NCERT Solutions for Class 9th Maths Chapter 6 Exercise Set 6.2 — Area of a Triangle and 6.9 Squaring a Rectangle
Book page 142–143 Updated on2026-09-08
Q1.
Find the area of triangle ADE in Fig. 6.31. (ABCD is a rectangle with AB = 10 cm and BC = 8 cm; A is the top-left corner, B the top-right, C the bottom-right and D the bottom-left; E is a point on side BC.)
Answer
Area = 40 cm² — and the position of E on BC makes no difference at all.
Take AD as the base of ΔADE. AD = BC = 8 cm (opposite sides of a rectangle)
The height is the perpendicular distance from E to the line AD. E lies on BC, and BC ∥ AD, so that distance is the width of the rectangle: height = AB = 10 cm
ar(ΔADE) = ½ × base × height = ½ × 8 × 10 = 40 cm²
AD is the base; every point of BC is 10 cm from the line AD, so sliding E up or down BC does not change the area.
Why it happens: the area of a triangle depends on its base and on the perpendicular distance of the apex from the line of that base — not on where the apex sits along a parallel line. Since BC ∥ AD, every choice of E gives the same height, so ΔADE always has area exactly half the rectangle: ½ × 80 = 40 cm². This "same base, same parallels" principle is the workhorse for Q6, Q7, Q9 and Q11 below.
Q2.
The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Answer
Area = 720 cm².
The trapezium is isosceles, so it is symmetric about the perpendicular bisector of the parallel sides. Drop a perpendicular from each end of the shorter side onto the longer one. This leaves a central rectangle 20 cm wide, and two congruent right triangles at the ends.
base of each end triangle = 40 − 20⁄2 = 10 cm hypotenuse of each end triangle = 26 cm
By the Baudhāyana–Pythagoras theorem: h² = 26² − 10² = 676 − 100 = 576 h = √576 = 24 cm
Area = ½ (sum of parallel sides) × height = ½ (40 + 20) × 24 = ½ × 60 × 24 = 720 cm²
The two 10–24–26 right triangles at the ends give the height at once.
Check it yourself: 10, 24, 26 is just the 5, 12, 13 triple doubled — worth recognising, because it saves the square root.
Q3.
Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Answer
Area = 8√30 ≈ 43.82 cm². No height is given, so this is Heron's formula.
third side = 32 − (8 + 11) = 13 cm s = ½ × 32 = 16 cm
s − a = 16 − 8 = 8 s − b = 16 − 11 = 5 s − c = 16 − 13 = 3
Why it happens: Heron's formula is the tool of choice exactly when you know the three sides but no height. Getting a height here would mean solving for it first — and the answer 8√30 is irrational, so no height would come out nicely either. Notice too that the perimeter alone tells you the third side; you never need to draw the triangle.
Tip: always simplify the surd by pulling out perfect squares: √1920 = √(64 × 30) = 8√30. An answer left as √1920 is correct but unfinished.
Q4.
The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.
Answer
Area = 1500√3 ≈ 2598.08 m².
Let the sides be 3k, 5k, 7k. 3k + 5k + 7k = 300 15k = 300 ⟹ k = 20 sides = 60 m, 100 m, 140 m
s = ½ × 300 = 150 m s − a = 150 − 60 = 90 s − b = 150 − 100 = 50 s − c = 150 − 140 = 10
Why it happens: a ratio does not fix the sides, only their shape; the perimeter fixes the scale. Once k is known the triangle is completely determined, because three sides determine a triangle rigidly. Check that they really do form a triangle: 60 + 100 = 160 > 140 ✓ — with 3 : 5 : 8 you would have got 60 + 100 = 160 = 160, a flat "triangle" of zero area, and Heron's formula would have returned 0.
Tip: factor before you take the square root. 150 × 90 × 50 × 10 = (15 × 9 × 5 × 1) × 10⁴ = 675 × 10⁴, and √675 = 15√3, so the area is 15√3 × 100 = 1500√3.
Q5.
One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Answer
Shorter diagonal = 8√2 ≈ 11.31 cm.
The diagonals of a rhombus bisect each other at right angles, so area = ½ d₁ d₂.
Let the shorter diagonal be d and the longer 2d. ½ × d × 2d = 128 d² = 128 d = √128 = √(64 × 2) = 8√2 ≈ 11.31 cm
(The longer diagonal is 16√2 ≈ 22.63 cm.)
Why it happens: the two diagonals cut a rhombus into four congruent right triangles with legs d/2 and d₂/2. Four of them give 4 × ½ × d⁄2 × d₂⁄2 = ½ d d₂. The result holds for any 4-gon whose diagonals are perpendicular — a kite, for example, which is Question 14 of the end-of-chapter set.
Check it yourself: ½ × 8√2 × 16√2 = ½ × 128 × 2 ÷ 2 … more carefully, 8√2 × 16√2 = 128 × 2 = 256, and half of 256 is 128 ✓.
Q6.
ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?
Answer
The ratio is 1 : 1 — the two triangles always have the same area, wherever P and Q sit on AB.
Both triangles have the same base CD. Both apexes P and Q lie on AB, and AB ∥ CD. ∴ both triangles have the same height h, the distance between AB and CD.
ar(ΔPCD) = ½ · CD · h ar(ΔQCD) = ½ · CD · h
ar(ΔPCD) : ar(ΔQCD) = 1 : 1
In fact each is half the parallelogram:
ar(parallelogram ABCD) = CD × h ar(ΔPCD) = ½ · CD · h = ½ ar(ABCD)
Why it happens: this is the "triangles on the same base and between the same parallels are equal in area" principle. Sliding the apex along a line parallel to the base is a shear, and a shear changes the shape of the triangle but not its height, so not its area. The triangles ΔPCD and ΔQCD generally are not congruent — one may be right-angled and the other obtuse — yet their areas are equal.
Q7.
O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Answer
Proof. Let the diagonals PR and QS of the parallelogram meet at M.
The diagonals of a parallelogram bisect each other, so M is the midpoint of QS, and M lies on PR.
Drop perpendiculars QX and SY from Q and S to the line PR. In ΔQXM and ΔSYM: QM = SM (M is the midpoint of QS) ∠QXM = ∠SYM = 90° ∠QMX = ∠SMY (vertically opposite) ∴ ΔQXM ≅ ΔSYM (AAS) ∴ QX = SY
Now ΔPQO and ΔPSO share the base PO, which lies along PR. Their heights are exactly QX and SY, which are equal.
ar(ΔPQO) = ½ · PO · QX = ½ · PO · SY = ar(ΔPSO) ∎
Q and S are the same perpendicular distance from the diagonal PR, so any triangle on PO has the same area whether its apex is Q or S.
Why it happens: the diagonal PR is a line of "balance" for the parallelogram: Q and S are reflections of each other in the midpoint M, so they are equally far from PR on opposite sides. Once that is established, the two triangles have the same base PO and the same height — and this is true for every position of O on PR, including O = P (both areas 0) and O = R (each area half the parallelogram).
Q8.
If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.
Answer
Let ABCD be the 4-gon and P, Q, R, S the midpoints of AB, BC, CD, DA. Draw the diagonal AC.
First, PQRS really is a parallelogram.
In ΔABC, P and Q are midpoints of AB and BC ⟹ PQ ∥ AC and PQ = ½AC (midpoint theorem) In ΔACD, S and R are midpoints of DA and CD ⟹ SR ∥ AC and SR = ½AC
∴ PQ ∥ SR and PQ = SR ⟹ PQRS is a parallelogram
Now the area. Four corner triangles are cut off. Each is similar to a big triangle with ratio ½, so each has one quarter of its area.
ΔBPQ: BP = ½BA, BQ = ½BC, same angle B ∴ ar(ΔBPQ) = ¼ ar(ΔABC) Similarly ar(ΔDSR) = ¼ ar(ΔACD)
The four corner triangles together make up exactly half the 4-gon, so the midpoint parallelogram is the other half.
Why it happens: the midpoint theorem does all the work twice — once with each diagonal. Halving both sides of an angle scales the enclosed triangle by ½ in every direction, and area scales by the square of the length factor, hence ¼. The result holds for any 4-gon at all, convex or not, and does not depend on its shape — which is why the answer is always exactly one half.
Q9.
In ∆ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).
Answer
Use the median theorem twice — once on ΔABC and once on ΔPBC — and subtract.
AD is a median of ΔABC (D is the midpoint of BC): ar(ΔABD) = ar(ΔACD) … (1)
P lies on AD, so PD is a median of ΔPBC (same midpoint D): ar(ΔPBD) = ar(ΔPCD) … (2)
Since P is on AD, ΔABD splits into ΔABP and ΔPBD; likewise on the other side. ∴ ar(ΔABP) = ar(ΔACP) ∎
Why it happens: both applications of the theorem rest on the same fact — triangles with equal bases BD = DC and a common apex have equal areas, because they share the height from that apex to the line BC. First the apex is A, then it is P. Subtracting the second pair of equal areas from the first pair leaves a third pair of equal areas. Nothing about where P sits on AD is used, so the result holds even for P = A (both areas 0) and P = D (each half the triangle).
Tip: the same subtraction trick — "equal minus equal is equal" — is the standard way to prove area results in this chapter. Look for it again in Question 11.
Q10.
Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?
Answer
The ratio is 1 : 1 — each pair is exactly half the square, wherever P is placed inside it.
Let the side of the square be a. Let h₁ be the distance from P to AB, and h₂ the distance from P to the opposite side CD. Since AB ∥ CD and they are a apart: h₁ + h₂ = a.
ar(ΔPAB) + ar(ΔPCD) = ½ a h₁ + ½ a h₂ = ½ a (h₁ + h₂) = ½ a · a = ½ a²
Exactly the same argument for the other pair, with the distances to BC and DA: ar(ΔPBC) + ar(ΔPDA) = ½ a²
ratio = ½a² : ½a² = 1 : 1
The two green triangles stand on the opposite sides AD and BC; their heights add up to the full side, so together they cover half the square. So do the two red ones.
Why it happens: the trick is to add the two triangles of a pair before computing anything. Individually each area depends on where P is; together their heights must add to the side of the square, and the dependence on P vanishes. Notice how little is used: only that AB and CD are equal, parallel, and a apart. The same result therefore holds in any parallelogram, and even in any rectangle.
Q11.
In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (∆BPQ) = ½ Area (∆ABC).
Answer
The plan: show ar(ΔBPQ) = ar(ΔBCD), and then note that CD is a median.
D is the midpoint of AB, so CD is a median of ΔABC. ∴ ar(ΔBCD) = ½ ar(ΔABC) … (1)
Now compare ΔBPQ with ΔBCD. On the line AB the order of points is B, D, Q, A, and P lies on BC. So ar(ΔBPQ) = ar(ΔBPD) + ar(ΔDPQ) ar(ΔBCD) = ar(ΔBPD) + ar(ΔPCD)
It is enough to show ar(ΔDPQ) = ar(ΔPCD).
ΔDPQ and ΔDPC stand on the same base DP. Their apexes Q and C lie on the line CQ, and CQ ∥ PD. ∴ Q and C are the same perpendicular distance from the line DP. ∴ ar(ΔDPQ) = ar(ΔPCD) … (2)
From (2): ar(ΔBPQ) = ar(ΔBCD) With (1): ar(ΔBPQ) = ½ ar(ΔABC) ∎
Why it happens: the condition CQ ∥ PD is doing one job only — it puts C and Q on a line parallel to the base DP, so that the shear which slides C to Q leaves the area untouched. This is the same "same base, between the same parallels" tool as in Q6 and Q7. The surprising part is the conclusion: the point P can be anywhere on BC, and Q moves to match, yet ΔBPQ always captures exactly half the triangle.
Check it yourself: take P = C. Then PD = CD, and CQ ∥ CD forces Q = D, so ΔBPQ becomes ΔBCD — half the triangle, as claimed.