ML Aggarwal Class-7 Triangles and its Properties ICSE Maths Solutions Chapter-11. We provide step by step Solutions of Exercise / lesson-11 Triangles and its Properties ICSE Class-7th  ML  Aggarwal Mathematics .

Our Solutions contain all type Questions with Exe-11.1 , Exe-11.2,  Exe-11.3 , Exe-11.4, Exe-11.5 ,Objective Type Questions ( including Mental Maths Multiple Choice Questions ,  , HOTS) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

ML Aggarwal Class-7 Triangles and its Properties ICSE Maths Solutions Chapter-11

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Exe-11.1 ,

Exe-11.2,

Exe-11.3,

Exe-11.4,

Exe-11.5

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

HOTS

ML Aggarwal Class-7 Triangles and its Properties ICSE Maths Solutions Exe-11.1

Question 1.

(i) Name the vertex opposite to side PQ.
(ii) Name the side opposite vertex Q.
(iii) Name the angle opposite to side QR.
(iv) Name the side opposite to ∠R.

In the given figure,
(i) Vertex opposite to side PQ is R.
(ii) The side opposite to the vertex Q is PR.
(iii) Angle opposite to the side QR is ∠R.
(iv) the side opposite to ∠R is PQ.

Question 2.
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles
(You may judge the nature of the angle by observation):

(i) Two sides are equal. So it is an isosceles triangle.
(ii) Three sides are unequal and one angle is 90°,
so it is a right-angled triangle end also it is a scalene triangle also.
(iii) Three sides are equal. So, it is an equilateral triangle.
(iv) Two sides are equal. So, it is an isosceles triangle.
It’s one angle is obtuse, therefore it obtuse angled triangle also.
(v) Three sides are not equal. So, it is a scalene triangle.
It’s one angle is obtuse. So, it is an obtuse angled triangle also.
(vi) Its two sides are equal and one angle is a right angle.
So it is a right-angled isosceles triangle.

Question 3.
In the given ∆PQR, if D is the mid-point of $\bar { QR }$, then
(i) $\bar { PM }$ is ……..
(ii) $\bar { PD }$ is ………
Is QM = MR?

If the given figure, in ∆PQR
D is mid-point of $\bar { QR }$, then
(i) PM is an altitude.
(ii) PQ is the median.
No, QM ≠ MR

Question 4.
Will an altitude always lie in the interior of a triangle? If no, draw a rough sketch to show such a case.

No, it is not necessary, it may lie outside the triangle also.
Here is given a rough sketch of the case AD is the altitude of ∆ABC.
Draw from A to the side CB (produced).

Question 5.
Can you think of a triangle in which two altitudes of the triangle is its sides? If yes, draw a rough sketch to show such a case.

Yes, it is a right-angled triangle.
Here, AB ⊥ BC and BC ⊥ AB

Question 6.
Draw rough sketches for the following:
(i) In ∆ABC, BE is a median of the triangle.
(ii) In ∆PQR, PQ and PR are altitudes of the triangle.
(iii) In ∆XYZ, YL is an altitude in the exterior of the triangle.

(i) In ∆ABC, BE is the median of the triangle.
E is the mid-point of AC. So, BE is the median.

(ii) In ∆PQR, PQ and PR are the altitudes of the triangle.
In ∆PQR, ∠P = 90°
So, PQ and PR are the altitudes.

(iii) In ∆XYZ, YL is the altitude in the exterior of the
triangle YL is altitude on ZX (produced).

Question 7.
Take an equilateral triangle and draw its medians and altitudes and check that the medians and altitudes are the same.

∆ABC is an equilateral triangle.
AD, BE and CF are altitudes of the triangle.
The altitudes of an equilateral triangle
divide the sides into two equal parts.
So, altitudes are also the medians of the triangle.

Ex 11.2, ML Aggarwal Class-7 Triangles and its Properties ICSE Maths Solutions

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

We know that the exterior angle of a triangle
is equal to the sum of its interior opposite angles.
Therefore,
(i) Ext. ∠x = 45° + 65° = 110°

(ii) Ext. ∠x = 55° + 40° = 95°

(iii) Ext. ∠x = 50° + 50° = 100°

Question 2.
Find the value of the unknown interior angle x in each of the following diagrams:

We know that the exterior angle of a triangle
is equal to the sum of its interior opposite angles.
Therefore,
(i) In the given triangle,
Ext. ∠115° = x + 50°
⇒ x = 115° – 50° = 65°
⇒ x = 65°

(ii) In given triangle,
Ext. ∠80° = 30° + x
⇒ x = 80° – 30° = 50°
⇒ x = 50°

(iii) In given triangle,
Ext. ∠70° = x + 36°
⇒ x = 70° – 36° = 34°
⇒ x = 34°

Question 3.
Find the value of x in each of the following diagrams:

We know that the exterior angle of a triangle
is equal to the sum of its interior opposite angles.
Therefore,
(i) In a given triangle,
Ext. ∠105° = 2x + x
⇒ 3x = 105°
⇒ x = 35°
x = 35°

(ii) In given triangle,
Ext. ∠125° = 2x + 3x
⇒ 5x = 125°
⇒ x = 25°
x = 25°

(iii) In given triangle,
∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
⇒ x + 60° + 50° = 180°
⇒ x + 110° = 180°
⇒ x = 180° – 110° = 70°
x = 70°

Question 4.
Find the value of unknown x in each of the following:

(i) In given triangle = Let ∆ABC
∠A + ∠B + ∠C = 180°
⇒ 50° + x + x = 180°
⇒ 2x = 180° – 50° = 130°
Hence, x = 65°

(ii) In the given figure,
Let the name of ∆ be ABC
∠ABC + ∠ABE = 180°
⇒ ∠ABC + 110° = 180°
⇒ ∠ABC = 180° – 110° = 70°
Similarly,
∠ACB + ∠ACD = 180°
⇒ ∠ACB + 120° = 180°
⇒ ∠ACB = 180° – 120° = 60°
Now in ∆ABC
∠BAC + ∠ABC + ∠ACB = 180°
⇒ x + 70° + 60° = 180°
⇒ x + 130° = 180°
⇒ x= 180° – 130° = 50°
⇒ x = 50°

(iii) Let the given triangle be named as ∆ABC,
where ∠C = 90°
In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 2x + x + 90° = 180°
⇒ 3x = 180° – 90° = 90°
⇒ x = 30°

Question 5.
Find the values of x and y in each of the following diagrams:

We know that an exterior angle of a triangle
is equal to the sum of its interior opposite angle.
Therefore,
(i) Let the ∆’s name = ∆ABC
In ∆ABC
Ext. ∠ACD = ∠A + ∠B
120° = x + 50°
⇒ x = 120° – 50° = 70°
But ∠ACD + ∠ABC = 180° (Linear pair)
120° + y = 180°
⇒ y = 180° – 120° = 60°
x = 70°, y = 60°

(ii) In the given figure,
∠ACB = ∠DCE (Vertically opposite angles)
x = 60°
But ∠A + ∠B + ∠ACB = 180°
(Sum of angles of a triangle)
⇒ y + 40° + x = 180°
⇒ y + 40° + 60° = 180°
⇒ y + 100° = 180
⇒ y = 180° – 100° = 80°
Hence, x = 60°, y = 80°

(iii) In the given figure,
∠BAC = ∠EAF (Vertically opposite angles)
y = 90°
In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ y + x + x = 180°
⇒ 90° + 2x = 180°
⇒ 2x = 180° – 90° = 90°
⇒ x = 45°
Hence, x = 45°

Question 6.
Find the values of x and y in each of the following diagrams:

(i) In the given figure,
In ∆ABC,
x = y (Vertically opposite angles)
Similarly,
∠BAC = y, ∠ABC = y, ∠BCA = y
But ∠BAC + ∠ABC + ∠BCA = 180°
(Angles of a triangle)
⇒ y + y + y = 180°
⇒ 3y = 180°
⇒ y = 60°
x = 60°, y = 60°

(ii) In ∆ABC,
∠ABC + ∠ABD = 180°
⇒ x + 125° = 180°
⇒ x = 180°- 125° = 55°
and Ext. ∠ABD = x + y
⇒ 125° = 55° + y
⇒ y = 125° – 55° = 70°
x = 55°, y = 70°

(iii) In ∆ABC,

Ext. ∠ABD = ∠A + ∠B = 50° + 70° = 120°
But ∠ABC + ∠ABD = 180° (Linear pair)
⇒ x + ∠ABD = 180°
⇒ x + 120° = 180°
⇒ x = 180° – 120° = 60°
But in ∆ABD
Ext. ∠ABC = ∠D + ∠DAB
⇒ x = y + 30°
⇒ 60° = y + 30°
⇒ y = 60° – 30° = 30°
x = 60°, y = 30°

Question 7.
In the adjoining figure, find the size of each lettefed angle.

In the given figure,

In ∆ABC
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ x + 80° + 52° = 180°
⇒ x + 132° = 180°
⇒ x = 180° – 132° = 48°
∠DEC + ∠DEA = 180° (Linear pair)
⇒ 143° + z = 180°
⇒ z = 180° – 143° = 37°
Ext. ∠DEC = ∠A + ∠ADE
⇒ 143° = x + y
⇒ 143° = 48° + y
⇒ y = 143° – 48° = 95°
x = 48°, y = 95°, z = 37°

Question 8.
One of the angles of a triangle measures 80° and the other two angles are equal. Find the measure of each of the equal angles.

One angle of an ∆ABC = 80°
Let ∠A = 80° and the other two angles are equal

Let ∠B = ∠C = x
In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 80° + x + x = 180°
⇒ 2x = 180° – 80° = 100°
⇒ x = 50°
∠A = 80°, ∠B = 50°, ∠C = 50°

Question 9.
If one angle of a triangle is 60° and the other two angles are in the ratio 2 : 3, find these angles.

One angle of a triangle = 60°
Other two angles are in the ratio 2 : 3
Sum of other two angles = 180° – 60° = 120°
Let one of other two angles = 2x
Then third angle = 3x
2x + 3x = 120°
⇒ 5x = 120°
⇒ x = 24
Other two angles are 2x = 2 × 24 = 48°
and 3x = 3 × 24 = 72°
Other two angles of the triangle are 48°, 72°

Question 10.
If the angles of a triangle are in the ratio 1 : 2 : 3, find the angles. Classify the triangle in two different ways.

Sum of angles of a triangle = 180°
Ratio in the angles of a triangle = 1 : 2 : 3
Let first angle = x
Second angle = 2x
Third angle = 3x
x + 2x + 3x = 180°
⇒ 6x = 180°
⇒ x = 30°
∴ First angle = 30°
Second angle = 30° × 2 = 60°
and third angle = 30° × 3 = 90°
∵ One angles is 90°
∴ It is a right angled triangle
∵ Sides an different
∴ It is a scalene triangle.

Question 11.
Can a triangle have three angles whose measures are
(i) 65°, 74°, 39°?
(ii) $\frac { 1 }{ 3 }$ right angle, 1 right angle, 60°?

We know that sum of angles of a triangle = 180°
(i) Angles are 65°, 74°, 39°
Sum of angles = 65° + 74° + 39° = 178°
178° ≠ 180°
There three angles can not be of triangle
(ii) $\frac { 1 }{ 3 }$ right angle = $\frac { 1 }{ 2 }$ × 90° = 30°
1 right angle = 90°
Third angle = 60°
Sum of angles = 30° + 90° + 60° = 180°
These angles are of a triangle.

ML Aggarwal Exe- 11.3, Class-7 Triangles and its Properties ICSE Maths Solutions

Question 1.
Find the value of x in each of the following figures:

(i) In ∆ABC,
AB = AC
∠B = ∠C (Angles opposite to equal sides)
x = 50°

(ii) In ∆ABC,
AB = AC
∠B = ∠C (Angles opposite to equal sides)
∠B = ∠C = x
and Ext. ∠BAD = ∠B + ∠C = x + x
110° = ∠B + ∠C
110° = x + x = 2x
x = 55°

(iii) In the given figure,
CA = CB
∠A = ∠ABC = x
∠EBD = 36°
∠ABC = ∠EBD = 36° (Vertically opposite angles)
x = 36°

Question 2.
Find the value of x in each of the following figures:

(i) In ∆ABC,
AC = BC, ∠C = 40°
∠A = ∠B (Angles opposite to equal sides)
∠A = ∠B = x
But A + B + C = 180° (Angles of a triangle)
⇒ x + x + 40° = 180°
⇒ 2x = 180° – 40° = 140°
⇒ x = 70°
x = 70°

(ii) In ∆ABC,
∠B = 45°
AC = BC
∠A = ∠B = 45°
But ∠A + ∠B + ∠C = 180°
⇒ 45° + 45° + x° = 180°
⇒ x + 90°= 180°
⇒ x = 180° – 90° = 90°
x = 90°

(iii) In the given figure, EF || BC
In ∆ABC, Ext. ∠DAF = 50°
AB = BC
∠A = ∠C = x
EF || BC
∠DAF = ∠ABC = 50°
Now in ∆ABC
∠BAC + ∠ABC + ∠BCA = 180°
⇒ x + 50° + x = 180°
⇒ 2x = 180° – 50° = 130°
⇒ x = 65°
x = 65°

Question 3.
Find the values of x and y in each of the following figures:

(i) In the given figure of ∆ABC
AB = AC
∠ABC = ∠ACB = y
But, Ext. ∠ACD + ∠ACB = 120° (Linear pair)
∠ACB = 180° – 120° = 60°
y = 60°
Now in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangle)
⇒ x + y + y = 180°
⇒ x + 60° + 60° = 180°
⇒ x + 120° = 180°
⇒ x = 180° – 120° = 60°
Here, x = 60°, y = 60°

(ii) Here, ∠A = ∠ACB
∠ACB = x (∵ angles opposite to equal sides)
⇒ x + 115° = 180°
⇒ x = 180° – 115° = 65
∠ACB = ∠A = 65°

Now, ∠A + ∠B + ∠ACB = 180° (Sum of ∠s of a ∆)
65° + 65° + ∠y = 180°
130° + ∠y = 180°
∠y = 180° – 130° = 50°
(iii) In ∆ABC,

∠ABC = ∠ACB (∠s opposite to equal sides)
∠ABC = x = ∠ACB
Now, In ∆ABC
48° + x + x = 180° (Sum of ∠s of a ∆)
⇒ 2x = 180° – 48°
⇒ 2x = 132°
⇒ x = 66°
In ∆ACD,
∠CAD = ∠CDA (∠s opposite to equal sides)
Now, x + ∠ACD = 180° (Linear pair ∠s)
66° + ∠ACD = 180°
∠ACD = 180° – 66° = 114°
Now, in ∆ACD
y + y + 114°= 180° (Sum of ∠s of a ∆)
⇒ 2y + 114° = 180°
⇒ 2y = 180° – 114°
⇒ 2y = 66°
⇒ y = 33°
Hence, x = 66° and y = 33°

Question 4.
Calculate the size of each lettered angle in the following figures:

(i) In ∆ABC,
AC = BC (Given)
∠ABC = ∠BAC (∠s opposite to equal sides)
∠ABC = 32°
⇒ x = 32°

Now, y = 32° + 32° = 64°
(∵ Exterior angle = Sum of two opposite interior ∠s)
In ∆BCD,
BC = BD
∠BDC = ∠BCD (∠s opposite to equal sides)
∠BDC = y = ∠BCD
∠BDC = 64° = ∠BCD
Now, z + 64° + 64° = 180° (Sum of ∠s of a ∆)
⇒ z + 128°= 180°
⇒ z = 180° – 128° = 52°
Hence, x = 32°, y = 64°, z = 52°
(iii) In ∆ABC,
AC = BC (Given)
∠ABC = ∠BAC (∠s opposite to equal sides)
∠ABC = 35° = ∠BAC

Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle = Sum)
= 35° + 35° = 70°.
In ∆ACD,
∠ADC = ∠ACD (∠s opposite to equal sides)
Now, ∠x = ∠ABD + ∠ADB
(Exterior angle = Sum of two opposite interior ∠s)
= 35° + 70° = 105°
Hence, ∠x = 105°

Question 5.
If the angles of a triangle are in the ratio 1 : 2 : 1, find all the angles of the triangle. Classify the triangle in two different ways.

Ratio in the angles of a triangle are 1 : 2 : 1
Sum of angles of a triangle = 180°
Let first angle = x
Then second = 2x
and third angle x
x + 2x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
Angles are 45°, 45° × 2 = 90° and 45°
Two angles are equal
Their opposite sides are also equal
It is an isosceles triangle
It’s one angle is 90°
It is a right-angled triangle.

Question 6.
In an isosceles triangle, a base angle is four times its vertical angle. Find all the angles of the triangle.