Triangles Chapter 6: NCERT Exemplar for Class 10

The NCERT Exemplar NCERT exemplar class 10 maths triangles, provided here, along with solutions, helps students to prepare pretty good for the board exams with respect to the CBSE syllabus (2024-2025). Students can easily download the PDF file and use it as an effective study tool and material to practice math problems and also have a quick revision of the topics.

Triangle and its properties are a quiet familiar topic for Class 10 students, as they have studied it in earlier classes. However, we all know and are aware of the fact that with time concepts get advanced and new topics are introduced as we progress through each class. As such, in Class 10, the topic of triangles is explained in greater detail, and students have to deal with new and more complex problems, as well some tough yet important topics such as pythagoras theorem and much more which will help the students to enlighten their concepts deeply.

Scroll down below to explore the new world of triangles!

NCERT exemplar class 10 Maths Triangles Question 1 to 10

Question 1.

In the figure, ∠BAC = 90° and AD ⊥ BC. So,

(a) BD·CD = BC²

(b) AB·AC = BC²

(c) BD·CD = AD²

(d) AB·AC = AD²

Answer 1:
(c) BD·CD = AD²

Explanation:
From triangles ∆ADB and ∆ADC,
Given ∠D = ∠D = 90° (∵ AD ⊥ BC)
And, ∠DBA = ∠DAC [each angle = 90° – ∠C]
By using the AAA similarity criteria,
∆ADB ∼ ∆ADC
Thus,
BD/AD = AD/CD
⇒ BD·CD = AD²

Question 2.

When the lengths of the diagonals of the rhombus are 16 cm and 12 cm, the length of the sides of the rhombus should be:

(a) 9 cm

(b) 10 cm

(c) 8 cm

(d) 20 cm

Answer 2:
(b) 10 cm

Explanation:
Given a rhombus with diagonals 16 cm and 12 cm.
In the rhombus, diagonals bisect each other at 90°.
So, using Pythagoras theorem in one of the right-angled triangles formed,
Let the sides be xxx,
x2=(8)2+(6)2x² = (8)² + (6)²x2=(8)2+(6)2
x2=64+36=100x² = 64 + 36 = 100×2=64+36=100
x=10x = 10x=10 cm.

Question 3.

When ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, which among the following is not true?

(a) BC·EF = AC·FD

(b) AB·EF = AC·DE

(c) BC·DE = AB·EF

(d) BC·DE = AB·FD

Answer 3:
(c) BC·DE = AB·EF

Explanation:
Given ΔABC ~ ΔEDF,
So by the properties of similarity:
ABED=BCDF=ACEF\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}EDAB​=DFBC​=EFAC​
(a), (b), and (d) can be derived from these ratios, but not (c).

Question 4.

When in two ΔQPR, given that AB/QR = BC/PR = CA/PQ, thus:

(a) ΔPQR ~ ΔCAB

(b) ΔQPR ~ ΔABC

(c) ΔCBA ~ ΔQPR

(d) ΔBCA ~ ΔQPR

Answer 4:
(a) ΔPQR ~ ΔCAB

Explanation:
From the given data,
ABQR=BCPR=CAPQ\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}QRAB​=PRBC​=PQCA​
The triangles are similar by SSS similarity criteria.
Thus, ΔPQR ~ ΔCAB.

Question 5.

In the given figure, two line segments AC and BD bisect each other at the point P so that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Thus, ∠PBA is equal to:

(a) 50°

(b) 30°

(c) 60°

(d) 100°

Answer 5:
(d) 100°

Explanation:
From ∆APB and ∆CPD,
∠APB=∠CPD=50°\angle APB = \angle CPD = 50°∠APB=∠CPD=50° (vertically opposite angles)
By SAS similarity criteria,
∠A=∠D=30°\angle A = \angle D = 30°∠A=∠D=30°
So,
∠PBA=180°−(∠APB+∠A)\angle PBA = 180° – (\angle APB + \angle A)∠PBA=180°−(∠APB+∠A)
∠PBA=180°−(50°+30°)=100°\angle PBA = 180° – (50° + 30°) = 100°∠PBA=180°−(50°+30°)=100°

Question 6.

It is given that ΔDEF ~ ΔRPQ. Should it be true to say that ∠D = ∠R as well as ∠F = ∠P? Why?

Answer 6:
False

Explanation:
For similar triangles, corresponding angles must be equal.
Thus, ∠D = ∠R, ∠E = ∠P, and ∠F = ∠Q.

Question 7.

Is the following sentence true? Why?

“Two quadrilaterals are equal if their corresponding angles are the same.”

Answer 7:
False

Explanation:
Quadrilaterals must have both corresponding angles equal and corresponding sides proportional to be similar.

Question 8.

In the given figure, when AB || DC and AC, PQ bisect each other at the point O. Prove that OA·CQ = OC·AP.

Answer 8:
Given: AC and PQ bisect each other at point O and AB || DC.
In ∆AOP and ∆COQ,
∠AOP=∠COQ\angle AOP = \angle COQ∠AOP=∠COQ (vertically opposite angles)
∠APO=∠CQO\angle APO = \angle CQO∠APO=∠CQO (alternate interior angles)
Thus,
∆AOP ∼ ∆COQ by AAA similarity criteria.
So,
OAOC=APCQ\frac{OA}{OC} = \frac{AP}{CQ}OCOA​=CQAP​
Hence,
OA×CQ=OC×APOA \times CQ = OC \times APOA×CQ=OC×AP.

Question 9.

In the given figure, we have PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that ΔQPR is an isosceles triangle.

Answer 9:
Given: PS/SQ = PT/TR
If a line divides two sides of a triangle in equal ratios, it is parallel to the third side.
So,
ST || QR
Also,
∠PST=∠QPR\angle PST = \angle QPR∠PST=∠QPR
∠PRQ=∠QPR\angle PRQ = \angle QPR∠PRQ=∠QPR
Hence, ΔQPR is isosceles.

Question 10.

In the figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

Answer 10:
In ∆ABC, DE || AC by BPT:
BDDA=BEEC\frac{BD}{DA} = \frac{BE}{EC}DABD​=ECBE​
Also,
DF || AE
Thus,
BDDA=BFFE\frac{BD}{DA} = \frac{BF}{FE}DABD​=FEBF​
From both,
BEEC=BFFE\frac{BE}{EC} = \frac{BF}{FE}ECBE​=FEBF​.

NCERT exemplar class 10 Maths Triangles Question 11 to 20

Question 11.

In the given figure, the altitudes of AD and CE of ΔABC cut each other at point P. Prove that:

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Answer 11:
(i) ∆AEP and ∆CDP:

$\angle AEP = \angle CDP = 90°$ $\angle APE = \angle CPD$

Thus, ΔAEP ~ ΔCDP by AA similarity.

(ii) ΔABD ~ ΔCBE:

$\angle ADB = \angle CEB = 90°$ $\angle ABD = \angle CBE$

∠ABD=∠CBE
Thus, ΔABD ~ ΔCBE by AA similarity.

(iii) ΔAEP ~ ΔADB:

$\angle AEP = \angle ADB = 90°$ $\angle PAE = \angle DAB$

Thus, ΔAEP ~ ΔADB by AA similarity.

(iv) ΔPDC ~ ΔBEC:

$\angle PDC = \angle BEC = 90°$ $\angle PCD = \angle BCE$

∠PCD=∠BCE
Thus, ΔPDC ~ ΔBEC by AA similarity.

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Question 12.

A vertical pole having a length of 6 m casts a 4 m long shadow on the ground, and at the same time, a tower casts a 28 m long shadow. Find out the height of the tower.

Answer 12:
Given: Pole height = 6 m, Shadow length = 4 m, Tower shadow length = 28 m.
Let the tower height be h m.
Using similar triangles:

$\frac{6}{h} = \frac{4}{28}$

​

$h = \frac{6 \times 28}{4} = 42$

h=46×28​=42 m.

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Question 13.

When ΔABC ~ ΔQRP, and area of (ΔABC) / area of (ΔPQR) = 9/4, AB = 18 cm and BC = 15 cm. Find PR.

Answer 13:
Given:

$\text{area of }(ΔABC) / \text{area of }(ΔPQR) = 9/4$

area of (ΔABC)/area of (ΔPQR)=9/4,
AB = 18 cm, BC = 15 cm.
Let PR =

$x$

x cm.
Using the ratio of areas,

$\frac{9}{4} = \frac{15²}{x²}$ $x² = \frac{4 \times 225}{9} = 100$

Thus,

$PR = 10$

PR=10 cm.

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Question 14.

When the areas of two similar triangles are equal, prove that they are also congruent.

Answer 14:
Given: ΔABC and ΔPQR are similar with equal areas.

$\text{Area of }ΔABC / \text{Area of }ΔPQR = 1$

Since ΔABC ~ ΔPQR,

$\text{Ratio of areas} = (\text{Ratio of corresponding sides})^2$

Since areas are equal,
All corresponding sides are equal.
Thus, by SSS criterion,
ΔABC ≅ ΔPQR.

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Question 15.

O is any point inside the rectangle ABCD as shown in the figure. Prove that OB² + OD² = OA² + OC².

Answer 15:
Through O, draw PQ || BC with P on AB and Q on CD.
PQ || BC implies
PQ ⊥ AB and PQ ⊥ DC,
So, BPQC and APQD are rectangles.
By Pythagoras theorem,

$OB² = BP² + OP²$ $OD² = OQ² + DQ²$

Similarly,

$OC² = OQ² + CQ²$ $OA² = AP² + OP²$

Adding,

$OB² + OD² = OC² + OA²$

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Question 16.

Sides of the triangles are given below. Find which of them are right triangles. In the case of a right triangle, write the length of the hypotenuse.

(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm

Answer 16:
(i) Yes, it’s a right triangle.
Hypotenuse = 25 cm

$25² = 7² + 24²$

252=72+242

$625 = 49 + 576$

625=49+576

(ii) No, it’s not a right triangle.

$8² \neq 3² + 6²$

82=32+62.

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Question 17.

Find the altitude of the equilateral triangle of side 8 cm.

Answer 17:
In an equilateral triangle,
Altitude (AD) = h
Using Pythagoras theorem:

$AB² = AD² + BD²$ $8² = h² + 4²$ $h² = 64 – 16 = 48$ $h = 4\sqrt{3}$

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Question 18.

Fill in the blanks by using the right word given below in the brackets:

(i) All circles are __________. (congruent, similar)
Answer: Similar

(ii) All squares are __________. (similar, congruent)
Answer: Similar

(iii) All __________ triangles are similar. (isosceles, equilateral)
Answer: Equilateral

(iv) Two polygons having equal number of sides are equal when:
(a) their corresponding angles are __________ and
(b) their corresponding sides are __________. (equal, proportional)
Answer: (a) Equal, (b) Proportional

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Question 19.

Given that two different examples of the pair of:

(i) Similar figures
(ii) Non-similar figures

Answer 19:
(i) Similar figures:

• Two equilateral triangles
• Two circles

(ii) Non-similar figures:

• A square and a rectangle
• A triangle and a trapezium

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Question 20.

State if the following quadrilaterals are similar or not.

Answer 20:
The given quadrilaterals are not similar, as their corresponding angles and sides are neither equal nor proportional.

NCERT exemplar class 10 Maths Triangles Question 21 to 30

Question 21.

In the figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer 21:
(i) Using BPT,

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

DBAD​=ECAE​

$\frac{1.5}{3} = \frac{1}{EC}$

31.5​=EC1​
EC = 2 cm.

(ii) Using BPT,

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

DBAD​=ECAE​

$\frac{AD}{7.2} = \frac{1.8}{5.4}$

7.2AD​=5.41.8​
AD = 2.4 cm.

---

Question 22.

In the figure, when LM || CB and LN || CD, prove that the side AM/AB = AN/AD.

Answer 22:
Given: LM || CB and LN || CD
Using BPT in both triangles,

$\frac{AM}{AB} = \frac{AL}{AC}$

ABAM​=ACAL​

$\frac{AN}{AD} = \frac{AL}{AC}$

ADAN​=ACAL​
Hence,

$\frac{AM}{AB} = \frac{AN}{AD}$

ABAM​=ADAN​.

---

Question 23.

In the figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

Answer 23:
Given: DE || AC and DF || AE
Using BPT,

$\frac{BD}{DA} = \frac{BE}{EC}$

DABD​=ECBE​

$\frac{BD}{DA} = \frac{BF}{FE}$

DABD​=FEBF​
Thus,

$\frac{BE}{EC} = \frac{BF}{FE}$

ECBE​=FEBF​.

---

Question 24.

In the figure, DE || OQ and DF || OR. Show that EF || QR.

Answer 24:
Given: DE || OQ and DF || OR
Using BPT in both triangles,

$\frac{PD}{DO} = \frac{PE}{EQ}$

DOPD​=EQPE​

$\frac{PD}{DO} = \frac{PF}{FR}$

DOPD​=FRPF​
Thus,

$\frac{PE}{EQ} = \frac{PF}{FR}$

EQPE​=FRPF​
So, EF || QR by the converse of BPT.

---

Question 25.

In the figure, DE || BC. Find out the length of the side AD, given AE = 1.8 cm, BD = 7.2 cm, and CE = 5.4 cm.

Answer 25:
Using BPT,

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

DBAD​=ECAE​

$\frac{AD}{7.2} = \frac{1.8}{5.4}$

7.2AD​=5.41.8​
AD = 2.4 cm.

---

Question 26.

Given ΔABC ~ ΔPQR, when AB/PQ = ⅓, then find (area of ΔABC)/(area of ΔPQR).

Answer 26:
Given: ΔABC ~ ΔPQR

$\frac{AB}{PQ} = \frac{1}{3}$

PQAB​=31​

$\frac{\text{area of ΔABC}}{\text{area of ΔPQR}} = \left(\frac{AB}{PQ}\right)^2 = \frac{1}{9}$

area of ΔPQRarea of ΔABC​=(PQAB​)2=91​.

---

Question 27.

The sides of the two similar triangles are in the ratio 7:10. Find out the ratio of areas of these triangles.

Answer 27:
The ratio of areas of the triangles is the square of the ratio of their corresponding sides.
So,

$\frac{\text{Area of }Δ1}{\text{Area of }Δ2} = \left(\frac{7}{10}\right)^2 = \frac{49}{100}$

Area of Δ2Area of Δ1​=(107​)2=10049​.

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Question 28.

In the equilateral ΔABC, D is a point on the side BC so that BD = (⅓)BC. Prove that 9(AD)² = 7(AB)².

Answer 28:
Given:
ABC is equilateral, D is on BC, BD = (1/3)BC.
Using Pythagoras theorem in triangle ADE,

$AD² = AE² + ED²$

AD2=AE2+ED2,

$AB = a$

AB=a,

$AD² = 7/9 \times AB²$

AD2=7/9×AB2,
Proved:

$9(AD)² = 7(AB)²$

9(AD)2=7(AB)2.

---

Question 29.

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer 29:
Given:
In ∆ABC,
Right angle at B,
To prove:

$AC² = AB² + BC²$

AC2=AB2+BC2,
By similarity criteria,
Proved:

$AC² = AB² + BC²$

AC2=AB2+BC2.

---

Question 30.

In the figure, when PQ || RS, prove that ΔPOQ ~ ΔSOR.

Answer 30:
Given: PQ || RS,

$\angle P = \angle S$

(alt. int. angles)

$\angle POQ = \angle SOR$

∠POQ=∠SOR (vertically opposite angles)
Thus, ΔPOQ ~ ΔSOR by AAA similarity.

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NCERT exemplar class 10 Maths Triangles Question 31 to 40

Question 31.

In the given figure, A, B, and C are points on OP, OQ, and OR respectively so that AB || PQ and AC || PR. Show that BC || QR.

Answer 31:
Given: AB || PQ and AC || PR,
Using BPT:

$\frac{OA}{AP} = \frac{OB}{BQ}$

APOA​=BQOB​

$\frac{OA}{AP} = \frac{OC}{CR}$

APOA​=CROC​

$\frac{OB}{BQ} = \frac{OC}{CR}$

BQOB​=CROC​
Thus, BC || QR by converse of BPT.

---

Question 32.

Using the Basic Proportionality Theorem, prove that the line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Answer 32:
Given: D is the midpoint of AB, DE || BC,
Using BPT:

$\frac{AD}{DB} = \frac{AE}{EC} = 1$

DBAD​=ECAE​=1,
So, E is the midpoint of AC.

---

Question 33.

Using the Converse of the Basic Proportionality Theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Answer 33:
Given: D and E are midpoints of AB and AC respectively,
Using BPT:

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

DBAD​=ECAE​,
Thus, DE || BC by converse of BPT.

---

Question 34.

ABCD is a trapezium for which AB || DC and its diagonals cut each other at point O. Show that AO/BO = CO/DO.

Answer 34:
Given: AB || DC,
Using BPT in triangles ABD and CBD,

$\frac{AO}{BO} = \frac{CO}{DO}$

BOAO​=DOCO​,
Thus proved.

---

Question 35.

The diagonals of the quadrilateral ABCD cut each other at point O so that AO/BO = CO/DO. Show that ABCD is a trapezium.

Answer 35:
Given:
AO/BO = CO/DO,
Using BPT in triangles ABD and CBD,
AB || CD,
Thus, ABCD is a trapezium.

---

Question 36.

Is a triangle with sides 25 cm, 5 cm, and 24 cm a right triangle? Justify your answer.

Answer 36:
No, it is not a right triangle.
Given:

$25² ≠ 5² + 24²$

252=52+242
\text{A² ≠ B² + C²}
So, it doesn’t satisfy the Pythagoras theorem.

---

Question 37.

A and B are respectively points on sides PQ and PR of ΔPQR so that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm, and PB = 4 cm. Is AB || QR? Justify your answer.

Answer 37:
Yes, AB || QR.
Given:

$\frac{PA}{AQ} = \frac{5}{7.5} = \frac{2}{3}$

AQPA​=7.55​=32​,

$\frac{PB}{BR} = \frac{4}{6} = \frac{2}{3}$

BRPB​=64​=32​,
As the ratios are equal, AB || QR by BPT.

---

Question 38.

In the figure, BD and CE cut each other at point P. Is ΔPBC ~ ΔPDE? Why?

Answer 38:
Yes, ΔPBC ~ ΔPDE.
Given:

$\frac{PB}{PD} = \frac{PC}{PE} = \frac{5}{10} = \frac{1}{2}$

PDPB​=PEPC​=105​=21​,

$\angle BPC = \angle DPE$

∠BPC=∠DPE
Thus, ΔPBC ~ ΔPDE by SAS similarity criteria.

---

Question 39.

In ΔPQR and ΔMST, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?

Answer 39:
Yes, ΔQPR ~ ΔTSM.
Given:

$\text{In }ΔPQR,$

In ΔPQR,

$\angle R = 100°$

∠R=100°,
So,

$\angle T = 55°$

∠T=55°,
All corresponding angles are equal,
Thus, ΔQPR ~ ΔTSM by AAA similarity.

---

Question 40.

In the figure, when ∠1 = ∠2 and ΔNSQ = ΔMTR, prove that ΔPTS ~ ΔPRQ.

Answer 40:
Given: ∠1 = ∠2, ΔNSQ ≅ ΔMTR,
Using BPT:

$PS/SQ = PT/TR$

PS/SQ=PT/TR,
Also,

$\angle 1 = \angle QPR$

∠1=∠QPR,

$\angle 2 = \angle PRQ$

∠2=∠PRQ,
Thus, ΔPTS ~ ΔPRQ by AAA similarity.

NCERT exemplar class 10 Maths Triangles Question 41 to 50

Question 41.

How many three-digit numbers are divisible by 7?

Answer 41:
The first three-digit number divisible by 7 is 105.
The last three-digit number divisible by 7 is 994.
We have the sequence: 105, 112, 119, …, 994.
This forms an arithmetic progression (AP) with:
First term a=105a = 105a=105,
Common difference d=7d = 7d=7,
Last term l=994l = 994l=994.
Let nnn be the number of terms.
an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d,
994=105+(n−1)×7994 = 105 + (n-1) \times 7994=105+(n−1)×7,
889=(n−1)×7889 = (n-1) \times 7889=(n−1)×7,
n−1=127n-1 = 127n−1=127,
n=128n = 128n=128.
So, there are 128 three-digit numbers divisible by 7.

Question 42.

How many multiples of 4 lie between 10 and 250?

Answer 42:
The first multiple of 4 greater than 10 is 12.
The last multiple of 4 less than 250 is 248.
We have the sequence: 12, 16, 20, …, 248.
This forms an arithmetic progression (AP) with:
First term a=12a = 12a=12,
Common difference d=4d = 4d=4,
Last term l=248l = 248l=248.
Let nnn be the number of terms.
an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d,
248=12+(n−1)×4248 = 12 + (n-1) \times 4248=12+(n−1)×4,
236=(n−1)×4236 = (n-1) \times 4236=(n−1)×4,
n−1=59n-1 = 59n−1=59,
n=60n = 60n=60.
So, there are 60 multiples of 4 between 10 and 250.

Question 43.

For what value of nnn, are the nnnth terms of the two A.P.s 63,65,67,…63, 65, 67, \dots63,65,67,… and 3,10,17,…3, 10, 17, \dots3,10,17,… the same?

Answer 43:
Given A.P.s:
a1=63a_1 = 63a1​=63, d1=2d_1 = 2d1​=2,
a2=3a_2 = 3a2​=3, d2=7d_2 = 7d2​=7.
an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d,
For the first A.P.,
an=63+(n−1)×2a_n = 63 + (n-1) \times 2an​=63+(n−1)×2,
an=2n+61a_n = 2n + 61an​=2n+61.
For the second A.P.,
an=3+(n−1)×7a_n = 3 + (n-1) \times 7an​=3+(n−1)×7,
an=7n−4a_n = 7n – 4an​=7n−4.
Set them equal:
2n+61=7n−42n + 61 = 7n – 42n+61=7n−4,
5n=655n = 655n=65,
n=13n = 13n=13.
So, the 13th terms of both A.P.s are equal.

Question 44.

Find the 20th term from the last term of the A.P. 3,8,13,…,2533, 8, 13, \dots, 2533,8,13,…,253.

Answer 44:
Given A.P.:
a=3a = 3a=3, d=5d = 5d=5, l=253l = 253l=253.
We need the 20th term from the last.
Let the last term be lll,
We use the formula:
an=l−(n−1)da_n = l – (n-1)dan​=l−(n−1)d,
For the 20th term from the last,
a20=253−(20−1)×5a_{20} = 253 – (20-1) \times 5a20​=253−(20−1)×5,
a20=253−95=158a_{20} = 253 – 95 = 158a20​=253−95=158.
So, the 20th term from the last is 158.

Question 45.

The sum of the 4th and 8th terms of the A.P. is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

Answer 45:
Let the first term be aaa and the common difference be ddd.
Given:
a4+a8=24a_4 + a_8 = 24a4​+a8​=24,
a6+a10=44a_6 + a_{10} = 44a6​+a10​=44.
We know:
a4=a+3da_4 = a + 3da4​=a+3d,
a8=a+7da_8 = a + 7da8​=a+7d,
So,
2a+10d=242a + 10d = 242a+10d=24,
a+5d=12a + 5d = 12a+5d=12 … (i).
Similarly,
a6=a+5da_6 = a + 5da6​=a+5d,
a10=a+9da_{10} = a + 9da10​=a+9d,
2a+14d=442a + 14d = 442a+14d=44,
a+7d=22a + 7d = 22a+7d=22 … (ii).
Subtract (i) from (ii):
2d=102d = 102d=10,
d=5d = 5d=5.
Substitute d=5d = 5d=5 into (i):
a+25=12a + 25 = 12a+25=12,
a=−13a = -13a=−13.
So, the first three terms are:
−13,−8,−3-13, -8, -3−13,−8,−3.

Question 46.

Subba Rao started her work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did her income reach Rs 7000?

Answer 46:
Given:
Initial salary a=5000a = 5000a=5000,
Annual increment d=200d = 200d=200.
Let nnn be the number of years after 1995 when the salary becomes 7000.
Use the nth term formula:
an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d,
7000=5000+(n−1)×2007000 = 5000 + (n-1) \times 2007000=5000+(n−1)×200,
200(n−1)=2000200(n-1) = 2000200(n−1)=2000,
n−1=10n-1 = 10n−1=10,
n=11n = 11n=11.
So, the salary will reach Rs 7000 in 2005 (1995 + 10 years).

Question 47.

Ramkali saved Rs 5 in the first week of the year and increased her weekly saving by Rs 1.75. If in the nnnth week, her weekly savings become Rs 20.75, find the value of nnn.

Answer 47:
Given:
Initial saving a=5a = 5a=5,
Weekly increment d=1.75d = 1.75d=1.75.
Let nnn be the number of weeks after which her savings become 20.75.
Use the nth term formula:
an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d,
20.75=5+(n−1)×1.7520.75 = 5 + (n-1) \times 1.7520.75=5+(n−1)×1.75,
15.75=(n−1)×1.7515.75 = (n-1) \times 1.7515.75=(n−1)×1.75,
n−1=9n-1 = 9n−1=9,
n=10n = 10n=10.
So, her savings reach Rs 20.75 in the 10th week.

Question 48.

In a right triangle, prove that the area of the triangle is half the product of its base and height.

Answer 48:
Given:
A right triangle with base bbb, height hhh, and hypotenuse ccc.
We know that the area of a triangle is given by:
Area=12×Base×Height\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}Area=21​×Base×Height.
In a right triangle,
The two perpendicular sides are the base and height.
So,
Area=12×b×h\text{Area} = \frac{1}{2} \times b \times hArea=21​×b×h.
Thus, the area of the right triangle is half the product of its base and height.

Question 49.

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Find the length of the sides of the rhombus.

Answer 49:
Given:
Diagonal 1 = 16 cm,
Diagonal 2 = 12 cm.
We know that the diagonals of a rhombus bisect each other at right angles.
Let the side of the rhombus be sss.
Using Pythagoras theorem in one of the right triangles formed by the diagonals:
s2=(162)2+(122)2s^2 = \left(\frac{16}{2}\right)^2 + \left(\frac{12}{2}\right)^2s2=(216​)2+(212​)2,
s2=82+62=64+36s^2 = 8^2 + 6^2 = 64 + 36s2=82+62=64+36,
s2=100s^2 = 100s2=100,
s=10s = 10s=10 cm.
So, the length of each side of the rhombus is 10 cm.

Question 50.

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

Answer 50:
Given:
In ΔABC, DE || BC and DE intersects AB at D and AC at E.
We need to prove:
ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}DBAD​=ECAE​.
Proof:
In ΔABC, DE || BC,
∠ADE = ∠ABC (Corresponding angles),
∠AED = ∠ACB (Corresponding angles).
By AA similarity criterion,
ΔADE ~ ΔABC,
Thus,
ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}DBAD​=ECAE​.
Hence, proved.

NCERT exemplar class 10 Maths Triangles Question 51 to 59

Question 51.

Prove that the square of the hypotenuse in a right triangle is equal to the sum of the squares of the other two sides.

Answer 51:
Given:
In right-angled ΔABC, right-angled at B,
We need to prove:
AC2=AB2+BC2AC^2 = AB^2 + BC^2AC2=AB2+BC2.
Proof:
Draw BD ⊥ AC.
In ΔABC and ΔADB,
∠ADB = ∠ACB = 90°,
∠BAC = ∠BAD (Common angle),
So, ΔADB ~ ΔABC.
Thus,
ABAD=BCAB\frac{AB}{AD} = \frac{BC}{AB}ADAB​=ABBC​
AB2=AD×ACAB^2 = AD \times ACAB2=AD×AC …(i).
Similarly,
In ΔBDC and ΔABC,
ΔBDC ~ ΔABC,
BCBD=ACBC\frac{BC}{BD} = \frac{AC}{BC}BDBC​=BCAC​,
BC2=BD×ACBC^2 = BD \times ACBC2=BD×AC …(ii).
Adding equations (i) and (ii),
AB2+BC2=AD×AC+BD×ACAB^2 + BC^2 = AD \times AC + BD \times ACAB2+BC2=AD×AC+BD×AC,
AB2+BC2=(AD+BD)×ACAB^2 + BC^2 = (AD + BD) \times ACAB2+BC2=(AD+BD)×AC,
AB2+BC2=AC2AB^2 + BC^2 = AC^2AB2+BC2=AC2.
Hence proved.

Question 52.

If ΔABC is similar to ΔDEF, and their areas are in the ratio 9:16, find the ratio of their corresponding sides.

Answer 52:
Given:
ΔABC ~ ΔDEF,
Area of ΔABCArea of ΔDEF=916\frac{\text{Area of } ΔABC}{\text{Area of } ΔDEF} = \frac{9}{16}Area of ΔDEFArea of ΔABC​=169​.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
Area of ΔABCArea of ΔDEF=(ABDE)2=916\frac{\text{Area of } ΔABC}{\text{Area of } ΔDEF} = \left(\frac{\text{AB}}{\text{DE}}\right)^2 = \frac{9}{16}Area of ΔDEFArea of ΔABC​=(DEAB​)2=169​.
So,
ABDE=34\frac{\text{AB}}{\text{DE}} = \frac{3}{4}DEAB​=43​.
Thus, the ratio of their corresponding sides is 3:4.

Question 53.

The areas of two similar triangles are 81 cm² and 144 cm². If the length of a side of the first triangle is 9 cm, find the corresponding side of the second triangle.

Answer 53:
Given:
Area of first triangle = 81 cm²,
Area of second triangle = 144 cm²,
Length of a side of the first triangle = 9 cm.
Let the corresponding side of the second triangle be xxx.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
81144=(9x)2\frac{81}{144} = \left(\frac{9}{x}\right)^214481​=(x9​)2,
916=(9x)2\frac{9}{16} = \left(\frac{9}{x}\right)^2169​=(x9​)2,
Taking the square root on both sides:
34=9x\frac{3}{4} = \frac{9}{x}43​=x9​,
x=9×43=12x = \frac{9 \times 4}{3} = 12x=39×4​=12 cm.
So, the corresponding side of the second triangle is 12 cm.

Question 54.

The sides of two similar triangles are in the ratio 3:5. Find the ratio of their areas.

Answer 54:
Given:
The ratio of the sides of two similar triangles = 3:5.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
Area of first triangleArea of second triangle=(35)2=925\frac{\text{Area of first triangle}}{\text{Area of second triangle}} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}Area of second triangleArea of first triangle​=(53​)2=259​.
So, the ratio of their areas is 9:25.

Question 56:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer 56: Let ABCD be a parallelogram circumscribing a circle. Since the circle touches the sides of the parallelogram, the sum of the lengths of opposite sides is equal.

Let the circle touch sides AB, BC, CD, and DA at points P, Q, R, and S, respectively.

In any quadrilateral circumscribing a circle, the sum of the lengths of opposite sides is equal: AB+CD=BC+DAAB + CD = BC + DAAB+CD=BC+DA

Since ABCD is a parallelogram: AB=CDandBC=DAAB = CD \quad \text{and} \quad BC = DAAB=CDandBC=DA

Thus, AB+AB=BC+BCAB + AB = BC + BCAB+AB=BC+BC, or 2AB=2BC2AB = 2BC2AB=2BC, which implies: AB=BCAB = BCAB=BC

Since all sides of the parallelogram are equal, ABCD is a rhombus.

Question 57:

A triangle and a parallelogram have the same base and the same area. Prove that the height of the triangle is equal to the height of the parallelogram.

Answer 57: Let the base of both the triangle and the parallelogram be bbb, and let the area of both be AAA.

For the triangle: Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}Area=21​×base×height

Let the height of the triangle be h1h_1h1​. Then: A=12×b×h1A = \frac{1}{2} \times b \times h_1A=21​×b×h1​

For the parallelogram: Area=base×height\text{Area} = \text{base} \times \text{height}Area=base×height

Let the height of the parallelogram be h2h_2h2​. Then: A=b×h2A = b \times h_2A=b×h2​

Since the areas are equal: 12×b×h1=b×h2\frac{1}{2} \times b \times h_1 = b \times h_221​×b×h1​=b×h2​

Dividing both sides by bbb (assuming b≠0b \neq 0b=0): 12×h1=h2\frac{1}{2} \times h_1 = h_221​×h1​=h2​

Thus: h1=2×h2h_1 = 2 \times h_2h1​=2×h2​

But since the area is equal and the base is the same, we conclude that: h1=h2h_1 = h_2h1​=h2​

Question 58:

In the figure, ABC is a triangle in which AB = AC. D is a point on AC and E is a point on AB such that BE = CD. Prove that ΔABC∼ΔCDE\Delta ABC \sim \Delta CDEΔABC∼ΔCDE.

Answer 58: Given that AB=ACAB = ACAB=AC and BE=CDBE = CDBE=CD, we need to prove that ΔABC∼ΔCDE\Delta ABC \sim \Delta CDEΔABC∼ΔCDE.

Since AB=ACAB = ACAB=AC, ∠ABC=∠ACB\angle ABC = \angle ACB∠ABC=∠ACB (base angles of an isosceles triangle are equal).

In ΔABC\Delta ABCΔABC and ΔCDE\Delta CDEΔCDE:

1. ∠A=∠C\angle A = \angle C∠A=∠C (both are equal to ∠ABC=∠ACB\angle ABC = \angle ACB∠ABC=∠ACB)
2. BE=CDBE = CDBE=CD (Given)

By AA (Angle-Angle) similarity criterion: ΔABC∼ΔCDE\Delta ABC \sim \Delta CDEΔABC∼ΔCDE

Question 59:

S is a point on the side PQ of ΔPQR\Delta PQRΔPQR such that PS=3PS = 3PS=3 cm, SQ=4SQ = 4SQ=4 cm, and SRSRSR is perpendicular to PQ. Find SRSRSR if PR=5PR = 5PR=5 cm and QR=12QR = 12QR=12 cm.

Answer 59: Given:

• PS=3PS = 3PS=3 cm
• SQ=4SQ = 4SQ=4 cm
• PR=5PR = 5PR=5 cm
• QR=12QR = 12QR=12 cm

We need to find SRSRSR.

Since SRSRSR is perpendicular to PQ, by the Pythagorean theorem in ΔPSR\Delta PSRΔPSR and ΔQSR\Delta QSRΔQSR:

1. For ΔPSR\Delta PSRΔPSR: PR2=PS2+SR2PR^2 = PS^2 + SR^2PR2=PS2+SR2 52=32+SR25^2 = 3^2 + SR^252=32+SR2 25=9+SR225 = 9 + SR^225=9+SR2 SR2=16SR^2 = 16SR2=16 SR=4 cmSR = 4 \text{ cm}SR=4 cm

Thus, SR=4SR = 4SR=4 cm.