NCERT Exemplar Solutions Class 10 Maths Triangles — Exercise 6.4

NCERT Exemplar Problems, Class 10 Mathematics — Triangles, Exercise 6.4. All 18 questions solved.

The long-answer exercise is half proof and half computation, and the proofs draw on a small stock of moves that are worth having in mind before starting.

• The Basic Proportionality Theorem and its converse are the engine of Questions 3, 4, 13, 15 and 16. Whenever a line is parallel to one side of a triangle, it cuts the other two in the same ratio — and conversely.
• The altitude to the hypotenuse creates two triangles similar to the original, giving $h^2 = pq$ and $\text{leg}^2 = (\text{adjacent segment}) \times (\text{hypotenuse})$. Questions 9 and 10 are these relations used twice each.
• Pythagoras in two right triangles sharing a leg is the whole of Questions 11 and 12: write the shared leg two ways and subtract.
• Similar figures on the sides of a right triangle (Questions 17, 18) all reduce to the same one-line argument, because the area of any such figure is a fixed multiple of the square of its side.

A note on the current syllabus. Pythagoras’ theorem and the area-ratio theorem have both been removed from the rationalised Class 10 chapter, which ends at the similarity criteria. Questions 5 to 12, 17 and 18 all use one or the other. Pythagoras is assumed knowledge from Class 9, so none of this is unreachable — but a board paper will not set it in this chapter.

Key insight. In the proportionality proofs, name the triangle you are working in before writing a single ratio. Almost every slip in this exercise is a ratio taken from one triangle and used in another.

Question 1

In Fig. 6.16, if $\angle A = \angle C$, $AB = 6$ cm, $BP = 15$ cm, $AP = 12$ cm and $CP = 4$ cm, then find the lengths of $PD$ and $CD$.

Solution. The two triangles $APB$ and $CPD$ face each other across $P$. In them:

• $\angle A = \angle C$ — given;
• $\angle APB = \angle CPD$ — vertically opposite.

So $\triangle APB \sim \triangle CPD$ by AA, with correspondence $A \leftrightarrow C$, $P \leftrightarrow P$, $B \leftrightarrow D$. Hence

$$\frac{AP}{CP} = \frac{PB}{PD} = \frac{AB}{CD}$$

The known pair fixes the ratio at $\dfrac{12}{4} = 3$, so

$$\frac{15}{PD} = 3 \implies PD = 5 \text{ cm}, \qquad \frac{6}{CD} = 3 \implies CD = 2 \text{ cm}$$

Question 2

It is given that $\triangle ABC \sim \triangle EDF$ such that $AB = 5$ cm, $AC = 7$ cm, $DF = 15$ cm and $DE = 12$ cm. Find the lengths of the remaining sides of the triangles.

Solution. Write the correspondence out: $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow F$, so

$$\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}$$

The pair $AB = 5$ and $ED = DE = 12$ fixes the ratio at $\dfrac{5}{12}$. Then

$$\frac{BC}{15} = \frac{5}{12} \implies BC = \frac{75}{12} = 6.25 \text{ cm}$$

$$\frac{7}{EF} = \frac{5}{12} \implies EF = \frac{84}{5} = 16.8 \text{ cm}$$

The trap here is assuming $AB$ pairs with $DE$ because both are written second-and-first; in fact $AB$ pairs with $ED$, which is the same segment — but $AC$ pairs with $EF$, not with $DF$. Writing the three vertex pairings first makes this automatic.

Question 3

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

Solution. This is the Basic Proportionality Theorem (Thales’ theorem). Let $\triangle ABC$ have a line $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. We must show $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Join $BE$ and $CD$, and drop $DM \perp AC$ and $EN \perp AB$.

Areas do the work. Taking $AD$ as base with height $EN$, and $DB$ as base with the same height:

$$\frac{\operatorname{ar}(ADE)}{\operatorname{ar}(BDE)} = \frac{\frac12 \times AD \times EN}{\frac12 \times DB \times EN} = \frac{AD}{DB} \tag{1}$$

Taking $AE$ and $EC$ as bases with the common height $DM$:

$$\frac{\operatorname{ar}(ADE)}{\operatorname{ar}(DEC)} = \frac{\frac12 \times AE \times DM}{\frac12 \times EC \times DM} = \frac{AE}{EC} \tag{2}$$

Now the key step. $\triangle BDE$ and $\triangle DEC$ stand on the same base $DE$ and lie between the same parallels $DE$ and $BC$, so they have equal areas:

$$\operatorname{ar}(BDE) = \operatorname{ar}(DEC) \tag{3}$$

Combining (1), (2) and (3), the two right-hand sides are equal:

$$\frac{AD}{DB} = \frac{AE}{EC}$$

Question 4

In Fig. 6.17, if $PQRS$ is a parallelogram and $AB \parallel PS$, then prove that $OC \parallel SR$.

In the figure, $\triangle ABC$ has $A$ at the top, $B$ at the bottom left and $C$ at the bottom right. The parallelogram $PQRS$ sits inside it with $Q$ on $AC$ and $R$ on $BC$; $PS$ and $QR$ are its two sides parallel to $AB$. The point $O$ is where $AP$ produced meets $BS$ produced, and the segments $OP$, $OS$ and $OC$ are drawn.

Solution. Since $PQRS$ is a parallelogram, $PS \parallel QR$; and $PS \parallel AB$ is given, so $QR \parallel AB$ as well. Both parallels give a pair of similar triangles.

In $\triangle OAB$, with $PS \parallel AB$, $P$ on $OA$ and $S$ on $OB$:

$$\triangle OPS \sim \triangle OAB \implies \frac{OP}{OA} = \frac{PS}{AB} \tag{1}$$

In $\triangle CAB$, with $QR \parallel AB$, $Q$ on $CA$ and $R$ on $CB$:

$$\triangle CQR \sim \triangle CAB \implies \frac{CQ}{CA} = \frac{QR}{AB} \tag{2}$$

Opposite sides of a parallelogram are equal, so $PS = QR$. Comparing (1) and (2), their right-hand sides are the same, hence

$$\frac{OP}{OA} = \frac{CQ}{CA}$$

Write $k$ for this common value. Then $\dfrac{AP}{AO} = 1 – k$ and $\dfrac{AQ}{AC} = 1 – k$, so

$$\frac{AP}{PO} = \frac{1-k}{k} = \frac{AQ}{QC}$$

Now look at $\triangle OAC$: $P$ lies on $AO$ and $Q$ on $AC$, and they divide those two sides in the same ratio. By the converse of the Basic Proportionality Theorem,

$$PQ \parallel OC$$

Finally $PQ \parallel SR$, being opposite sides of the parallelogram. Two lines parallel to the same line are parallel to each other, so

$$OC \parallel SR$$

Question 5

A $5$ m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point $4$ m high. If the foot of the ladder is moved $1.6$ m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Solution. The ladder keeps its length, so each position is a right triangle with hypotenuse $5$ m.

Before. Height $4$ m, so the foot is

$$\sqrt{5^2 – 4^2} = \sqrt{9} = 3 \text{ m from the wall}$$

After. The foot moves $1.6$ m closer, to $3 – 1.6 = 1.4$ m. The new height is

$$\sqrt{5^2 – 1.4^2} = \sqrt{25 – 1.96} = \sqrt{23.04} = 4.8 \text{ m}$$

The slide up is the difference of the two heights:

$$4.8 – 4 = 0.8 \text{ m}$$

Note that the top rises by only $0.8$ m while the foot moves $1.6$ m — the two do not move by the same amount, which is the point of the question.

Question 6

For going to a city $B$ from city $A$, there is a route via city $C$ such that $AC \perp CB$, $AC = 2x$ km and $CB = 2(x+7)$ km. It is proposed to construct a $26$ km highway which directly connects the two cities $A$ and $B$. Find how much distance will be saved in reaching city $B$ from city $A$ after the construction of the highway.

Solution. $AC \perp CB$ makes $\triangle ACB$ right-angled at $C$, with $AB = 26$ as the hypotenuse. By Pythagoras,

$$(2x)^2 + \left[2(x+7)\right]^2 = 26^2$$ $$4x^2 + 4\left(x^2 + 14x + 49\right) = 676$$ $$8x^2 + 56x + 196 = 676$$ $$8x^2 + 56x – 480 = 0 \implies x^2 + 7x – 60 = 0$$

Factorising, $(x + 12)(x – 5) = 0$, so $x = 5$ (a distance cannot be negative).

$$AC = 2(5) = 10 \text{ km}, \qquad CB = 2(12) = 24 \text{ km}$$

The old route is $10 + 24 = 34$ km and the new one is $26$ km, so

$$\text{distance saved} = 34 – 26 = 8 \text{ km}$$

Question 7

A flag pole $18$ m high casts a shadow $9.6$ m long. Find the distance of the top of the pole from the far end of the shadow.

Solution. The pole, its shadow and the line from the top of the pole to the end of the shadow form a right triangle, right-angled at the foot of the pole. The distance asked for is the hypotenuse:

$$d = \sqrt{18^2 + 9.6^2} = \sqrt{324 + 92.16} = \sqrt{416.16} = 20.4 \text{ m}$$

(The check: $20.4^2 = 416.16$ ✓. The triangle is $9.6$–$18$–$20.4$, which is the $8$–$15$–$17$ triangle scaled by $1.2$.)

Question 8

A street light bulb is fixed on a pole $6$ m above the level of the street. If a woman of height $1.5$ m casts a shadow of $3$ m, find how far she is away from the base of the pole.

Solution. Let the woman stand $d$ m from the base of the pole. The light at the top of the pole, the top of her head and the tip of her shadow are in line, giving two similar right triangles: the big one formed by the pole and the whole distance from its base to the shadow tip, and the small one formed by the woman and her shadow.

$$\frac{\text{woman’s height}}{\text{pole’s height}} = \frac{\text{her shadow}}{\text{shadow tip’s distance from the pole}}$$

$$\frac{1.5}{6} = \frac{3}{d + 3}$$

The shadow tip is $d + 3$ from the pole, because her shadow stretches away from the pole. Cross-multiplying,

$$1.5(d + 3) = 18 \implies d + 3 = 12 \implies d = 9 \text{ m}$$

Question 9

In Fig. 6.18, $ABC$ is a triangle right angled at $B$ and $BD \perp AC$. If $AD = 4$ cm and $CD = 5$ cm, find $BD$ and $AB$.

Solution. The altitude from the right angle to the hypotenuse gives two relations, and this question uses both.

$BD$ — the geometric-mean relation. $\triangle ADB \sim \triangle BDC$ (each has a right angle at $D$, and $\angle DAB = \angle DBC$ as complements of $\angle ABD$), so

$$\frac{AD}{BD} = \frac{BD}{DC} \implies BD^2 = AD \times DC = 4 \times 5 = 20$$

$$BD = \sqrt{20} = 2\sqrt5 \text{ cm}$$

$AB$ — the leg relation. $\triangle ADB \sim \triangle ABC$ (common $\angle A$, right angles at $D$ and $B$), so

$$\frac{AD}{AB} = \frac{AB}{AC} \implies AB^2 = AD \times AC = 4 \times (4+5) = 36$$

$$AB = 6 \text{ cm}$$

Question 10

In Fig. 6.19, $PQR$ is a right triangle right angled at $Q$ and $QS \perp PR$. If $PQ = 6$ cm and $PS = 4$ cm, find $QS$, $RS$ and $QR$.

Solution. Work in the order the relations allow.

$PR$ first, from the leg relation on $PQ$:

$$PQ^2 = PS \times PR \implies 36 = 4 \times PR \implies PR = 9 \text{ cm}$$

$RS$ is then the rest of the hypotenuse:

$$RS = PR – PS = 9 – 4 = 5 \text{ cm}$$

$QS$, from the geometric-mean relation:

$$QS^2 = PS \times SR = 4 \times 5 = 20 \implies QS = 2\sqrt5 \text{ cm}$$

$QR$, from the leg relation on the other side:

$$QR^2 = RS \times RP = 5 \times 9 = 45 \implies QR = 3\sqrt5 \text{ cm}$$

A final check with Pythagoras on the whole triangle: $PQ^2 + QR^2 = 36 + 45 = 81 = 9^2 = PR^2$ ✓.

Question 11

In $\triangle PQR$, $PD \perp QR$ such that $D$ lies on $QR$. If $PQ = a$, $PR = b$, $QD = c$ and $DR = d$, prove that $(a+b)(a-b) = (c+d)(c-d)$.

Solution. The altitude $PD$ is a leg of two right triangles, and writing it twice is the whole proof.

In right $\triangle PDQ$: $\ PD^2 = PQ^2 – QD^2 = a^2 – c^2$.

In right $\triangle PDR$: $\ PD^2 = PR^2 – DR^2 = b^2 – d^2$.

Both equal $PD^2$, so

$$a^2 – c^2 = b^2 – d^2$$

Rearranging to group the letters,

$$a^2 – b^2 = c^2 – d^2$$

and factorising each side as a difference of squares,

$$(a+b)(a-b) = (c+d)(c-d)$$

Question 12

In a quadrilateral $ABCD$, $\angle A + \angle D = 90^\circ$. Prove that $AC^2 + BD^2 = AD^2 + BC^2$.

[Hint: Produce $AB$ and $DC$ to meet at $E$.]

Solution. Follow the hint: produce $AB$ and $DC$ to meet at $E$.

In $\triangle ADE$, the angles at $A$ and $D$ sum to $90^\circ$ by hypothesis, so the third angle is

$$\angle AED = 180^\circ – 90^\circ = 90^\circ$$

Everything now happens in right angles at $E$. Note that $B$ lies on $AE$ and $C$ on $DE$, so four right triangles share the vertex $E$:

$$\triangle AEC: \quad AC^2 = AE^2 + EC^2$$ $$\triangle BED: \quad BD^2 = BE^2 + ED^2$$ $$\triangle AED: \quad AD^2 = AE^2 + ED^2$$ $$\triangle BEC: \quad BC^2 = BE^2 + EC^2$$

Add the first two, and add the last two:

$$AC^2 + BD^2 = AE^2 + EC^2 + BE^2 + ED^2$$ $$AD^2 + BC^2 = AE^2 + ED^2 + BE^2 + EC^2$$

The two right-hand sides contain exactly the same four terms, so

$$AC^2 + BD^2 = AD^2 + BC^2$$

Question 13

In Fig. 6.20, $l \parallel m$ and line segments $AB$, $CD$ and $EF$ are concurrent at point $P$. Prove that $\dfrac{AE}{BF} = \dfrac{AC}{BD} = \dfrac{CE}{FD}$.

In the figure $A$, $C$ and $E$ lie on the line $l$ and $B$, $D$ and $F$ on the parallel line $m$; the three segments $AB$, $CD$ and $EF$ all pass through the single point $P$ between the two lines.

Solution. Each pair of the three segments gives a pair of similar triangles, because the parallel lines turn every crossing into equal alternate angles.

$\triangle APC$ and $\triangle BPD$. $\angle APC = \angle BPD$ (vertically opposite) and $\angle PAC = \angle PBD$ (alternate angles, $l \parallel m$ with transversal $AB$). So by AA,

$$\triangle APC \sim \triangle BPD \implies \frac{AC}{BD} = \frac{PA}{PB} \tag{1}$$

$\triangle APE$ and $\triangle BPF$. The same two reasons apply, with $EF$ in place of $CD$:

$$\triangle APE \sim \triangle BPF \implies \frac{AE}{BF} = \frac{PA}{PB} \tag{2}$$

$\triangle CPE$ and $\triangle DPF$. Again vertically opposite angles at $P$, and $\angle PCE = \angle PDF$ (alternate angles with transversal $CD$):

$$\triangle CPE \sim \triangle DPF \implies \frac{CE}{DF} = \frac{PC}{PD} \tag{3}$$

Finally, (1) also gives $\dfrac{PC}{PD} = \dfrac{PA}{PB}$, so the ratio in (3) is the same as the other two. Combining,

$$\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD} \ \left(= \frac{PA}{PB}\right)$$

Question 14

In Fig. 6.21, $PA$, $QB$, $RC$ and $SD$ are all perpendiculars to a line $l$, $AB = 6$ cm, $BC = 9$ cm, $CD = 12$ cm and $SP = 36$ cm. Find $PQ$, $QR$ and $RS$.

Solution. All four perpendiculars are parallel to one another, and $P$, $Q$, $R$, $S$ are collinear. A set of parallel lines cutting two transversals divides them in the same ratio, so the points $Q$ and $R$ split $PS$ exactly as $B$ and $C$ split $AD$.

First the whole base:

$$AD = AB + BC + CD = 6 + 9 + 12 = 27 \text{ cm}$$

Then each piece of $SP = 36$ cm is the matching fraction of $36$:

$$PQ = 36 \times \frac{6}{27} = 8 \text{ cm}$$ $$QR = 36 \times \frac{9}{27} = 12 \text{ cm}$$ $$RS = 36 \times \frac{12}{27} = 16 \text{ cm}$$

Check: $8 + 12 + 16 = 36$ ✓.

Question 15

$O$ is the point of intersection of the diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$. Through $O$, a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = QO$.

Solution. Two applications of the Basic Proportionality Theorem, plus one fact about the diagonals.

The diagonals. $AB \parallel DC$ gives $\triangle AOB \sim \triangle COD$ (alternate angles both ways), so

$$\frac{AO}{OC} = \frac{BO}{OD} \tag{1}$$

In $\triangle ADC$, $PO \parallel DC$ (since $PO \parallel AB \parallel DC$) with $P$ on $AD$ and $O$ on $AC$:

$$\frac{PO}{DC} = \frac{AO}{AC} \tag{2}$$

In $\triangle BDC$, $OQ \parallel DC$ with $O$ on $BD$ and $Q$ on $BC$:

$$\frac{OQ}{DC} = \frac{BQ}{BC} = \frac{BO}{BD} \tag{3}$$

It remains to show the two fractions on the right are equal. From (1), $\dfrac{AO}{OC} = \dfrac{BO}{OD}$; adding $1$ to both sides,

$$\frac{AO + OC}{OC} = \frac{BO + OD}{OD} \implies \frac{AC}{OC} = \frac{BD}{OD}$$

Inverting and rearranging gives $\dfrac{AO}{AC} = \dfrac{BO}{BD}$. Substituting into (2) and (3),

$$\frac{PO}{DC} = \frac{OQ}{DC} \implies PO = QO$$

Question 16

In Fig. 6.22, line segment $DF$ intersects the side $AC$ of a triangle $ABC$ at the point $E$ such that $E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$. Prove that $\dfrac{BD}{CD} = \dfrac{BF}{CE}$.

[Hint: Take point $G$ on $AB$ such that $CG \parallel DF$.]

Solution. Follow the hint and take $G$ on $AB$ with $CG \parallel DF$.

Step 1 — the isosceles triangle. $\angle AEF = \angle AFE$ makes $\triangle AEF$ isosceles, so

$$AE = AF \tag{1}$$

Step 2 — $F$ is the midpoint of $AG$. In $\triangle ACG$, the segment $EF$ is parallel to $CG$ with $E$ on $AC$ and $F$ on $AG$. By the Basic Proportionality Theorem,

$$\frac{AE}{EC} = \frac{AF}{FG}$$

But $E$ is the mid-point of $CA$, so $AE = EC$ and the left side is $1$. Hence $AF = FG$, and with (1),

$$FG = AF = AE = EC \tag{2}$$

Step 3 — the ratio. In $\triangle BCG$, the segment $DF$ is parallel to $CG$ with $D$ on $BC$ and $F$ on $BG$. By the Basic Proportionality Theorem again,

$$\frac{BD}{DC} = \frac{BF}{FG}$$

and substituting $FG = CE$ from (2),

$$\frac{BD}{CD} = \frac{BF}{CE}$$

Question 17

Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

Solution. Let the right triangle have legs $a$ and $b$ and hypotenuse $c$, so that by Pythagoras

$$c^2 = a^2 + b^2 \tag{1}$$

A semicircle drawn on a side of length $s$ has that side as its diameter, so its radius is $\frac{s}{2}$ and its area is

$$\frac12 \pi \left(\frac{s}{2}\right)^2 = \frac{\pi}{8}\,s^2$$

The area is therefore a fixed multiple of the square of the side — which is exactly what Pythagoras relates. Summing the two smaller semicircles:

$$\frac{\pi}{8}a^2 + \frac{\pi}{8}b^2 = \frac{\pi}{8}\left(a^2 + b^2\right) \overset{(1)}{=} \frac{\pi}{8}c^2$$

and that is the area of the semicircle on the hypotenuse.

Question 18

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Solution. Identical in structure to Question 17, with a different constant. An equilateral triangle of side $s$ has area

$$\frac{\sqrt3}{4}\,s^2$$

again a fixed multiple of $s^2$. With legs $a$, $b$ and hypotenuse $c$ satisfying $c^2 = a^2 + b^2$,

$$\frac{\sqrt3}{4}a^2 + \frac{\sqrt3}{4}b^2 = \frac{\sqrt3}{4}\left(a^2 + b^2\right) = \frac{\sqrt3}{4}c^2$$

which is the area of the equilateral triangle on the hypotenuse.

The general result. Both questions are instances of one fact: if similar figures are drawn on the three sides of a right triangle, the one on the hypotenuse has area equal to the sum of the other two. Any family of similar figures has area $k\,s^2$ for some constant $k$ depending only on the shape, and the $k$ cancels. Semicircles take $k = \frac{\pi}{8}$, equilateral triangles $k = \frac{\sqrt3}{4}$, and squares $k = 1$ — the last being Pythagoras’ theorem itself.

Common mistakes

• Matching sides by position rather than by correspondence in Question 2. $\triangle ABC \sim \triangle EDF$ pairs $AC$ with $EF$, not with $DF$. Writing $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow F$ first removes the guesswork.
• Assuming the ladder’s top moves as far as its foot in Question 5. The foot moves $1.6$ m and the top rises $0.8$ m; the relation between them is not linear.
• Measuring the shadow tip from the woman in Question 8. The similar-triangle proportion needs the distance from the pole to the shadow tip, which is $d + 3$, not $3$.
• Using the wrong segment in the leg relation. $AB^2 = AD \times AC$ uses the whole hypotenuse, while $BD^2 = AD \times DC$ uses the two pieces. Mixing them is the commonest error in Questions 9 and 10.
• Forgetting to produce the lines in Question 12. Without the point $E$ there is no right angle to apply Pythagoras to, and the hint is doing essential work.
• Not drawing $CG$ in Question 16. The hint supplies the auxiliary line that turns one unusable configuration into two standard BPT applications.
• Treating the semicircle’s radius as the side in Question 17. The side is the diameter, so the radius is $\frac{s}{2}$ and the area carries a $\frac{\pi}{8}$, not a $\frac{\pi}{2}$.
• Proving Questions 17 and 18 separately from scratch. They are the same argument; noticing that any similar figures have area $k s^2$ answers both at once and is worth a sentence.

Practise next

• Exemplar Exercise 6.3 — shorter versions of the same proportionality and area work.
• Exemplar Exercise 6.2 for the criteria themselves, and Exercise 6.1 for the objective forms.
• NCERT textbook Exercise 6.2 — the Basic Proportionality Theorem, proved in Question 3 above and used throughout.
• NCERT textbook Exercise 6.3 for the similarity criteria on the book’s own problems, including shadow questions of the shape of Questions 7 and 8.