NCERT Solutions Ganita Prakash (Part 2) Chapter 7 – 159Triangles — Figure it Out

Book page 157 Updated on2026-09-05

Q1.

Find the areas of the following triangles: [Figure (i): ∆ABC with AE ⊥ BC, AE = 3 cm, BC = 4 cm. Figure (ii): ∆DEF with DN ⊥ EF, DN = 3.2 cm, EF = 5 cm. Figure (iii): ∆NAT, right-angled at A, with AN = 4 cm, AT = 3 cm.]

Answer

In each case pick the matching base–height pair and halve the product.

Triangle	Base	Height	Area
(i) ∆ABC	BC = 4 cm	AE = 3 cm	½ × 4 × 3 = 6 cm²
(ii) ∆DEF	EF = 5 cm	DN = 3.2 cm	½ × 5 × 3.2 = 8 cm²
(iii) ∆NAT	AT = 3 cm	AN = 4 cm	½ × 3 × 4 = 6 cm²

(i) 6 cm² (ii) 8 cm² (iii) 6 cm²

Why it happens: in (i) the foot E lies inside BC, in (ii) the height DN falls on the side EF, and in (iii) the two perpendicular sides are themselves a base–height pair — no extra line is needed. The formula does not care which case it is; it only needs a side and the perpendicular distance from the opposite vertex to that side's line.

Tip: in (i) the whole base BC is 4 cm. Where the foot E sits along it makes no difference to the area.

Q2.

Find the length of the altitude BY.

Answer

In the figure AX ⊥ BC (with X outside the segment), AX = 4 units, BC = 6 units, AC = 8 units, and BY ⊥ AC.

Using base BC: Area (∆ABC) = ½ × BC × AX = ½ × 6 × 4 = 12 sq. units

Using base AC: Area (∆ABC) = ½ × AC × BY = ½ × 8 × BY = 4 BY

So 4 BY = 12
BY = 3 units

Why it happens: the triangle is obtuse at B, so the foot X of the altitude from A lands outside BC. That does not matter — as shown on page 154, ½ × base × height is still the area. Once the area is known, the second base–height pair gives the second altitude immediately.

Q3.

Find the area of ∆SUB, given that it is isosceles, SE is perpendicular to UB, and the area of ∆SEB is 24 sq. units.

Answer

48 sq. units.

∆SUB is isosceles with SU = SB, and SE is the perpendicular from the apex S to the base UB. In an isosceles triangle that perpendicular also bisects the base, so UE = EB.

∆SEU and ∆SEB have equal bases (UE = EB) and the same height SE
So Area (∆SEU) = Area (∆SEB) = 24 sq. units

Area (∆SUB) = 24 + 24 = 48 sq. units

Why it happens: ∆SEU ≅ ∆SEB by RHS — right angle at E, hypotenuses SU = SB, and SE common — so UE = EB and the two halves are congruent. Congruent pieces have equal areas, so the altitude from the apex of an isosceles triangle splits it into two equal halves.

Q4.

[Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

Answer

Take the rectangle ABCD with DC as base and height h.

Extend DC beyond C to a point E so that CE = DC. Now DE = 2 × DC.
Mark any point P on the line AB (the side opposite DC) — the vertex A itself will do.
Join PD and PE. Then ∆PDE is the required triangle.

base DE = 2 × DC, height of ∆PDE = distance from AB to DC = h
Area (∆PDE) = ½ × (2 × DC) × h = DC × h
= Area of rectangle ABCD

Why it happens: the triangle keeps the rectangle's height but is given twice its base, and the ½ in the triangle formula cancels the doubling exactly. P may be anywhere on the line AB because the apex may slide along a line parallel to the base without changing the area.

Q5.

[Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

Answer

Take ∆ABC with base BC and height h, and cut it at half the height.

Mark M, the midpoint of AB, and N, the midpoint of AC. Join MN — this line is parallel to BC and is at height h/2.
Cut along MN. The top piece is the small triangle ∆AMN.
Rotate ∆AMN through half a turn about M. The vertex A lands on B, and the piece fills the gap on the left. Do the same on the right with the other half — or simply cut ∆AMN along the altitude and swing the two halves out to the sides.

The result is a rectangle of base BC and height h/2
Area = BC × h/2 = ½ × BC × h = Area (∆ABC) ✓

Why it happens: M is the midpoint of AB, so turning ∆AMN about M carries A onto B and MN onto a segment of the same line — nothing is stretched, so no area is created or lost. This is a dissection, and dissection is exactly the operation that preserves area.

Try This: cut a paper triangle along the line joining the midpoints of two sides and fold the top down. It lands flat on the base and the figure becomes a rectangle of half the height.

Q6.

ABCD, BCEF, and BFGH are identical squares. (i) If the area of the red region is 49 sq. units, then what is the area of the blue region? (ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square? [Math Talk]

Answer

Let each square have side s. Place D at the corner, so that D, C, B, H run as in the figure: H is directly above B, and B is directly above C.

The line DH cuts off the blue triangle; C, B and H lie on one vertical line, which makes the red region a single triangle.

Red region. C, B and H are collinear, so the red region is the triangle DCH with base CH = CB + BH = 2s and height DC = s.

Area (red) = ½ × 2s × s = s² = the area of one square

Blue region. DH rises 2s across a horizontal run of s, so at half the run it has risen half the way. It therefore crosses AB at its midpoint P, with AP = s/2.

Area (blue) = Area (∆ADP) = ½ × AP × AD = ½ × (s/2) × s = s²/4

(i) Red = 49, so s² = 49.

Area (blue) = 49 ÷ 4 = 12.25 sq. units

(ii) Blue + red = 180.

s²/4 + s² = 180
(5/4) s² = 180
s² = 180 × 4/5 = 144 sq. units — the area of each square

Why it happens: the red triangle's area equals exactly one square, however big the squares are, because its base is two sides long and its height is one side long — and the ½ halves the product back. Once both regions are expressed as multiples of s², a single given number fixes s².

Q7.

If M and N are the midpoints of XY and XZ, what fraction of the area of ∆XYZ is the area of ∆XMN? [Hint: Join NY] [Try This]

Answer

One quarter.

Join NY, as the hint suggests, and use the midpoint fact twice.

In ∆XYZ, N is the midpoint of XZ, so YN is a median
Area (∆XNY) = ½ × Area (∆XYZ)

In ∆XNY, M is the midpoint of XY, so NM is a median
Area (∆XMN) = ½ × Area (∆XNY) = ½ × ½ × Area (∆XYZ)

Area (∆XMN) = ¼ × Area (∆XYZ)

Why it happens: a median cuts a triangle into two triangles with equal bases and the same height, so it halves the area every time. Halving twice gives a quarter. Notice the answer does not depend on the shape of ∆XYZ at all.

Did you know? MN is parallel to YZ and half as long. Since ∆XMN has half the base and half the height of ∆XYZ, its area is ½ × ½ = ¼ of it — the scaling rule of Q5 on page 152, seen again.

Q8.

Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path. [Math Talk]

Answer

Reflect the water tank in the river, join the house to that image by a straight line, and fetch the water where that line meets the river.

Let H be the house, T the water tank, and let the river be the line r. (Both H and T are on the same side of r.)
Draw T′, the mirror image of T in r.
Join HT′ by a straight line. Let it cut r at P.
The shortest journey is H → P → T.

For any point Q on the river, QT = QT′ (mirror)
So HQ + QT = HQ + QT′, a path from H to T′ through Q
The shortest path from H to T′ is the straight segment HT′
That segment meets the river at P, so HP + PT is the least possible

Why it happens: this is the same mirror argument used on page 157 for the triangle of least perimeter. The reflection does not change any distance on the far side of the river, but it straightens a bent path into a single line — and between two points the straight line is the shortest.

Did you know? Light does exactly this. A ray reflecting off a mirror takes the shortest path, which is why the angle of incidence equals the angle of reflection at P.

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