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NCERT Solutions Ganita Prakash (Part 2) Chapter 7 Area of any Polygon — Figure it Out

Book page 160 Updated on2026-09-05

Q1.
Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.
Answer

The diagonal AC splits ABCD into ∆ABC and ∆ACD, and both stand on AC.

Area (∆ABC) = ½ × AC × BM = ½ × 22 × 3 = 33 cm²
Area (∆ACD) = ½ × AC × DN = ½ × 22 × 3 = 33 cm²

Area (ABCD) = 33 + 33 = 66 cm²

Or in one step:

Area (ABCD) = ½ × AC × (BM + DN) = ½ × 22 × 6 = 66 cm²
Why it happens: B and D lie on opposite sides of AC, so the two triangles do not overlap and their areas add. Both use the same base AC, so the two heights can be collected together — 22 × 6 ÷ 2 is quicker than doing two separate multiplications.
Q2.
Find the area of the shaded region given that ABCD is a rectangle.
Answer

ABCD is 18 cm by 10 cm. E lies on AB with AE = 10 cm and EB = 8 cm; F lies on AD with AF = 6 cm and FD = 4 cm. The segments FE and EC cut off two triangles at the corners A and B, and the shaded region is what remains — the quadrilateral DFEC.

Area of rectangle ABCD = 18 × 10 = 180 cm²

Corner at A: Area (∆AFE) = ½ × AE × AF = ½ × 10 × 6 = 30 cm²
Corner at B: Area (∆EBC) = ½ × EB × BC = ½ × 8 × 10 = 40 cm²

Shaded area = 180 − 30 − 40 = 110 cm²
Why it happens: each corner triangle is right-angled, so its two perpendicular sides are already a base–height pair. Subtracting is far quicker than trying to split the four-sided shaded region into triangles, because the two pieces being removed are the simple ones.
Check it yourself: split DFEC instead — join FC. Then ∆FDC = ½ × 18 × 4 = 36 cm² and ∆FEC has base FC and is harder to handle. Subtracting from the rectangle is the better route.
Q3.
What measurements would you need to find the area of a regular hexagon? [Math Talk]
Answer

Two are enough: the sidelength and the distance from the centre to a side. In fact the sidelength alone determines the hexagon completely.

Join the centre O to all six vertices. The six triangles are congruent, each with base s (a side) and height h (the perpendicular from O to that side).

Area = 6 × (½ × s × h) = 3sh

For a regular hexagon those six triangles are equilateral, so h is fixed once s is known:

h = √(s² − (s/2)²) = (√3/2) s   (Baudhāyana–Pythagoras)
Area = 3s × (√3/2)s = (3√3/2) s²
Why it happens: "regular" means all sides equal and all angles equal, so the shape is pinned down by a single number. Measuring one side is enough — everything else, including h, follows from it.
Q4.
What fraction of the total area of the rectangle is the area of the blue region? [Math Talk]
Answer

Exactly one half — and it does not matter where the meeting point sits.

The blue region is two triangles that meet at a point P inside the rectangle. One has the whole top side as its base; the other has the whole bottom side as its base.

h₁ h₂ H base = w P
Both blue triangles span the full width. Their heights add up to the height of the rectangle.
Let the rectangle be w wide and H high, and let P be at heights h₁ and h₂ from the two sides, so h₁ + h₂ = H

Blue area = ½ × w × h₁ + ½ × w × h₂
= ½ × w × (h₁ + h₂)
= ½ × w × H
= ½ × area of the rectangle
Why it happens: both triangles use the full width as their base, so their areas depend only on their heights — and those two heights must together make up the height of the rectangle, wherever P is placed. Slide P anywhere inside and one triangle grows by exactly as much as the other shrinks. The fraction stays ½.
Q5.
Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral. [Math Talk]
Answer

Take the given quadrilateral ABCD.

  1. Draw the diagonal AC and mark M, the midpoint of AC.
  2. Join BM and DM.
  3. ABMD is the required quadrilateral.
In ∆ABC, BM is a median (M is the midpoint of AC), so
Area (∆ABM) = ½ × Area (∆ABC)

In ∆ACD, DM is a median, so
Area (∆AMD) = ½ × Area (∆ACD)

Adding: Area (ABMD) = ½ [Area (∆ABC) + Area (∆ACD)] = ½ × Area (ABCD)
Why it happens: the diagonal splits the quadrilateral into two triangles, and one median halves each of them. Halving both halves of a whole halves the whole.
Another way: join the midpoints of the four sides in order. The quadrilateral you get always has exactly half the area of the original — each of the four corner triangles cut off is a quarter of the triangle it sits in, and the four together come to half.
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