The angles of a quadrilateral are in ratio 1:2:3:4 — find each angle

hard CBSE JEE-MAIN 3 min read

Question

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find all four angles.

Solution — Step by Step

The sum of all interior angles of any quadrilateral is always 360°. This comes from the fact that any quadrilateral can be split into two triangles, each with an angle sum of 180°.

So: A+B+C+D=360°\angle A + \angle B + \angle C + \angle D = 360°

The angles are in ratio 1 : 2 : 3 : 4. Let the common factor be kk. Then:

A=k,B=2k,C=3k,D=4k\angle A = k, \quad \angle B = 2k, \quad \angle C = 3k, \quad \angle D = 4k
k+2k+3k+4k=360°k + 2k + 3k + 4k = 360° 10k=360°10k = 360° k=36°k = 36° A=1×36°=36°\angle A = 1 \times 36° = 36° B=2×36°=72°\angle B = 2 \times 36° = 72° C=3×36°=108°\angle C = 3 \times 36° = 108° D=4×36°=144°\angle D = 4 \times 36° = 144°

Verification: 36°+72°+108°+144°=360°36° + 72° + 108° + 144° = 360°

Why This Works

When angles are given in a ratio, the ratio tells us the relative sizes — not the actual sizes. The variable kk acts as the “scale factor.” Since we have an additional constraint (the sum must equal 360°), we can find kk uniquely.

This technique applies to any polygon: set angles as multiples of kk, use the angle sum property to find kk, then scale up.

Alternative Method

You can also think of it as: the four angles take up fractions 110\frac{1}{10}, 210\frac{2}{10}, 310\frac{3}{10}, 410\frac{4}{10} of the total 360°.

A=110×360°=36°,B=210×360°=72°\angle A = \frac{1}{10} \times 360° = 36°, \quad \angle B = \frac{2}{10} \times 360° = 72° C=310×360°=108°,D=410×360°=144°\angle C = \frac{3}{10} \times 360° = 108°, \quad \angle D = \frac{4}{10} \times 360° = 144°

To quickly find individual angles from a ratio when the total is known: each angle = (its ratio part ÷ sum of all ratio parts) × total. Here: each angle = (ratio part ÷ 10) × 360°.

Common Mistake

Students sometimes use the wrong angle sum property. This question is about a quadrilateral (4 sides) — angle sum = 360°. If the problem were about a triangle, the sum would be 180°. Always identify the polygon first. For a polygon with nn sides, the interior angle sum is (n2)×180°(n-2) \times 180°.

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