ABC is an equilateral triangle. P, Q and R are the mid-points of sides AB, BC and CA, respectively. If the length of the side of the triangle ABC is 11 cm, then the area of is:

Correct — B. Joining the three midpoints cuts out the medial triangle, which is equilateral with half the parent's side.

Side of triangle PQR = 11/2 = 5.5 cm, since a midsegment is half the side it runs parallel to

Area of an equilateral triangle = (√3/4) × side²
= (√3/4) × (11/2)²
= (√3/4) × 121/4
= 121√3/16 cm² → option (b)

The short route agrees: the medial triangle holds one quarter of triangle ABC, and (√3/4 × 121) ÷ 4 = 121√3/16.

Why the others are wrong

• (a)
  21√3/16 has the right shape and the wrong square: 11² is 121, not 21. Every other part of this option matches the keyed value, which is exactly what makes a hurried squaring so expensive.
• (c)
  111√3/16 comes from writing 11 × 11 as 111 — repeating the digit instead of squaring it. The true numerator, 121, is 10 higher.
• (d)
  11√3/16 keeps the side where the formula wants side². Substituting 11 in place of 121 makes the area eleven times too small.

Joining the midpoints of a triangle's three sides produces the medial triangle. Each of its sides is a midsegment, parallel to a side of the parent and exactly half as long, so the two triangles are similar in the ratio 1 : 2.

Areas of similar figures scale as the square of the length ratio, so triangle PQR holds one quarter of triangle ABC. That much is true of any triangle.

Being equilateral only supplies the area formula (√3/4)a², which turns the quarter into a number.

Key facts

• The medial triangle of any triangle has one quarter of the parent's area.
• A midsegment is parallel to the third side and half its length.
• The area of an equilateral triangle of side a is (√3/4)a², so side 11 cm gives 121√3/4 cm².

Common traps

• Halving the area because the side was halved, instead of quartering it
• Working out the area of ABC and stopping there
• Writing 11² as 111

No directly related past PYQ was found.