THE TRIANGLE AND ITS PROPERTIES - Testlabz · 6 THE TRIANGLE AND ITS PROPERTIES Exercise 6.1 ......

CBSEPracticalSkills.com ©Edulabz International


1

6 THE TRIANGLE AND ITS PROPERTIES

Exercise 6.1

Q.1. In ΔPQR, D is the mid-point of QR . PM is ___________. PD is ____________. Is QM = MR?

Ans. PM is altitude. PD is median. Hence, no, QM ≠ MR.

Q.2. Draw rough sketches for the following : (a) In ΔABC, BE is a median. (b) In ΔPQR, PQ and PR are altitudes of the triangle. (c) In ΔXYZ, YL is an altitude in the exterior of the

triangle. Ans.

Q.3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

Ans. Draw an isosceles triangle ABC. Find out the mid-point of BC and mark

it by D. Most the vertex A to this



2

mid-point D. AD is the median of the triangle ABC.

Now, measure the angle ADC, which comes out 90°. It means AD is perpendicular to the side BC. AD is also the altitude of ΔABC.

Exercise 6.2

Q.1. Find the value of the unknown exterior angle x in the following diagrams :

Ans. (i) x is an exterior angle, an exterior angle is the sum of its two interior opposite angles.

So, x = 50° + 70° (by property) or x = 120° Hence, exterior angle is 120°. (ii) x is an exterior angle, an exterior angle is the sum of its

two interior opposite angles. So, x = 65° + 45°



3

or x = 110° (by property) Hence, exterior angle is 110°. (iii) x is an exterior angle, an exterior angle is the sum of its

two interior opposite angles. So, x = 30° + 40° (by property) or x = 70° Hence, exterior angle is 110°. (iv) x is an exterior angle. ∴ x = 60° + 60° (by property) or x = 120° Hence, exterior angle is 120°. (v) x is an exterior angle. ∴ x = 50° + 50° (by property) or x = 100° Hence, exterior angle is 100°. (vi) x is an exterior angle. ∴ x = 60° + 30° (by property) or x = 90° Hence, exterior angle is 90°.

Q.2. Find the value of the unknown interior angle x in the following figures :



4

Ans. Exterior angle = Sum of interior opposite angles (i) 115° = x + 50° ⇒ 115° – 50° = x ⇒ 65° = x So, x = 65° (ii) 70° + x = 100° ⇒ x = 100° – 70° So, x = 30° (iii) x + 90° = 125° ⇒ x = 125° – 90° So, x = 35° (iv) 60° + x = 120° ⇒ x = 120° – 60° So, x = 60° (v) 30° + x = 80° ⇒ x = 80° – 30° So, x = 50° (vi) x + 35° = 75° ⇒ x = 75° – 35° So, x = 40°

Exercise 6.3



5

Q.1. Find the value of the unknown x in the following diagrams :

Ans. (i) In ΔABC, ∠A + ∠B + ∠C = 180° (by angle sum property of a triangle)

∴ x + 50° + 60° = 180° or x + 110° = 180° or x = 180° – 110° or x = 70° (ii) In ΔPQR, ∠P + ∠Q + ∠R = 180° (by angle sum

property of a triangle) ∴ 90° + 30° + x = 180° or 120° + x = 180° or x = 180° – 120° = 60° (iii) In ΔXYZ, ∠X + ∠Y + ∠Z = 180° (by angle sum

property of a triangle) ∴ 30° + 110° + x = 180°



6

or 140° + x = 180° or x = 180° – 140° = 40° (iv) ∴ x + x + 50° = 180° (by property) or 2x + 50° = 180° or 2x = 180° – 50° = 130°

or x = 1302

° = 65°

(v) ∴ x + x + x = 180° (by property) or 3x = 180°

or x = 1803

= 60°

(vi) x + 2x + 90° = 180° (by property) ∴ 3x = 180° – 90° = 90°

or x = 903

° = 30°

Q.2. Find the values of the unknowns x and y in the following diagrams :

Ans. (i) x + y + 50° = 180° (by the angle sum property of a triangle)

x + y = 180° – 50° x + y = 130°



7

120° = x + 50° (by property of an exterior angle)

⇒ x = 120° – 50° = 70° ∴ 70° + y = 130° ⇒ y = 130° – 70° = 60° (ii) y = 80° (vertically opposite angles) x + y + 50° = 180° (by the angle sum property

of a triangle) or x + 80° + 50° = 180° or x + 130° = 180° or x = 180° – 130° Hence, x = 50° and y = 80°. (iii) y + 50° + 60° = 180° (by the angle sum property

of a triangle) or y + 110° = 180° or y = 180° – 110° or y = 70° x + y = 180° (Linear pair) ⇒ x + 70° = 180° or x = 180° – 70° or x = 110°

(iv) x = 60° (vertically opposite angles) x + y + 30° = 180° (by the angle sum property

of a triangle) or 60° + y + 30° = 180°



8

y + 90° = 180° or y = 180° – 90° or y = 90° Hence, x = 60° and y = 90° (v) y = 90° (vertically opposite angles) ∴ x + y + z = 180° (by the angle sum property

of a triangle) ⇒ x + 90° + x = 180° or 2x + 90° = 180° or 2x = 90° or x = 45° Hence, x = 45° and y = 90°. (vi) y = x (vertically opposite angles) ∴ x + x + y = 180° (by angle sum property of

a triangle) ⇒ x + x + x = 180° or 3x = 180° or x = 60° So, y = 60° Hence, x = 60° and y = 60°.

Exercise 6.4

Q.1. Is it possible to have a triangle with the following sides? (i) 2 cm, 3 cm, 5 cm (ii) 3 cm, 6 cm, 7 cm (iii) 6 cm, 3 cm, 2 cm



9

Ans. We know that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

(i) 2 cm + 3 cm = 5 cm = 5 cm 3 cm + 5 cm = 8 cm > 2 cm Hence, it is not possible to have a triangle. (ii) 3 cm + 6 cm = 9 cm > 7 cm 3 cm + 7 cm = 10 cm > 6 cm 6 cm + 7 cm = 13 cm > 3 cm Hence, it is possible to have a triangle. (iii) 6 cm + 3 cm = 9 cm > 2 cm 6 cm + 2 cm = 8 cm > 3 cm 2 cm + 3 cm = 5 cm < 6 cm Hence, it is not possible to have a triangle.

Q.2. Take any point O in the interior of a triangle PQR. Is (i) OP + OQ > PQ? (ii) OQ + OR > QR? (iii) OR + OP > RP?

Ans. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

(i) In ΔOPQ, yes, OP + OQ > PQ

(ii) In ΔORQ, yes, OR + OQ > QR (iii) In ΔPOR, yes, OR + OP > PR

Q.3. AM is a median of a triangle ABC.



10

Is AB + BC + CA > 2 AM? (Consider the sides of triangles ΔABM and ΔAMC).

Ans. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

In ΔAMB, AB + BM > AM ----- (i) In ΔAMC AC + CM > AM ----- (ii) Adding (i) and (ii), we have AB + BM + AC + CM > AM + AM or AB + AC + (BM + CM) > 2 AM or AB + AC + BC > 2 AM or AB + BC + CA > 2AM

Q.4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

Ans. Given : A quadrilateral ABCD in which AC and BD are diagonals. To prove : AB + BC + CD + DA > AC + BD

Proof : In ΔABD, AB + DA > BD ----- (i) In ΔBDC,



11

BC + CD > BD ----- (ii) In ΔADC, DA + CD > AC -----(iii) In ΔABC, AB + BC > AC ----- (iv) Adding (i), (ii), (iii) and (iv), we have AB + DA + BC + CD + DA + CD + AB + BC

> BD + AC + AC + BD or 2 [AB + BC + CD + DA] > 2[BD + AC] or AB + BC + CD + DA > BD + AC

Q.5. ABCD is a quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Ans. Given : A quadrilateral ABCD in which AC and BD are

diagonals. To prove : AB + BC + CD + DA < 2(AC + BD) Proof : In ΔAOB, OB + OA > AB ------ (i) In ΔBOC, OB + OC > BC ------ (ii) In ΔCOD, OD + OC > CD ------ (iii) In ΔAOD, OA + OD > DA ------ (iv)



12

Adding (i), (ii), (iii) and (iv), we have OB + OA + OB + OC + OD + OC + OA + OD >

AB + BC + CD + DA or 2OB + 2OD + 2OC + 2OA > AB + BC + CD

+ DA or 2 (OB + OD) + 2 (OC + OA) > AB + BC + CD

+ DA or 2 (BD + AC) > AB + BC + CD + DA or AB + BC + CD + AD < 2 (BD + AC)

Q.6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Ans. Let x cm be the length of the third side. We know that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∴ 12 + 15 > x, so, 27 > x x + 12 > 15, so, x > 3 x + 15 > 12, so, x > – 3 The numbers between 3 and 27 satisfy these. ∴ The length of the third side could be any length between

3 cm and 27 cm.

Exercise 6.5

Q.1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.



13

Ans. In ΔPQR, by Pythagoras Property, (QR)2 = (PR)2 + (PQ)2

⇒ (QR)2 = (24)2 + (10)2

⇒ (QR)2 = 576 + 100 ⇒ (QR)2 = 676

⇒ QR = 676 = 26 cm

Q.2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Ans. In ΔABC, according to Pythagoras Property, (AB)2 = (AC)2 + (BC)2

⇒ (25)2 = 72 + (BC)2 ⇒ 625 – 49 = (BC)2

⇒ 576 = (BC)2

⇒ 576 = BC ⇒ 24 = BC ∴ BC = 24 cm.

Q.3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Ans. Let the distance of the ladder from the wall is a m. Now, in given right triangle, (15)2 = (12)2 + a2 (By Pythagoras property) ⇒ 225 – 144 = a2

⇒ 81 = a2



14

or 81 = a or 9 = a or a = 9 m Hence, the distance of the foot of the ladder from the

wall is 9 m.

Q.4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm (ii) 2 cm, 2 cm, 5 cm (iii) 1.5 cm, 2 cm, 2.5 cm In the case of right-angled triangles, identify the

right angles.

Ans. (i) Let a = 2.5 cm, c = 6.5 cm, b = 6 cm c2 = a2 + b2 (by Pythagoras property) ⇒ (6.5)2 = (2.5)2 + (6)2

⇒ 42.25 = 6.25 + 36 = 42.25 Hence, Pythagoras property holds, we can say it will be

right triangle and right angle will be the angle between side a and b.

(ii) 2 cm, 2 cm, 5 cm Let a = 2 cm, b = 2 cm, c = 5 cm c2 = a2 + b2 (By Pythagoras property) ⇒ (5)2 = (2)2 + (2)2

⇒ 25 ≠ 4 + 4 ⇒ 25 ≠ 8 Hence, Pythagoras property does not hold, we can say it is

not right triangle. (iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm c2 = a2 + b2 (By Pythagoras property) ⇒ (2.5)2 = (1.5)2 + (2)2



15

⇒ 6.25 = 2.25 + 4 ⇒ 6.25 = 6.25 Hence, Pythagoras property holds, we can say it is right

triangle and right angle is formed between sides a and b.

Q.5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Ans. AC = CD (Given) In right angled triangle DBC, DC2 = BC2 + BD2 (by Pythagoras Property) = 52 + 122

= 25 + 144 = 169

⇒ DC = 169 = 13 m ⇒ AC = 13 m ∴ AB = AC + BC ⇒ AB = 13 m + 5 m = 18 m Hence, the original height of the tree = 18 m

Q.6. Angles Q and R of a ΔPQR are 25° and 65°. Write which of the following is true :

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

Ans. In ΔPQR, ∠P + ∠Q + ∠R = 180° (Sum of the angles of a triangle is 180°) ⇒ ∠P + 25° + 65° = 180° ⇒ ∠P + 90° = 180°



16

⇒ ∠P = 180° – 90° = 90° ∴ Triangle PQR is a right angled triangle. Now, in right angled triangle PQR, by Pythagoras Property,

QR2 = PQ2 + PR2

Hence, (ii) is true.

Q.7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Ans. Let ABCD be a rectangle Length (l) = 40 cm (Given) Diagonal = 41 cm (Given) We know that perimeter of rectangle = 2 (l + b) -------- (i) Where, l = length and b = breadth In ΔABC, according to Pythagoras property (AC)2 = (AB)2 + (BC)2 ⇒ (41)2 = (40)2 + (BC)2

⇒ (BC)2 = (41)2 – (40)2

⇒ (BC)2 = 1681 – 1600 ⇒ (BC)2 = 81 So, BC = 9 cm (where b = BC) Thus, b = breadth = 9 cm Putting the values of l and b in equation (i), we have = 2[40 + 9] = 2 × 49 = 98 cm Hence, Perimeter of rectangle is 98 cm.

Q.8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.



17

Ans. Since diagonals of a rhombus bisect each other perpendicularly.

∴ OB = 12

BD = 12

× 16 = 8 cm

and OC = 12

AC = 12

× 30 = 15 cm

Now, in right ΔBOC, BC2 = OB2 + OC2

= 82 + 152 = 64 + 225 = 289

∴ BC = 289 = 17 cm ∴ All sides of rhombus are equal ∴ AB = BC = CD = AD = 17 cm Now, perimeter = 4 × 17 cm = 68 cm

THE TRIANGLE AND ITS PROPERTIES - Testlabz · 6 THE TRIANGLE AND ITS PROPERTIES Exercise 6.1 ......

Documents

Transcript of THE TRIANGLE AND ITS PROPERTIES - Testlabz · 6 THE TRIANGLE AND ITS PROPERTIES Exercise 6.1 ......