^2\)

Now, substitute these expressions into the given relationship between their areas:

\(\text{Area(PQR)} = 16 \times \text{Area(ABC)}\)

\(\frac{\sqrt{3}}{4} \times (s_{PQR})^2 = 16 \times \left(\frac{\sqrt{3}}{4} \times (28)^2\right)\)

We can cancel out the common term \(\frac{\sqrt{3}}{4}\) from both sides of the equation:

\((s_{PQR})^2 = 16 \times (28)^2\)

To find \(s_{PQR}\), take the square root of both sides:

\(s_{PQR} = \sqrt{16 \times (28)^2}\)

\(s_{PQR} = \sqrt{16} \times \sqrt{(28)^2}\)

\(s_{PQR} = 4 \times 28\)

Calculate the final value:

\(s_{PQR} = 112\)

So, the side of the equilateral triangle PQR is 112 cm.

Relationship Between Side and Area Ratio

Notice that if the area of one equilateral triangle is \(k\) times the area of another, then the ratio of their areas is \(k\). The ratio of the areas of similar figures (which all equilateral triangles are) is equal to the square of the ratio of their corresponding sides.

Let Area(PQR) = \(k \times\) Area(ABC).

Then \(\frac{\text{Area(PQR)}}{\text{Area(ABC)}} = k\).

Also, \(\frac{\text{Area(PQR)}}{\text{Area(ABC)}} = \left(\frac{s_{PQR}}{s_{ABC}}\right)^2\).

So, \(\left(\frac{s_{PQR}}{s_{ABC}}\right)^2 = k\).

Taking the square root, \(\frac{s_{PQR}}{s_{ABC}} = \sqrt{k}\).

Therefore, \(s_{PQR} = \sqrt{k} \times s_{ABC}\).

In this problem, \(k = 16\).

\(s_{PQR} = \sqrt{16} \times s_{ABC} = 4 \times 28 = 112\) cm.

This confirms our previous calculation.

Final Answer

The side of the equilateral triangle PQR is 112 cm.

Revision Table: Equilateral Triangle Formulas

Property Formula (side = s)
Area \(\frac{\sqrt{3}}{4} s^2\)
Perimeter \(3s\)
Height \(\frac{\sqrt{3}}{2} s\)

Additional Information: Properties of Equilateral Triangles

An equilateral triangle is a special type of triangle with several unique properties:

  • All three sides are equal in length.
  • All three interior angles are equal, each measuring 60 degrees.
  • It is a regular polygon with 3 sides.
  • The altitude, median, angle bisector, and perpendicular bisector from any vertex are all the same line segment.
  • The centroid, orthocenter, incenter, and circumcenter all coincide at the same point.
  • It has three lines of symmetry.

Understanding these properties is helpful when solving geometry problems involving equilateral triangles.

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Important Questions from Plane Figures

  1. If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:

  2. The width of the path around a square field is 4.5 m and its area is 105.75 m 2. Find the cost of fencing the field at the rate of Rs. 100 per meter.

  3. What is the area of the square (in cm 2) whose vertices lie on a circle of radius 5 cm?

  4. The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?

  5. The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:

    Take \(\left(\pi=\frac{22}{7}\right)\)

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