**Important questions for Class 10 Maths Chapter 6** Triangles are provided here to help the students in their exam preparation based on the new pattern of CBSE for the 2022-2023 academic session. Students preparing for the board exam are advised to practice these important Triangles questions to score full marks for the questions from this chapter. In the exam, students may come across some of these questions. So, they should not leave any stone unturned for their board exam and must practice these questions. They can also access important questions for 10th maths all chapters at BYJUâ€™S.

The chapter Triangles contains many topics related to a triangle, such as criteria for similarity, congruency, areas of similar triangles and the Pythagoras theorem etc., Students can also get the answers to all the questions in Class 10 Maths of NCERT solutions.

**Read more:**

We have provided important questions of this chapter, along with the detailed solutions. After that, we have also provided some questions for students to practice which does not have solutions, and they must solve all of them to gain command over the Triangles topic.

## Important Questions & Answers For Class 10 Maths Chapter 6 Triangles

**Q. 1: In the given figure, PS/SQ = PT/TR and âˆ PST = âˆ PRQ. Prove that PQR is an isosceles triangle.**

**Solution:**

Given,

PS/SQ = PT/TR

We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore, ST // QR

And âˆ PST = âˆ PQR (Corresponding angles) â€¦â€¦..(i)

Also, given,

âˆ PST = âˆ PRQâ€¦â€¦…(ii)

From (i) and (ii),

âˆ PRQ = âˆ PQR

Therefore, PQ = PR (sides opposite the equal angles)

Hence, PQR is an isosceles triangle.

**Q. 2:** **In the figure, DE // AC and DF // AE. Prove that BF/FE = BE/EC.**

**Solution:**

Given that,

In triangle ABC, DE // AC.

By Basic Proportionality Theorem,

BD/DA = BE/ECâ€¦â€¦â€¦.(i)

Also, given that DF // AE.

Again by Basic Proportionality Theorem,

BD/DA = BF/FEâ€¦â€¦â€¦.(ii)

From (i) and (ii),

BE/EC = BF/FE

Hence proved.

**Q. 3:** **In the given figure, altitudes AD and CE of âˆ† ABC intersect each other at the point P. Show that:**

**(i) âˆ†AEP ~ âˆ† CDP**

**(ii) âˆ†ABD ~ âˆ† CBE**

**(iii) âˆ†AEP ~ âˆ†ADB**

**(iv) âˆ† PDC ~ âˆ† BEC**

**Solution:**

Given that AD and CE are the altitudes of triangle ABC and these altitudes intersect each other at P.

(i) In Î”AEP and Î”CDP,

âˆ AEP = âˆ CDP (90Â° each)

âˆ APE = âˆ CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

Î”AEP ~ Î”CDP

(ii) In Î”ABD and Î”CBE,

âˆ ADB = âˆ CEB ( 90Â° each)

âˆ ABD = âˆ CBE (Common Angles)

Hence, by AA similarity criterion,

Î”ABD ~ Î”CBE

(iii) In Î”AEP and Î”ADB,

âˆ AEP = âˆ ADB (90Â° each)

âˆ PAE = âˆ DAB (Common Angles)

Hence, by AA similarity criterion,

Î”AEP ~ Î”ADB

(iv) In Î”PDC and Î”BEC,

âˆ PDC = âˆ BEC (90Â° each)

âˆ PCD = âˆ BCE (Common angles)

Hence, by AA similarity criterion,

Î”PDC ~ Î”BEC

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

**Solution:**

Given,

Length of the vertical pole = 6 m

Shadow of the pole = 4 m

Let the height of the tower be h m.

Length of the shadow of the tower = 28 m

In Î”ABC and Î”DFE,

âˆ C = âˆ E (angle of elevation)

âˆ B = âˆ F = 90Â°

By AA similarity criterion,

Î”ABC ~ Î”DFE

We know that the corresponding sides of two similar triangles are proportional.

AB/DF = BC/EF

6/h = 4/28

h = (6 Ã—28)/4

h = 6 Ã— 7

h = 42

Hence, the height of the tower = 42 m.

**Q. 5:** **If Î”ABC ~ Î”QRP, ar (Î”ABC) / ar (Î”PQR) =9/4 , AB = 18 cm and BC = 15 cm, then find PR.**

**Solution:**

Given that Î”ABC ~ Î”QRP.

ar (Î”ABC) / ar (Î”QRP) =9/4

AB = 18 cm and BC = 15 cm

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

ar (Î”ABC) / ar (Î”QRP) = BC^{2}/RP^{2}

9/4 = (15)^{2}/RP^{2}

RP^{2} = (4/9) Ã— 225

PR^{2} = 100

Therefore, PR = 10 cm

**Q. 6:** **If the areas of two similar triangles are equal, prove that they are congruent.**

**Solution:**

Let Î”ABC and Î”PQR be the two similar triangles with equal area.

To prove Î”ABC â‰… Î”PQR.

Proof:

Î”ABC ~ Î”PQR

âˆ´ Area of (Î”ABC)/Area of (Î”PQR) = BC^{2}/QR^{2} = AB^{2}/PQ^{2} = AC^{2}/PR^{2}

â‡’ BC^{2}/QR^{2} = AB^{2}/PQ^{2} = AC^{2}/PR^{2} = 1 [Since, ar (Î”ABC) = ar (Î”PQR)]
â‡’ BC^{2}/QR^{2} = 1

â‡’ AB^{2}/PQ^{2} = 1

â‡’ AC^{2}/PR^{2} = 1

BC = QR

AB = PQ

AC = PR

Therefore, Î”ABC â‰… Î”PQR [SSS criterion of congruence]

**Q. 7:** **O is any point inside a rectangle ABCD as shown in the figure. Prove that OB ^{2} + OD^{2} = OA^{2} + OC^{2}.**

**Solution:**

Through O, draw PQ || BC so that P lies on AB and Q lies on DC.

PQ || BC

Therefore, PQ âŠ¥ AB and PQ âŠ¥ DC (âˆ B = 90Â° and âˆ C = 90Â°)

So, âˆ BPQ = 90Â° and âˆ CQP = 90Â°

Hence, BPQC and APQD are both rectangles.

By Pythagoras theorem,

In âˆ† OPB,

OB^{2} = BP^{2} + OP^{2}â€¦..(1)

Similarly,

In âˆ† OQD,

OD^{2} = OQ^{2} + DQ^{2}â€¦..(2)

In âˆ† OQC,

OC^{2} = OQ^{2} + CQ^{2}â€¦..(3)

In âˆ† OAP,

OA^{2} = AP^{2} + OP^{2}â€¦..(4)

Adding (1) and (2),

OB^{2}+ OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2}

= CQ^{2} + OP^{2} + OQ^{2} + AP^{2}

(since BP = CQ and DQ = AP)

= CQ^{2 }+ OQ^{2} + OP^{2} + AP^{2}

= OC^{2} + OA^{2} [From (3) and (4)]

Hence proved that OB^{2} + OD^{2} = OA^{2} + OC^{2}.

**Q. 8: Sides of triangles are given below. Determine which of them are right triangles.**

**In case of a right triangle, write the length of its hypotenuse.**

**(i) 7 cm, 24 cm, 25 cm**

**(ii) 3 cm, 8 cm, 6 cm**

**Solution:**

(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)^{2} + (24)^{2} = (25)^{2}

Therefore, the above equation satisfies the Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 â‰ 64

Or, 3^{2} + 6^{2} â‰ 8^{2}

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfy the Pythagoras theorem.

**Q.9: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.**

**Solution:**

Consider a triangle Î”ABCand draw a line PQ parallel to the side BC of Î”ABC and intersect the sides AB and AC in P and Q, respectively.

To prove: AP/PB = AQ/QC

Cosntruction:

Join the vertex B of Î”ABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and draw PMâŠ¥AC as shown in the given figure.

Proof:

Now the area of âˆ†APQ = 1/2 Ã— AP Ã— QN (Since, area of a triangle= 1/2Ã— Base Ã— Height)

Similarly, area of âˆ†PBQ= 1/2 Ã— PB Ã— QN

area of âˆ†APQ = 1/2 Ã— AQ Ã— PM

Also,area of âˆ†QCP = 1/2 Ã— QC Ã— PM â€¦â€¦â€¦â€¦ (1)

Now, if we find the ratio of the area of triangles âˆ†APQand âˆ†PBQ, we have

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

Therefore, we can say that âˆ†PBQ and QCP have the same area.

area of âˆ†PBQ = area of âˆ†QCP â€¦â€¦â€¦â€¦..(3)

Therefore, from the equations (1), (2) and (3), we can say that,

AP/PB = AQ/QC

**Q.10: In the figure, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.**

**Solution:**

Given,

DE || BC

AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm

By basic proportionality theorem,

AD/DB = AE/EC

AD/7.2 = 1.8/5.4

AD = (1.8 Ã— 7.2)/5.4

= 7.2/4

= 2.4

Therefore, AD = 2.4 cm.

**Q.11: Given Î”ABC ~ Î”PQR, if AB/PQ = â…“, then find (ar Î”ABC)/(ar Î”PQR).**

**Solution:**

Given,Â

Î”ABC ~ Î”PQR

And

AB/PQ = â…“,Â

We know that The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(ar Î”ABC)/(ar Î”PQR) = AB^{2}/PQ^{2} = (AB/PQ)^{2} = (â…“)^{2} = 1/9

Therefore, (ar Î”ABC)/(ar Î”PQR) = 1/9

Or

(ar Î”ABC) : (ar Î”PQR) = 1 : 9

**Q.12: The sides of two similar triangles are in the ratio 7 : 10.Â Find the ratio of areas of these triangles.**

**Solution:**

Given,

The ratio of sides of two similar triangles = 7 : 10

We know thatÂ the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

The ratio of areas of these triangles = (Ratio of sides of two similar triangles)^{2}

= (7)^{2} : (10)^{2}

= 49 : 100

Therefore, the ratio of areas of the given similar triangles is 49 : 100.

**Q.13: In an equilateral Î”ABC, D is a point on side BC such that BD = (â…“) BC. Prove that 9(AD) ^{2} = 7(AB)^{2}.**

**Solution:**

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = (1/3)BC.

Let a be the side of the equilateral triangle and AE be the altitude of Î”ABC.

âˆ´ BE = EC = BC/2 = a/2

And, AE = aâˆš3/2

Given, BD = 1/3BC

âˆ´ BD = a/3

DE = BE â€“ BD = a/2 â€“ a/3 = a/6

In Î”ADE, by Pythagoras theorem,

AD^{2} = AE^{2} + DE^{2}

= [(aâˆš3)/2]^{2} + (a/6)^{2}

= (3a^{2}/4) + (a^{2}/36)

= (37a^{2} + a^{2})/36

= (28a^{2})/36

= (7/9)a^{2}

= (7/9) (AB)^{2}

Therefore, 9(AD)^{2} = 7(AB)^{2}.

**Q.14: Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.**

**Solution:**

Given: A right-angled triangle ABC, right-angled at B.

To Prove: AC^{2} = AB^{2} + BC^{2}

Construction: Draw a perpendicular BD meeting AC at D.

Proof:

In Î”ABC and Î”ADB,

âˆ ABC = âˆ ADB = 90Â°

Â Â âˆ A = âˆ A â†’ common

Using the AA criterion for the similarity of triangles,Â

Î”ABC ~ Î”ADBÂ

Therefore, AD/AB = AB/AC

â‡’ AB^{2} = AC x AD â€¦â€¦(1)

ConsideringÂ Î”ABC and Î”BDC from the figure.

C = âˆ CÂ â†’ common

Â âˆ CDB = âˆ ABC = 90Â°

Using the Angle Angle(AA) criterion for the similarity of triangles, we conclude that,

Î”BDC ~ Î”ABC

Therefore, CD/BC = BC/AC

â‡’ BC^{2} = AC x CD â€¦..(2)

By adding equation (1) and equation (2), we get:

AB^{2} + BC^{2} = (AC x AD) + (AC x CD)

AB^{2} + BC^{2} = AC (AD + CD) â€¦..(3)

AB^{2} + BC^{2} = AC (AC) {sinceÂ AD + CD = AC}

AB^{2} + BC^{2} = AC^{2}

Hence proved.

**Q.15: In the figure, if PQ || RS, prove that âˆ† POQ ~ âˆ† SOR.**

**Solution:**

Given,

PQ || RS

âˆ P = âˆ S (Alternate angles)Â

and âˆ Q = âˆ RÂ

Also, âˆ POQ = âˆ SOR (Vertically opposite angles)Â

Therefore, âˆ† POQ ~ âˆ† SOR (by AAA similarity criterion)

Hence proved.

## Video Lesson on BPT and Similar Triangles

**Class 10 Maths Chapter 6 Triangles Questions for Practice**

- Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
- Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP Ã— PC = BP Ã— DP.
- Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.
- Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.
- Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
- Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm
^{2}, find the area of the larger triangle. - A foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
- An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 1/2 hours?
- Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.
- Prove that if in a triangle square on one side is equal to the sum of the squares on the other two sides, then the angle opposite the first side is a right angle.

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