Table of Contents
Exercise 15.1
Question: 1
(i) The side opposite to ∠B
(ii) The angle opposite to side AB
(iii) The vertex opposite to side BC
(iv) The side opposite to vertex B.
Solution:
(i) AC
(ii) ∠B
(iii) A
(iv) AC
Question: 2
Take three collinear points A, B and C on a page of your note book. Join AB. BC and CA. Is the figure a triangle? If not, why?
Solution:
No, the figure is not a triangle. By definition a triangle is a plane figure formed by three non-parallel line segments
Question: 3
Distinguish between a triangle and its triangular region.
Solution:
A triangle is a plane figure formed by three non-parallel line segments, whereas, its triangular region includes the interior of the triangle along with the triangle itself.
Question: 4
D is a point on side BC of a ∆CAD is joined. Name all the triangles that you can observe in the figure. How many are they?
Solution:
We can observe the following three triangles in the given figure
(i) ∆ABC
(ii) ∆ACD
(iii) ∆ADB
Question: 5
A, B. C and D are four points, and no three points are collinear. AC and BD intersed at O. There are eight triangles that you can observe. Name all the triangles
Solution:
(i) ∆ABC
(ii) ∆ABD
(iii) ∆ABO
(iv) ∆BCD
(v) ∆DCO
(vi) ∆AOD
(vii) ∆ACD
(viii) ∆BCD
Question: 6
What is the difference between a triangle and triangular region?
Solution:
Plane of figure formed by three non-parallel line segments is called a triangle where as triangular region is the interior of triangle ABC together with the triangle ABC itself is called the triangular region ABC
Question: 7
Explain the following terms:
(i) Triangle
(a) Parts or elements of a triangle
(iii) Scalene triangle
(iv) Isosceles triangle
(v) Equilateral triangle
(vi) Acute triangle
(vii) Right triangle
(viii) Obtuse triangle
(ix) Interior of a triangle
(x) Exterior of a triangle
Solution:
(i) A triangle is a plane figure formed by three non-parallel line segments.
(ii) The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.
(iii) A scalene triangle is a triangle in which no two sides are equal.
(iv) An isosceles triangle is a triangle in which two sides are equal. Isosceles triangle
(v) An equilateral triangle is a triangle in which all three sides are equal. Equilateral triangle
(vi) An acute triangle is a triangle in which all the angles are acute (less than 90°).
(vii) A right angled triangle is a triangle in which one angle is right angled, i.e. 90°.
(viii) An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).
(ix) The interior of a triangle is made up of all such points that are enclosed within the triangle.
(x) The exterior of a triangle is made up of all such points that are not enclosed within the triangle.
Question: 8
In Figure, the length (in cm) of each side has been indicated along the side. State for each triangle angle whether it is scalene, isosceles or equilateral:
Solution:
(i) This triangle is a scalene triangle because no two sides are equal.
(ii) This triangle is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.
(iii) This triangle is an equilateral triangle because all its three sides are equal.
(iv) This triangle is a scalene triangle because no two sides are equal.
(v) This triangle is an isosceles triangle because two of its sides are equal.
Question: 9
There are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.
Solution:
(i) This is a right triangle because one of its angles is 90°.
(ii) This is an obtuse triangle because one of its angles is 120°, which is greater than 90°.
(iii) This is an acute triangle because all its angles are acute angles (less than 90°).
(iv) This is a right triangle because one of its angles is 90°.
(v) This is an obtuse triangle because one of its angles is 110°, which is greater than 90°.
Question: 10
Fill in the blanks with the correct word/symbol to make it a true statement:
(i) A triangle has _____ sides.
(ii) A triangle has ______ vertices.
(iii) A triangle has _____ angles.
(iv) A triangle has ________ parts.
(v) A triangle whose no two sides are equal is known as ________
(v0 A triangle whose two sides are equal is known as ________
(vii) A triangle whose all the sides are equal is known as ________
(viii) A triangle whose one angle is a right angle is known as _________
(ix) A triangle whose all the angles are of measure less than 90′ is known as _________
(x) A triangle whose one angle is more than 90′ is known as _________
Solution:
(i) three
(ii) three
(iii) three
(iv) six (three sides + three angles)
(v) a scalene triangle
(vi) an isosceles triangle
(vii) an equilateral triangle
(viii) a right triangle
(ix) an acute triangle
(x) an obtuse triangle
Question: 11
In each of the following, state if the statement is true (T) or false (F):
(i) A triangle has three sides.
(ii) A triangle may have four vertices.
(iii) Any three line-segments make up a triangle.
(iv) The interior of a triangle includes its vertices.
(v) The triangular region includes the vertices of the corresponding triangle.
(vi) The vertices of a triangle are three collinear points.
(vii) An equilateral triangle is isosceles also.
(viii) Every right triangle is scalene.
(ix) Each acute triangle is equilateral.
(x) No isosceles triangle is obtuse.
Solution:
(i) True.
(ii) False. A triangle has three vertices.
(iii) False. Any three non-parallel line segments can make up a triangle.
(iv) False. The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.
(v) True. The triangular region includes the interior region and the triangle itself.
(vi) False. The vertices of a triangle are three non-collinear points.
(vii) True. In an equilateral triangle, any two sides are equal.
(viii) False. A right triangle can also be an isosceles triangle.
(ix) False. Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.
(x) False. An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle
Exercise 15.2
Question: 1
Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.
Solution:
Let the third angle be x
Sum of all the angles of a triangle = 180°
105° + 30° + x = 180°
135° + x = 180°
x = 180° – 135°
x = 45°
Therefore the third angle is 45°
Question: 2
One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?
Solution:
Let the second and third angle be x
Sum of all the angles of a triangle = 180°
130° + x + x = 180°
130° + 2x = 180°
2x = 180°– 130°
2x = 50°
x = 50/2
x = 25°
Therefore the two other angles are 25° each
Question: 3
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
Solution:
Let the each angle be x
Sum of all the angles of a triangle =180°
x + x + x = 180°
3x = 180°
x = 180/3
x = 60°
Therefore angle is 60° each
Question: 4
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Solution:
If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’
Sum of all the angles of a triangle=180°
x + 2x + 3x = 180°
6x = 180°
x = 180/6
x = 30°
2x = 30° × 2 = 60°
3x = 30° × 3 = 90°
Therefore the first angle is 30°, second angle is 60° and third angle is 90°
Question: 5
The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.
Solution:
Question: 6
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.
Solution:
Let the first angle be x
Second angle be x + 10°
Third angle be x + 10° + 10°
Sum of all the angles of a triangle = 180°
x + x + 10° + x + 10° +10° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
First angle is 50°
Second angle x + 10° = 50 + 10 = 60°
Third angle x + 10° +10° = 50 + 10 + 10 = 70°
Question: 7
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle
Solution:
Let the first and second angle be x
The third angle is greater than the first and second by 30° = x + 30°
The first and the second angles are equal
Sum of all the angles of a triangle = 180°
x + x + x + 30° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
Third angle = x + 30° = 50° + 30° = 80°
The first and the second angle is 50° and the third angle is 80°
Question: 8
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
One angle of a triangle is equal to the sum of the other two
x = y + z
Let the measure of angles be x, y, z
x + y + z = 180°
x + x = 180°
2x = 180°
x = 180/2
x = 90°
If one angle is 90° then the given triangle is a right angled triangle
Question: 9
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Each angle of a triangle is less than the sum of the other two
Measure of angles be x, y and z
x > y + z
y < x + z
z < x + y
Therefore triangle is an acute triangle
Question: 10
In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°
Solution:
(i) 63°, 37°, 80° = 180°
Angles form a triangle
(ii) 45°, 61°, 73° is not equal to 180°
Therefore not a triangle
(iii) 59°, 72°, 61° is not equal to 180°
Therefore not a triangle
(iv) 45°, 45°, 90° = 180°
Angles form a triangle
(v) 30°, 20°, 125° is not equal to 180°
Therefore not a triangle
Question: 11
The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle
Solution:
Given that
Angles of a triangle are in the ratio: 3: 4: 5
Measure of the angles be 3x, 4x, 5x
Sum of the angles of a triangle =180°
3x + 4x + 5x = 180°
12x = 180°
x = 180/12
x = 15°
Smallest angle = 3x
=3 × 15°
= 45°
Question: 12
Two acute angles of a right triangle are equal. Find the two angles.
Solution:
Given acute angles of a right angled triangle are equal
Right triangle: whose one of the angle is a right angle
Measured angle be x, x, 90°
x + x + 180°= 180°
2x = 90°
x = 90/2
x = 45°
The two angles are 45° and 45°
Question: 13
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
Solution:
Angle of a triangle is greater than the sum of the other two
Measure of the angles be x, y, z
x > y + z or
y > x + z or
z > x + y
x or y or z > 90° which is obtuse
Therefore triangle is an obtuse angle
Question: 14
AC, AD and AE are joined. Find
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
Solution:
We know that sum of the angles of a triangle is 180°
Therefore in ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° — (i)
In ∆ACD, we have
∠DAC + ∠ACD + ∠CDA = 180° — (ii)
In ∆ADE, we have
∠EAD + ∠ADE + ∠DEA =180° — (iii)
In ∆AEF, we have
∠FAE + ∠AEF + ∠EFA = 180° — (iv)
Adding (i), (ii), (iii), (iv) we get
∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720°
Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°
Question: 15
Find x, y, z (whichever is required) from the figures given below:
Solution:
(i) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (corresponding angles)
x = 40°
∠AED = ∠ACB (corresponding angles)
y = 30°
We know that the sum of all the three angles of a triangle is equal to 180°
x + y + z = 180° (Angles of ∆ADE)
Which means: 40° + 30° + z = 180°
z = 180° – 70°
z = 110°
Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.
(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.
(∆ADC is a right triangle)
We also know that the sum of all the angles of a triangle is equal to 180°.
Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)
135° + y = 180°
y = 180° – 135°.
y = 45°.
We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.
(Sum of the angles of ∆ABC)
40° + y + (x + 45°) = 180°
40° + 45° + x + 45° = 180° (y = 45°)
x = 180° –130°
x = 50°
Therefore, we can say that the required angles are 45° and 50°.
(iii) We know that the sum of all the angles of a triangle is equal to 180°.
Therefore, for △ABD:
∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)
50° + x + 50° = 180°
100° + x = 180°
x = 180° – 100°
x = 80°
For ∆ABC:
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° – 130°
z = 50°
Using the same argument for ∆ADC:
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)
y +z + 30° =180°
y + 50° + 30° = 180° (z = 50°)
y = 180° – 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°.
(iv) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (Corresponding angles)
y = 50°
Also, ∠AED = ∠ACB (Corresponding angles)
z = 40°
We know that the sum of all the three angles of a triangle is equal to 180°.
Which means: x + 50° + 40° = 180° (Angles of ∆ADE)
x = 180° – 90°
x = 90°
Therefore, we can conclude that the required angles are 50°, 40° and 90°.
Question: 16
If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles
Solution:
We know that one of the angles of the given triangle is 60°. (Given)
We also know that the other two angles of the triangle are in the ratio 1: 2.
Let one of the other two angles be x.
Therefore, the second one will be 2x.
We know that the sum of all the three angles of a triangle is equal to 180°.
60° + x + 2x = 180°
3x = 180° – 60°
3x = 120°
x = 120/3
x = 40°
2x = 2 × 40
2x = 80°
Hence, we can conclude that the required angles are 40° and 80°.
Question: 17
It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.
Solution:
We know that one of the angles of the given triangle is 100°.
We also know that the other two angles are in the ratio 2: 3.
Let one of the other two angles be 2x.
Therefore, the second angle will be 3x.
We know that the sum of all three angles of a triangle is 180°.
100° + 2x + 3x = 180°
5x = 180° – 100°
5x = 80°
x = 80/5
2x = 2 ×16
2x = 32°
3x = 3×16
3x = 48°
Thus, the required angles are 32° and 48°.
Question: 18
In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.
Solution:
We know that for the given triangle, 3∠A = 6∠C
∠A = 2∠C — (i)
We also know that for the same triangle, 4∠B = 6∠C
∠B = (6/4)∠C — (ii)
We know that the sum of all three angles of a triangle is 180°.
Therefore, we can say that:
∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)
On putting the values of ∠A and ∠B in equation (iii), we get:
2∠C + (6/4)∠C +∠C = 180°
(18/4) ∠C = 180°
∠C = 40°
From equation (i), we have:
∠A = 2∠C = 2 × 40
∠A = 80°
From equation (ii), we have:
∠B = (6/4)∠C = (6/4) × 40°
∠B = 60°
∠A = 80°, ∠B = 60°, ∠C = 40°
Therefore, the three angles of the given triangle are 80°, 60°, and 40°.
Question: 19
Is it possible to have a triangle, in which
(i) Two of the angles are right?
(ii) Two of the angles are obtuse?
(iii) Two of the angles are acute?
(iv) Each angle is less than 60°?
(v) Each angle is greater than 60°?
(vi) Each angle is equal to 60°
Solution:
Give reasons in support of your answer in each case.
(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C.
As per the given information,
∠A < 60° … (i)
∠B< 60° … (ii)
∠C < 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C < 60°+ 60°+ 60°
∠A + ∠B + ∠C < 180°
We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.
(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A > 60° … (i)
∠B > 60° … (ii)
∠C > 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C > 60°+ 60°+ 60°
∠A + ∠B + ∠C > 180°
We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.
(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A = 60° … (i)
∠B = 60° …(ii)
∠C = 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C = 60°+ 60°+ 60°
∠A + ∠B + ∠C =180°
We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.
Question: 20
In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B
Solution:
Consider ∆ABD
∠BAD = 100/2 (AD bisects ∠A)
∠BAD = 50°
∠ADB = 90° (AD perpendicular to BC)
We know that the sum of all three angles of a triangle is 180°.
Thus,
∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)
Or,
∠ABD + 50° + 90° = 180°
∠ABD =180° – 140°
∠ABD = 40°
Question: 21
In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC
Solution:
We know that the sum of all three angles of a triangle is equal to 180°.
Therefore, for the given △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
50° + 70° + ∠C = 180°
∠C= 180° –120°
∠C = 60°
∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )
∠ACD = ∠BCD = 60/2= 30°
Using the same logic for the given ∆ACD, we can say that:
∠DAC + ∠ACD + ∠ADC = 180°
50° + 30° + ∠ADC = 180°
∠ADC = 180°– 80°
∠ADC = 100°
If we use the same logic for the given ∆BCD, we can say that
∠DBC + ∠BCD + ∠BDC = 180°
70° + 30° + ∠BDC = 180°
∠BDC = 180° – 100°
∠BDC = 80°
Thus,
For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°
∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°
Question: 22
In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C, meet at O. Find
(i) ∠C
(ii) ∠BOC
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
60° + 80° + ∠C= 180°.
∠C = 180° – 140°
∠C = 140°.
For △OBC,
∠OBC = ∠B2 = 80/2 (OB bisects ∠B)
∠OBC = 40°
∠OCB =∠C2 = 40/2 (OC bisects ∠C)
∠OCB = 20°
If we apply the above logic to this triangle, we can say that:
∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)
20° + 40° + ∠BOC = 180°
∠BOC = 180° – 60°
∠BOC = 120°
Question: 23
The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠A + ∠B + ∠C = 180°
∠A + 90° + ∠C = 180°
∠A + ∠C = 180° – 90°
∠A + ∠C = 90°
For ∆OAC:
∠OAC = ∠A2 (OA bisects LA)
∠OCA = ∠C2 (OC bisects LC)
On applying the above logic to △OAC, we get:
∠AOC + ∠OAC + ∠OCA = 180° (Sum of angles of ∆AOC)
∠AOC + ∠A2 + ∠C2 = 180°
∠AOC + ∠A + ∠C2 = 180°
∠AOC + 90/2 = 180°
∠AOC = 180° – 45°
∠AOC = 135°
Question: 24
In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.
Solution:
In the given triangle,
∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)
We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we can say that:
Question: 25
In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for ∆ABC, we can say that:
Question: 26
Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.
Solution:
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
∠CAB = ∠DBA (Alternate interior angles)
∠DBA = 35°
We also know that the sum of all three angles of a triangle is 180°.
Hence, for △OBD, we can say that:
∠DBO + ∠ODB + ∠BOD = 180°
35° + 55° + ∠BOD = 180° (∠DBO = ∠DBA and ∠ODB = ∠CDB)
∠BOD = 180° – 90°
∠BOD = 90°
Question: 27
In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P
Solution:
In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.
∠QCA = ∠CQP (Alternate interior angles)
Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,
∠ABC = ∠PRQ (alternate interior angles).
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠ABC + ∠ACB + ∠BAC = 180°
∠ABC + ∠ACB + 90° = 180° (Right angled at A)
∠ABC + ∠ACB = 90°
Using the same logic for △PQR, we can say that:
∠PQR + ∠PRQ + ∠QPR = 180°
∠ABC + ∠ACB + ∠QPR = 180° (∠ABC = ∠PRQ and ∠QCA = ∠CQP)
Or,
90°+ ∠QPR =180° (∠ABC+ ∠ACB = 90°)
∠QPR = 90°
Exercise 15.3
Question: 1
∠CBX is an exterior angle of ∆ABC at B. Name
(i) the interior adjacent angle
(ii) the interior opposite angles to exterior ∠CBX
Also, name the interior opposite angles to an exterior angle at A.
Solution:
(i) ∠ABC
(ii) ∠BAC and ∠ACB
Also the interior angles opposite to exterior are ∠ABC and ∠ACB
Question: 2
In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?
Solution:
In ∆ABC, ∠A = 50° and ∠B = 55°
Because of the angle sum property of the triangle, we can say that
∠A + ∠B + ∠C = 180°
50°+ 55°+ ∠C = 180°
Or
∠C = 75°
∠ACB = 75°
∠ACX = 180°− ∠ACB = 180°−75° = 105°
Question: 3
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.
Solution:
We know that the sum of interior opposite angles is equal to the exterior angle.
Hence, for the given triangle, we can say that:
∠ABC+ ∠BAC = ∠BCO
55° + ∠BAC = 95°
Or,
∠BAC= 95°– 95°
= ∠BAC = 40°
We also know that the sum of all angles of a triangle is 180°.
Hence, for the given △ABC, we can say that:
∠ABC + ∠BAC + ∠BCA = 180°
55° + 40° + ∠BCA = 180°
Or,
∠BCA = 180° –95°
= ∠BCA = 85°
Question: 4
One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?
Solution:
Let us assume that A and B are the two interior opposite angles.
We know that ∠A is equal to ∠B.
We also know that the sum of interior opposite angles is equal to the exterior angle.
Hence, we can say that:
∠A + ∠B = 80°
Or,
∠A +∠A = 80° (∠A= ∠B)
2∠A = 80°
∠A = 40/2 =40°
∠A= ∠B = 40°
Thus, each of the required angles is of 40°.
Question: 5
The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In the given figure, ∠ABE and ∠ABC form a linear pair.
∠ABE + ∠ABC =180°
∠ABC = 180°– 136°
∠ABC = 44°
We can also see that ∠ACD and ∠ACB form a linear pair.
∠ACD + ∠ACB = 180°
∠AUB = 180°– 104°
∠ACB = 76°
We know that the sum of interior opposite angles is equal to the exterior angle.
Therefore, we can say that:
∠BAC + ∠ABC = 104°
∠BAC = 104°– 44° = 60°
Thus,
∠ACE = 76° and ∠BAC = 60°
Question: 6
In Fig, the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC
Solution:
In a ∆ABC, ∠BAC and ∠EAF are vertically opposite angles.
Hence, we can say that:
∠BAC = ∠EAF = 45°
Considering the exterior angle property, we can say that:
∠BAC + ∠ABC = ∠ACD = 105°
∠ABC = 105°– 45° = 60°
Because of the angle sum property of the triangle, we can say that:
∠ABC + ∠ACS +∠BAC = 180°
∠ACB = 75°
Therefore, the angles are 45°, 65° and 75°.
Question: 7
In Figure, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.
Solution:
In the given triangle, the angles are in the ratio 3: 2: 1.
Let the angles of the triangle be 3x, 2x and x.
Because of the angle sum property of the triangle, we can say that:
3x + 2x + x = 180°
6x = 180°
Or,
x = 30° … (i)
Also, ∠ACB + ∠ACE + ∠ECD = 180°
x + 90° + ∠ECD = 180° (∠ACE = 90°)
∠ECD = 60° [From (i)]
Question: 8
A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?
Solution:
Here,
Internal angle at A + External angle at A = 180°
Internal angle at A + 103° =180°
Internal angle at A = 77°
Internal angle at B + External angle at B = 180°
Internal angle at B + 74° = 180°
Internal angle at B = 106°
Sum of internal angles at A and B = 77° + 106° =183°
It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.
Question: 9
In Figure, AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆FCB, we can say that:
∠FCB + ∠CBF + ∠BFC = 180°
50° + ∠CBF + 90°= 180°
Or,
∠CBF = 180° – 50°– 90° = 40° … (i)
Using the above rule for ∆ABD, we can say that:
∠ABD + ∠BDA + ∠BAD = 180°
∠BAD = 180° – 90°– 40° = 50° [from (i)]
Question: 10
In Figure, measures of some angles are indicated. Find the value of x.
Solution:
Here,
∠AED + 120° = 180° (Linear pair)
∠AED = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for △ADE, we can say that:
∠ADE + ∠AED + ∠DAE = 180°
60°+ ∠ADE + 30° =180°
Or,
∠ADE = 180°– 60°– 30° = 90°
From the given figure, we can also say that:
∠FDC + 90° = 180° (Linear pair)
∠FDC = 180°– 90° = 90°
Using the above rule for △CDF, we can say that:
∠CDF + ∠DCF + ∠DFC = 180°
90° + ∠DCF + 60° =180°
∠DCF = 180°−60°− 90°= 30°
Also,
∠DCF + x = 180° (Linear pair)
30° + x = 180°
Or,
x = 180°– 30° = 150°
Question: 11
In Figure, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130°, find
(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC
Solution:
(i) Here,
∠BAF + ∠FAD = 180° (Linear pair)
∠FAD = 180°- ∠BAF = 180°– 90° = 90°
Also,
∠AFE = ∠ADF + ∠FAD (Exterior angle property)
∠ADF + 90° = 130°
∠ADF = 130°− 90° = 40°
(ii) We know that the sum of all the angles of a triangle is 180°.
Therefore, for ∆BDE, we can say that:
∠BDE + ∠BED + ∠DBE = 180°.
∠DBE = 180°– ∠BDE ∠BED = 180°− 90°− 40°= 50° — (i)
Also,
∠FAD = ∠ABC + ∠ACB (Exterior angle property)
90° = 50° + ∠ACB
Or,
∠ACB = 90°– 50° = 40°
(iii) ∠ABC = ∠DBE = 50° [From (i)]
Question: 12
ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°. Find ∠ACB.
Solution:
Here,
∠CAX = ∠DAX (AX bisects ∠CAD)
∠CAX =70°
∠CAX +∠DAX + ∠CAB =180°
70°+ 70° + ∠CAB =180°
∠CAB =180° –140°
∠CAB =40°
∠ACB + ∠CBA + ∠CAB = 180° (Sum of the angles of ∆ABC)
∠ACB + ∠ACB+ 40° = 180° (∠C = ∠B)
2∠ACB = 180°– 40°
∠ACB = 140/2
∠ACB = 70°
Question: 13
The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30° and ∠ACD = 115°, find ∠ALC
Solution:
∠ACD and ∠ACL make a linear pair.
∠ACD+ ∠ACB = 180°
115° + ∠ACB =180°
∠ACB = 180°– 115°
∠ACB = 65°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
30° + ∠BAC + 65° = 180°
Or,
∠BAC = 85°
∠LAC = ∠BAC/2 = 85/2
Using the above rule for ∆ALC, we can say that:
∠ALC + ∠LAC + ∠ACL = 180°
Question: 14
D is a point on the side BC of ∆ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find
(i) ∠AQD
(ii) ∠APD
Solution:
∠ABD and ∠QBD form a linear pair.
∠ABC + ∠QBC =180°
60° + ∠QBC = 180°
∠QBC = 120°
∠PDC = ∠BDQ (Vertically opposite angles)
∠BDQ = 75°
In ∆QBD:
∠QBD + ∠QDB + ∠BDQ = 180° (Sum of angles of ∆QBD)
120°+ 15° + ∠BQD = 180°
∠BQD = 180°– 135°
∠BQD = 45°
∠AQD = ∠BQD = 45°
In ∆AQP:
∠QAP + ∠AQP + ∠APQ = 180° (Sum of angles of ∆AQP)
80° + 45° + ∠APQ = 180°
∠APQ = 55°
∠APD = ∠APQ
Question: 15
Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y
Solution:
The interior angles of a triangle are the three angle elements inside the triangle.
The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Using these definitions, we will obtain the values of x and y.
(I) From the given figure, we can see that:
∠ACB + x = 180° (Linear pair)
75°+ x = 180°
Or,
x = 105°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠BAC+ ∠ABC +∠ACB = 180°
40°+ y +75° = 180°
Or,
y = 65°
(ii) x + 80°= 180° (Linear pair)
= x = 100°
In ∆ABC:
x+ y+ 30° = 180° (Angle sum property)
100° + 30° + y = 180°
= y = 50°
(iii) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ACD, we can say that:
30° + 100° + y = 180°
Or,
y = 50°
∠ACB + 100° = 180°
∠ACB = 80° … (i)
Using the above rule for ∆ACD, we can say that:
x + 45° + 80° = 180°
= x = 55°
(iv) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆DBC, we can say that:
30° + 50° + ∠DBC = 180°
∠DBC = 100°
x + ∠DBC = 180° (Linear pair)
x = 80°
And,
y = 30° + 80° = 110° (Exterior angle property)
Question: 16
Compute the value of x in each of the following figures
Solution:
(i) From the given figure, we can say that:
∠ACD + ∠ACB = 180° (Linear pair)
Or,
∠ACB = 180°– 112° = 68°
We can also say that:
∠BAE + ∠BAC = 180° (Linear pair)
Or,
∠BAC = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠BAC + ∠ACB = 180°
x = 180°– 60°– 68° = 52°
= x = 52°
(ii) From the given figure, we can say that:
∠ABC + 120° = 180° (Linear pair)
∠ABC = 60°
We can also say that:
∠ACB+ 110° = 180° (Linear pair)
∠ACB = 70°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠ABC + ∠ACB = 180°
x = 50°
(iii) From the given figure, we can see that:
∠BAD = ∠ADC = 52° (Alternate angles)
We know that the sum of all the angles of a triangle is 180°.
Therefore, for ∆DEC:
x + 40°+ 52° = 180°
= x = 88°
(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.
Thus,
35° + 45° + 50° + reflex ∠ADC = 360°
Or,
reflex ∠ADC = 230°
230° + x = 360° (A complete angle)
= x = 130°
Exercise 15.4
Question: 1
In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:
(i) 5, 7, 9
(ii) 2, 10.15
(iii) 3, 4, 5
(iv) 2, 5, 7
(v) 5, 8, 20
Solution:
(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side. Here, 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5
(ii) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.
(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side. Here, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3
(iv) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 2 + 5 = 7
(v) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 5 + 8 < 20
Question: 2
In Fig, P is the point on the side BC. Complete each of the following statements using symbol ‘=’,’ > ‘or ‘< ‘so as to make it true:
(i) AP… AB+ BP
(ii) AP… AC + PC
(iii) AP…. 1/2(AB + AC + BC)
Solution:
(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.
(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.
(iii) AP < 12(AB + AC + BC) In triangles ABP and ACP, we can see that:
AP < AB + BP…(i) (Because the sum of any two sides of a triangle is greater than the third side)
AP < AC + PC…(ii) (Because the sum of any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we have:
AP + AP < AB + BP + AC + PC
2AP < AB + AC + BC (BC = BP + PC)
AP < (AB – FAC + BC)
Question: 3
P is a point in the interior of △ABC as shown in Fig. State which of the following statements are true (T) or false (F):
(i) AP + PB < AB
(ii) AP + PC > AC
(iii) BP + PC = BC
Solution:
(i) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
(ii) True
We know that the sum of any two sides of a triangle is greater than the third side: it is true for the given triangle.
(iii) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
Question: 4
O is a point in the exterior of △ABC. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that
OA + OB + OC > 1/2(AB + BC +CA)
Solution:
Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:
OA + OB > AB — (i)
OB + OC > BC — (ii)
OA + OC > CA — (iii)
On adding equations (i), (ii) and (iii) we get:
OA + OB + OB + OC + OA + OC > AB + BC + CA
2(OA + OB + OC) > AB + BC + CA
OA + OB + OC > (AB + BC + CA)/2
Question: 5
In ∆ABC, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.
Solution:
Because the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.
Exercise 15.5
Question: 1
State Pythagoras theorem and its converse.
Solution:
The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.
Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.
Question: 2
In right ∆ABC, the lengths of the legs are given. Find the length of the hypotenuse
(i) a = 6 cm, b = 8 cm
(ii) a = 8 cm, b = 15 cm
(iii) a = 3 cm, b = 4 cm
(iv) a = 2 cm, b =1.5 cm
Solution:
According to the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
(i) c2 = a2 + b2
c2 = 62 + 82
c2 = 36 + 64 = 100
c = 10 cm
(ii) c2 = a2 + b2
c2 = 82 + 152
c2 = 64 + 225 = 289
c = 17cm
(iii) c2 = a2 + b2
c2 = 32 + 42
c2 = 9 + 16 = 25
c = 5 cm
(iv) c2 = a2 + b2
c2 = 22 + 1.52
c2 = 4 + 2.25 = 6.25
c = 2.5 cm
Question: 3
The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side.
Solution:
Let the hypotenuse be ” c ” and the other two sides be ” b ” and ” c”.
Using the Pythagoras theorem, we can say that:
c2 = a2 + b2
2.52 = 1.52 + b2
b2 = 6.25 −2.25 = 4
c = 2 cm
Hence, the length of the other side is 2 cm.
Question: 4
A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.
Solution:
Let the hypotenuse be h.
Using the Pythagoras theorem, we get:
3.72 = 1.22 + h2
h2 = 13.69 – 1.44 = 12.25
h = 3.5 m
Hence, the height of the wall is 3.5 m.
Question: 5
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.
Solution:
In the given triangle, the largest side is 6 cm.
We know that in a right angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side.
Therefore,
32 + 42 = 9 + 16 = 25
But,
62 = 36
32 + 42 not equal to 62
Hence, the given triangle is not a right angled triangle.
Question: 6
The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c= 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
Solution:
(i) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 25 cm.
c2 = 625
We have:
a2+ b2 = 72 + 242 = 49 + 576 = 625 = c2
Thus, the given triangle is a right triangle.
(ii) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 18 cm.
c2 = 324
We have:
a2+ b2 = 92+162 = 81 + 256 = 337 not equal to c2
Thus, the given triangle is not a right triangle.
Question: 7
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.
(Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 m and 12 m)
Solution:
The distance between the tops of the poles is the distance between points A and B.
We can see from the given figure that points A, B and C form a right triangle, with AB as the hypotenuse.
On using the Pythagoras Theorem in ∆ABC, we get:
(11−6)2 + 122 = AB2
AB2 = 25 + 144
AB2 = 169
AB = 13
Hence, the distance between the tops of the poles is 13 m.
Question: 8
A man goes 15 m due west and then 8 m due north. How far is he from the starting point?
Solution:
Let O be the starting point and P be the final point.
By using the Pythagoras theorem, we can find the distance OP.
OP2 = 152 + 82
OP2 = 225 + 64
OP2 = 289
OP = 17
Hence, the required distance is 17 m.
Question: 9
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
Solution:
Given Let the length of the ladder be L m.
By using the Pythagoras theorem, we can find the length of the ladder.
62 + 82 = L2
L2 = 36 + 64 =100
L = 10
Thus, the length of the ladder is 10 m.
When the ladder is shifted:
Let the height of the ladder after it is shifted be H m.
By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.
82 + H2 = 102
H2 = 100 – 64 = 36
H = 6
Thus, the height of the ladder is 6 m.
Question: 10
A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?
Solution:
Let the distance of the lower end of the ladder from the wall be x m.
On using the Pythagoras theorem, we get:
x2 + 482 = 502
x2 = 502 − 482 = 2500 – 2304 = 196
H = 14 dm
Hence, the distance of the lower end of the ladder from the wall is 14 dm.
Question: 11
The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg.
Solution:
Let the length of each leg of the given triangle be x units.
Using the Pythagoras theorem, we get:
x2 + x2 = (Hypotenuse)2
x2 + x2 = 50
2x2 = 50
x2 = 25
x = 5
Hence, we can say that the length of each leg is 5 units.
Question: 12
Verity that the following numbers represent Pythagorean triplet:
(i) 12, 35, 37
(ii) 7, 24, 25
(iii) 27, 36, 45
(iv) 15, 36, 39
Solution:
We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.
(i) 372 =1369
122 + 352 = 144 + 1225 = 1369
122 + 352 = 372
Yes, they represent a Pythagorean triplet.
(ii) 252 = 625
72 + 242 = 49 + 576 = 625
72 + 242 = 252
Yes, they represent a Pythagorean triplet.
(iii) 452 = 2025
272 + 362 = 729 + 1296 = 2025
272 + 362 = 452
Yes, they represent a Pythagorean triplet.
(iv) 392 = 1521
152 + 362 = 225 + 1296 = 1521
152 + 362 = 392
Yes, they represent a Pythagorean triplet.
Question: 13
In ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD perpendicular to AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
100° + 35° + ∠ACB = 180°
∠ACB = 180° –135°
∠ACB = 45°
∠C = 45°
If we apply the above rule on ∆BCD, we can say that:
∠BCD + ∠BDC + ∠CBD = 180°
45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)
∠CBD = 180°– 135°
∠CBD = 45°
We know that the sides opposite to equal angles have equal length.
Thus, BD = DC
DC = 2 cm
Question: 14
In a ∆ABC, AD is the altitude from A such that AD = 12 cm. BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.
Solution:
In ∆ADC,
∠ADC = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 162 = AC2
AC2 = 144 + 256 = 400
AC = 20 cm
In ∆ADB,
∠ADB = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 92 = AB2
AB2 = 144 + 81 = 225
AB = 15 cm
In ∆ABC,
BC2 = 252 = 625
AB2 + AC2 = 152 + 202 = 625
AB2 + AC2 = BC2
Because it satisfies the Pythagoras theorem, we can say that ∆ABC is right angled at A.
Question: 15
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 105° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB, which will be approximately 5.5 cm.
AC2 + BC2 = 42 + 32 = 9 +16 = 25
AB2 = 5.52 = 30.25
AB2 not equal to AC2+ BC2
Here,
AB2 > AC2 + BC2
Question: 16
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 80° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB; it will be approximately 4.5 cm.
AC2 + BC2 = 42 + 32 = 9 +16 = 25
AB2 = 4.52 = 20.25
AB2 not equal to AC2 + BC2
Here,
AB2 < AC2 + BC2
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