Important Questions & Answers For Class 10 Maths Chapter 6 Triangles

Important Questions & Answers For Class 10 Maths Chapter 6 Triangles

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Important Questions for Class 10 Maths Chapter 6 Triangles

Q. 1: In the given figure, PS/SQ = PT/TR and ∠PST = ∠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 ∠PST = ∠PQR (Corresponding angles) .....(i)

∠PST = ∠PRQ (given( ...(ii)

From (i) and (ii),

∠PRQ = ∠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/EC....(i)

Also, given that DF // AE.

Again by Basic Proportionality Theorem,

BD/DA = BF/FE....(ii)

From (i) and (ii),

BE/EC = BF/FE

Hence proved.

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

(i) ∇AEP ~ ∇ CDP

(ii) ∇ABD ~ ∇ CBE

(iii) ∇AEP ~ ∇ADB

(iv) ∇ PDC ~ ∇ BEC

Solution:

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

(i) In ∇AEP and ∇CDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

∇AEP ~ ∇CDP

(ii) In ∇ABD and ∇CBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

∇ABD ~ ∇CBE

(iii) In ∇AEP and ∇ADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

∇AEP ~ ∇ADB

(iv) In ∇PDC and ∇BEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

∇PDC ~ ∇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 ∇ABC and ∇DFE,

∠C = ∠E (angle of elevation)

∠B = ∠F = 90°

By AA similarity criterion,

∇ABC ~ ∇DFE

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

AB/DF = BC/EF

6/h = 4/28

h = (6 ×28)/4

h = 6 × 7

h = 72

Hence, the height of the tower = 42 m.

Q. 5: If ∇ABC ~ ∇QRP, ar (∇ABC) / ar (∇PQR) =9/4 , AB = 18 cm and BC = 15 cm, then PR is equal to

(A) 10 cm (B) 12 cm (C) 20/3 cm (D) 8 cm

Solution:

Given that ∇ABC ~ ∇QRP.

ar (∇ABC) / ar (∇PQR) =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 (∇ABC) / ar (∇PQR) = BC²/RP²

9/4 = (15)²/RP²

RP² = (4/9) ×225

PR² = 100

Therefore, PR = 10 cm

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

Solution:

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

To prove ∇ABC ≅ ∇PQR.

Proof:

∇ABC ~ ∇PQR

∴ Area of (∇ABC)/Area of (∇PQR) = BC2/QR2

⇒BC2/QR2 =1 [Since, ar (∇ABC) = ar (∇PQR)]

⇒BC2/QR2

⇒BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Therefore, ∇ABC ≅ ∇PQR [SSS criterion of congruence]

Q. 7: O is any point inside a rectangle ABCD as shown in the figure. Prove that OB² + OD² = OA² + OC².

Solution:

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

PQ || BC

Therefore, PQ ⊥ AB and PQ ⊥ DC (∠B = 90° and ∠C = 90°)

So, ∠BPQ = 90° and ∠CQP = 90°

Hence, BPQC and APQD are both rectangles.

By Pythagoras theorem,

In ∇ OPB,

OB² = BP² + OP²â€¦..(1)

Similarly,

In ∇ OQD,

OD² = OQ² + DQ²â€¦..(2)

In ∇ OQC,

OC² = OQ² + CQ²â€¦..(3)

In ∇ OAP,

OA² = AP² + OP²â€¦..(4)

Adding (1) and (2),

OB²+ OD² = BP² + OP² + OQ² + DQ²

= CQ² + OP² + OQ² + AP²

(since BP = CQ and DQ = AP)

= CQ² + OQ² + OP² + AP²

= OC² + OA² [From (3) and (4)]

Hence proved that OB² + OD² = OA² + OC².

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 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 ≠ 64

Or, 32 + 62 ≠ 82

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 Pythagoras theorem.

Practice Questions For Class 10 Maths Chapter 6 Triangles

1. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², find the area of the larger triangle.
2. Find the value of x for which DE AB in the adjoint figure.
   
3. 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.
4. 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?
5. 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.
6. 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.
7. In the figure, if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.