## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Check Your Progress

Question 1.

Find the value of x in each of the following diagrams:

Solution:

(i) In the given figure,

Ext. angle of triangle = Sum of its interior opposite angles.

⇒ 115° = 2x – 50 + x

⇒ 3x – 50 = 115°

⇒ 3x = 115° + 5° = 120°

⇒ x = 40°

x = 40°

(ii) In the given figure,

∠1 = 45° (Alternate angles)

Now in triangle,

80° + x + ∠1 = 180° (Sum of angles of a triangle)

⇒ 80°+ x + 45° = 180°

⇒ x + 125°= 180°

⇒ x = 180° – 125° = 55°

(iii) In ∆ABC

∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)

⇒ 50° + 72° + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180°- 122° = 58°

Now in ∆ACD

Ext. ∠ACB = ∠CAB + ∠CDA

⇒ 58° = x + 45°

⇒ x = 58° – 45° = 13°

Question 2.

In the given figure, ∠B = 70° and ∠A = 50°. If the bisector of ∠C meets AB in D, then find ∠ADC.

Solution:

In the given figure,

∠A = 50°, ∠B = 70°, ∠ADC = x

CD is the bisector of ∠C

In ∆ABC,

∠A + ∠B + ∠ACB = 180° (Angles of a triangle)

⇒ 50° + 70° + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180°- 120° = 60°

But CD is the bisector of ∠C

∠DCB = \(\frac { 60 }{ 2 }\) = 30°

Now in ∆BCD,

Ext. ∠ADC = ∠B + ∠DCB (Interior opposite angles)

= 70° + 30° = 100°

Question 3.

Find the values of x and y in each of the following figures:

Solution:

(i) In the given figure, AC = AE

Ext. ∠DAC = ∠B + ∠C (Interior opposite angles)

⇒ x = 30° + 55° = 85°

In ∆AEC,

AC = AE

∠AEC = ∠ACE = 55°

In ∆ABE,

Now, Ext. ∠AEC = y + 30°

⇒ 55° = y + 30°

⇒ y = 55° – 30° = 25°

x = 85° and y = 25°

(ii) In the given figure,

PQ || EF

In ∆ABC,

AC = BC

∠PAB = 66°

∠ABC = ∠PAB (Alternate angles)

x = 66°

AC = BC

∠BAC = ∠ABC = 66°

But ∠ABC + ∠ACB + ∠BAC = 180° (Angles of a triangle)

⇒ 66°+ y + 66° = 180°

⇒ 132° + y = 180°

⇒ y = 180° – 132° = 48°

x = y = 48°

(iii) In the given figure,

BD = AD = AC

∠B = 35°

In ∆ABD

AD = BD

∠BAD = ∠ABD = 35°

Now, Ext. ∠ADC = B + BAD

y = 35° + 35° = 70°

In ∆ADC,

AD = AC

∠ACD = ∠ADC = y = 70°

and Ext. ∠EAC = ∠ABC + ∠ACB = 35° + 70° = 105°

Hence, x = 105°, y = 70°

Question 4.

If the two acute angles of a right-angled triangle are in the ratio 7 : 8, find these angles.

Solution:

In a right-angled triangle.

Sum of two acute angles = 90°

Ratio in two angles = 7 : 8

First angle = \(\frac { 90 }{ 7+8 }\) × 7

= \(\frac { 90 }{ 15 }\) × 7 = 42°

and second angle = \(\frac { 90 }{ 15 }\) × 8 = 48°

Question 5.

If the angles of a triangle are (3x)°, (2x – 7)° and (4x – 11)°, then find the value of x.

Solution:

Angles of a triangle are (3x)°, (2x – 7)° and (4x – 11)°

But sum of three angles of a triangle = 180°

3x + 2x – 7 + 4x – 11° = 180°

⇒ 9x – 18° = 180°

⇒ 9x = 180° + 18° = 198°

⇒ x = 22°

Question 6.

In an isosceles triangle, the vertical angle is 15° greater than each of its base angles. Find all the angles of the triangle.

Solution:

In an isosceles triangle,

Vertical angle = 15° greater than each base angles

Let each base angle = x

Then vertical angle = x + 15°

Now sum of angles of a triangle = 180°

x + 15° + x + x = 180°

⇒ 3x + 15° = 180°

⇒ 3x = 180° – 15° = 165°

⇒ x = 55°

Each base angle = 55°

and vertical angle = 55° + 15° = 70°

Question 7.

Can a triangle have three sides whose lengths are

(i) 4.5 cm, 3.8 cm, 7.2 cm?

(ii) 3.2 cm, 5.3 cm, 9.4 cm?

Solution:

(?) Sides are 4.5 cm, 3.8 cm, 7.2 cm

Sum of two sides = 4.5 + 3.8 = 8.3 cm

8.3 > 7.2 cm

The triangle can have there sides.

(ii) Sides are 3.2 cm, 5.3 cm, 9.4 cm

Sum of sides 3.2 and 5.3 cm = 3.2 + 5.3 = 8.5 cm

8.5 cm < 9.4 cm

The triangle does not have there sides.

Question 8.

If the lengths of two sides of a triangle are 5 cm and 12 cm, then what can be the length of the third side?

Solution:

Length of two sides of a triangle is 5 cm and 12 cm

Sum of there two sides = 5 + 12 = 17 cm

and difference = 12 – 5 = 7 cm

The third side will be greater than 7 cm but less than 17 cm.

Question 9.

In the given figure, all measurements are in centimeters. If AD is perpendicular to BC, find the length of AB.

Solution:

In the given figure,

ABC is a triangle in which AC = 25 cm, BC = 28 cm

AD ⊥ BC and AD = 15 cm

To find the length of AB

In right ∆ADC

AC^{2} = AD^{2} + DC^{2} (Pythagoras Theorem)

⇒ (25)^{2} = 15^{2} + DC^{2}

⇒ 625 = 225 + DC^{2}

⇒ DC^{2} = 625 – 225 = 400 = (20)^{2}

⇒ DC = 20 cm

But BC = 28 cm

BD = 28 – 20 = 8 cm

Now in right ∆ADB

AB^{2} = AD^{2} + BD^{2}

= 15^{2} + 8^{2} = 225 + 64 = 289 = (17)^{2}

⇒ AB = 17 cm

Question 10.

In the given figure, AB and CD are two vertical poles of height 19 m and 11 m respectively. If the shortest distance between their tops is 17 m, find how far apart they are?

Solution:

In the given figure,

Pole AB = 19 m, Pole CD = 11 m

Distance between their tops AC = 17 cm

Draw EC || BD, then

Let BD = CE = x

EB = CD = 11 m

AE = AB – EB = 19 – 11 = 8m

Now in right ∆AEC,

AC^{2} = AE^{2} + EC^{2}

⇒ (17)^{2 }= (8)^{2 }+ x^{2}

⇒ 289 = 64 + x^{2}

⇒ x^{2} = 289 – 64 = 225 = (15)^{2}

⇒ x = 15

BD = EC = x = 15m