This MCQ module is based on: Areas of Quadrilaterals and Combined Figures
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Areas of Quadrilaterals and Combined Figures
🎓 Class 9
Mathematics
CBSE
Theory
Ch 6 — Measuring Space: Perimeter and Area
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Areas of Quadrilaterals and Combined Figures
Targeting Class 9 level in Mensuration, with Intermediate difficulty.
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6.9 From Triangle to Quadrilateral
Heron's formula gives the area of any triangle from its three sides. For a quadrilateral, the four side-lengths alone are NOT enough — many shapes exist with the same four sides! We need at least one extra piece of information, usually a diagonal or an angle. The standard trick: split the quadrilateral with a diagonal into two triangles, then apply Heron's formula to each.
Quadrilateral ABCD split by diagonal AC into triangles ABC and ACD.
Worked Examples
Example 10. Find the area of quadrilateral ABCD with AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m and AC = 15 m.
△ABC: sides 9, 12, 15. \(s = 18\). Area \(= \sqrt{18\cdot 9\cdot 6\cdot 3} = \sqrt{2916} = 54\) m². △ACD: sides 15, 5, 8. \(s = 14\). Area \(=\sqrt{14\cdot(-1)\cdot 9 \cdot 6}\) — negative ⇒ NO triangle (15 > 5+8). The data is inconsistent. This shows you must verify the triangle inequality first! A consistent example: AC = 11 m. Then △ACD: \(s = 12\), area \(= \sqrt{12\cdot 1\cdot 7\cdot 4} = \sqrt{336} \approx 18.33\). Total ≈ 54 + 18.33 ≈ 72.33 m².
Example 11. A field is in the shape of a parallelogram with sides 26 m and 28 m, and one of its diagonals is 30 m. Find its area.
A diagonal of a parallelogram divides it into two congruent triangles. So total area = 2 × area of one triangle. Triangle: sides 26, 28, 30. \(s = 42\). Area \(= \sqrt{42\cdot 16\cdot 14\cdot 12} = \sqrt{112{,}896} = 336\) m². Total = 672 m².
Example 12. The diagonals of a rhombus measure 16 cm and 12 cm. Find the area and the side length.
Area = ½ × d₁ × d₂ = ½ × 16 × 12 = 96 cm². Diagonals bisect at right angles, so half-diagonals 8, 6 and side = √(8² + 6²) = √100 = 10 cm.
Example 13. The parallel sides of a trapezium are 25 cm and 11 cm and the non-parallel sides are 15 cm and 13 cm. Find the area.
Drop perpendiculars from the shorter parallel side. The horizontal "overhangs" satisfy x + y = 25 − 11 = 14. With heights equal: \(15^2 - x^2 = 13^2 - y^2 \Rightarrow x^2 - y^2 = 56\). With \(x+y=14\): \(x-y=4\), so \(x=9, y=5\). Height \(h = \sqrt{15^2 - 81} = \sqrt{144} = 12\) cm. Area \(= \tfrac12 (25+11)(12) = 216\) cm².
6.10 Brahmagupta's Formula (Cyclic 4-gon)
Brahmagupta (598–668 CE) discovered a stunning generalisation of Heron's formula for cyclic quadrilaterals — quadrilaterals whose four vertices lie on a circle.
Brahmagupta's Formula
For a cyclic quadrilateral with sides \(a, b, c, d\) and semi-perimeter \(s = \tfrac{a+b+c+d}{2}\),
\[\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}.\]
Notice the elegant symmetry — Heron's formula is the special case when one side is zero (i.e. the "quadrilateral" collapses to a triangle).Cyclic quadrilateral inscribed in a circle.
Example 14. A cyclic quadrilateral has sides 5, 5, 5 and 5 cm. Find the area.
It is a square. \(s = 10\). Area \(= \sqrt{5\cdot 5\cdot 5\cdot 5} = 25\) cm². Same as side² ✓.
Example 15. A cyclic quadrilateral with sides 3, 5, 7, 9 cm. Find the area.
\(s = 12\). Area \(= \sqrt{9\cdot 7\cdot 5\cdot 3} = \sqrt{945} \approx 30.74\) cm².
Historical Note
Brahmagupta's Brāhmasphuṭa-siddhānta (628 CE) presents this formula and many algebraic gems — including rules for negative numbers, zero, and quadratic equations. He worked at the astronomical observatory in Bhinmal, in present-day Rajasthan.Activity: Two Triangles, One Quadrilateral
L3 ApplyMaterials: graph paper, ruler, scissors.
Predict: If two triangles share a common side, the total area equals the sum of their individual areas. Will the same four side-lengths always give the same area?
- Cut two paper triangles with sides (3, 4, 5) and (5, 5, 6).
- Glue them along the common side of length 5.
- Now flex the joint — does the outer perimeter change? Does the area change?
The four outer side lengths stay 3, 4, 5, 6 (in order) but the diagonal — and therefore the area — depends on the angle at which the two triangles are joined. So four sides alone do NOT determine the area; you need the diagonal too. Brahmagupta's formula applies only when ABCD is cyclic.
Competency-Based Questions
Scenario: A landowner has a four-sided plot with sides AB = 40 m, BC = 60 m, CD = 50 m, DA = 70 m. He measures the diagonal AC = 80 m. He plans to install fencing along the boundary and grass on the interior.
Q1. Find the area of triangle ABC by Heron's formula.
L3 Apply\(s = 90\). Area \(= \sqrt{90\cdot 50\cdot 30\cdot 10} = \sqrt{1{,}350{,}000} = 300\sqrt{15}\) m². None of the listed options match exactly — the correct value is approximately 1162.0 m². (Some options have arithmetic errors; the verified answer is \(300\sqrt{15}\) m².)
Q2. Compute the area of △ACD and the total area of the plot.
L4 Analyse△ACD: sides 80, 50, 70. \(s = 100\). Area \(= \sqrt{100\cdot 20\cdot 50\cdot 30} = \sqrt{3{,}000{,}000} = 1000\sqrt{3}\) ≈ 1732.05 m². Total ≈ 1162 + 1732 ≈ 2894 m².
Q3. Evaluate: a neighbour says, "I have a plot with the same four side-lengths in the same order; therefore my plot has the same area." Critique.
L5 EvaluateFalse. Four side lengths do NOT determine a quadrilateral. The shape can be deformed (think of a pin-jointed frame). Only when an additional measurement (a diagonal or that the quadrilateral is cyclic) is given does the area become uniquely determined.
Q4. Design: among ALL quadrilaterals with sides 40, 60, 50, 70 m (in order), which one has MAXIMUM area? State the answer and justify.
L6 CreateBy a classical theorem, for a fixed sequence of side lengths the area is maximised when the quadrilateral is cyclic. Apply Brahmagupta: \(s = 110\). Area \(= \sqrt{(110-40)(110-60)(110-50)(110-70)} = \sqrt{70\cdot 50\cdot 60\cdot 40} = \sqrt{8{,}400{,}000} \approx 2898 m²\). Slightly larger than our 2894 m² estimate ⇒ the landowner's plot is nearly optimal.
Assertion–Reason Questions
A: The area of a quadrilateral is uniquely determined by its four side lengths.
R: Two quadrilaterals with the same side lengths must be congruent.
R: Two quadrilaterals with the same side lengths must be congruent.
Both false. A quadrilateral can flex with the sides fixed — counter-example: a square deforming into a rhombus. Hence neither A nor R holds.
A: Brahmagupta's formula reduces to Heron's formula when one of the four sides shrinks to zero.
R: A quadrilateral with one zero side is essentially a triangle.
R: A quadrilateral with one zero side is essentially a triangle.
(a) Set d=0, then s=(a+b+c)/2 and Area = √((s−a)(s−b)(s−c)·s) — exactly Heron's formula.
A: Among all quadrilaterals with four given side lengths (in the same order), the cyclic one has the largest area.
R: The maximum-area constraint and the cyclic property are equivalent in this setting.
R: The maximum-area constraint and the cyclic property are equivalent in this setting.
(a) A is the classical isoperimetric-type result; R restates it.
Frequently Asked Questions
What is the area formula for a parallelogram in Class 9?
The area of a parallelogram is Area = base x height, where the height is the perpendicular distance between the chosen base and the opposite side. The slanted side length is not used in the area formula.
How do you find the area of a trapezium?
Use Area = (1/2) x (sum of parallel sides) x distance between them. For instance, a trapezium with parallel sides 8 cm and 12 cm separated by height 5 cm has area (1/2) x (8 + 12) x 5 = 50 cm^2.
What is the area of a rhombus when both diagonals are known?
If the diagonals of a rhombus have lengths d1 and d2, then Area = (1/2) x d1 x d2. The diagonals of a rhombus bisect each other at right angles, which is why this product/2 formula works.
How do you find the area of an irregular quadrilateral?
Draw a diagonal to split it into two triangles. Compute the area of each triangle using (1/2) x base x height when an altitude is given, or Heron's formula when only sides are known, and add the two areas.
How is the area of an L-shaped or combined figure calculated?
Decompose the figure into rectangles, triangles or trapeziums whose areas are easy to compute, then add or subtract the parts. Alternatively, surround the figure with a rectangle and subtract the missing pieces.
How do you find the area of a path of uniform width around a rectangle?
If the inner rectangle is L x B and the path has uniform width w, the outer rectangle is (L + 2w) x (B + 2w). Path area = outer area - inner area = (L + 2w)(B + 2w) - L x B. The same idea works for circular tracks using pi(R^2 - r^2).
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