Free The Triangle and Its Properties Subjective Test 01 Practice Test - 7th grade

Question 1

In a triangle, all the medians meet at point A. Similarly, all the altitudes meet at point N. For which type of triangle is point A same as point N? [1 MARK]

SOLUTION

Solution :
Equilateral Triangle; Only in an equilateral triangle, all the altitudes and median are same lines. So, point A will be same as point N.

Question 2

Can we say that there exists only one triangle with a particular set of sides? What are such triangles known as? [2 MARKS]

SOLUTION

Solution :
Name and Explanation: 2 Marks

Yes, there exists a unique triangle for a particular set of sides. This is because 2 triangles having the same set of sides will be exactly identical. You can realize this by drawing 2 triangles with the same set of sides and then trying to superimpose one over the other. You will find that they are exactly identical.

If the two triangles have equal sides, then the two triangles are said to be congruent as equal sides are one of the criteria for congruency of triangles.

Question 3

Answer the following: [2 MARKS]
(i) Angle opposite to CA in $Δ A B C$ .
(ii) Side opposite to $∠ A$ in $Δ A O D$ .

SOLUTION

Solution : Each part: 1 Mark each

(i) In $Δ A B C$ , opposite to side CA is the $∠ B$ .
(ii) In $Δ A O D$ , $∠ A$ gets formed with AO and AD. Hence, the side opposite to $∠ A$ will be OD.

Question 4

If there is a scalene ∆ABC in which all its angles have integral values, what would be the maximum possible value for ∠ABC? Which type of triangle would ∆ABC be, on the basis of its angles? [2 MARKS]

SOLUTION

Solution :
Calculation of angle: 1 Mark
Type of triangle: 1 Mark

In scalene triangles, all the angles are different. For $∠ A B C$ to be maximum, the other two angles must be minimum, but different (since scalene triangle).

The minimum value of the other angles will, therefore, be $1^{\circ}$ and $2^{\circ}$ as the angles have an integral value. Hence, for a scalene triangle the maximum integral value of $∠ A B C = 177^{\circ}$ [Angle sum property of a triangle].
The following triangle will be an Obtuse angled triangle , because one of the angle is greater than $90 °$

Question 5

If we draw a median AD in ∆ABC, are the altitudes of ∆ADB and ∆ADC equal in length? Justify. [2 MARKS]

SOLUTION

Solution :
Answer: 1 Mark
Justification: 1 Mark

Yes.

The median AD will divide the ∆ABC into 2 triangles ∆ADB and ∆ADC. These two triangles, as shown in the figure will have the same altitude AN as the altitude is the perpendicular distance from the vertex to the base of the triangle.The foot of the perpendicular can be inside as well as outside the triangle as shown in the figure. (Note: The median of the triangle divides it into 2 triangles with equal area.)

Question 6

Can the following triangles exist? [3 MARKS]
(i) AB = 8 cm, BC = 6 cm, AC = 10cm
(ii) MN = 9 cm, NO = 5 cm, OM = 4 cm
(iii) $∠ A$ = $175^{\circ}$ , $∠ B$ = $5^{\circ}$ and $∠ C$ = $0^{\circ}$

SOLUTION

Solution : Each part: 1 Mark each

(i) Since AB + BC > AC, AB + AC > BC and BC + AC > AB, the following triangle can exist.
(ii) The following triangle cannot exist since OM + NO is not greater than MN.
(iii) The following triangle cannot exist since one of the angles of the triangles is 0 $^{o}$ . All angles of a triangle should be greater than 0 $^{o}$ .

Question 7

(a) Find all the angles of a right-angled isosceles triangle.
(b) Is a triangle ABC with side length AB = 12 cm, BC = 8 cm and AC = 20 cm possible? Explain why. [3 MARKS]

SOLUTION

Solution :
(a) Answer: 1 Mark
(b) Answer: 1 Mark
Reason: 1 Mark

(a) In a right-angled isosceles triangle, sides other than the hypotenuse are equal in length.

In ABC, AB = BC $\Rightarrow ∠ 1 = ∠ 3$ -------------------I

By angle sum property of triangle,

$∠ 1 + ∠ 2 + ∠ 3 = 180^{\circ}$

$∠ 1 + 90 + ∠ 1 = 180^{\circ}$

$2 ∠ 1 = 90^{\circ}$

$\Rightarrow ∠ 1 = 45^{\circ} = ∠ 2$

So, angles are $45^{\circ}, 45^{\circ} a n d 90^{\circ}$

(b) A triangle ABC with the given side lengths of AB = 12 cm, BC = 8 cm and AC = 20 cm is not possible because in a triangle the sum of any two sides should be greater than the third side but in this triangle, AB + BC = 20 cm which is equal to the third side AC = 20 cm. AB + BC should have been greater than AC.

Question 8

Find the third side in the given triangle. Name the theorem used to find the third side. What are such triangles called? [3 MARKS]

SOLUTION

Solution : Name: 1 Mark
Theorem: 1 Mark
Answer: 1 Mark

Since $∠ C$ is $90^{\circ}$ , the following is a right-angled triangle.
The theorem used is known as Pythagoras Theorem.
In a right angled triangle,
Base $^{2}$ + perpendicular $^{2}$ = hypotenuse $^{2}$ (Pythagaros's Theorem)
Or, AB $^{2}$ = BC $^{2}$ + AC $^{2}$
Or, AB $^{2}$ = 16 + 9 = 25
Or, AB = $\sqrt{25}$ = 5

Question 9

(a) Find x in the triangle given below if it is right-angled.

(b) Find x in the triangle given below if it is a right-angled triangle.

[3 MARKS]

SOLUTION

Solution : Formula: 1 Mark
Each part: 1 Mark

(a) Suppose, the above triangle is right-angled such that 10 is the longest side.
Then, according to Pythagoras Theorem,
Base $^{2}$ + Perpendicular $^{2}$ = Hypotenuse $^{2}$
Or, (16x) $^{2}$ + (12x) $^{2}$ = 10 $^{2}$
Or, 256x $^{2}$ + 144x $^{2}$ = 100
Or, 400x $^{2}$ = 100
Or, x = $\sqrt{\frac{100}{400}}$ = 0.5

(b) According to Pythagoras Theorem,

Base $^{2}$ + Perpendicular $^{2}$ = Hypotenuse $^{2}$
x $^{2}$ + 12 $^{2}$ = 13 $^{2}$
x $^{2}$ + 144 = 169
x $^{2}$ = 169 - 144
x $^{2}$ = 25
x = 5

Question 10

(a) Derive the angle sum property of the triangle using the exterior angle property.

(b) AE is the angular bisector. $∠ 4 = 140^{\circ}, ∠ 2 = 60^{\circ}, ∠ E A C = x$ . Find $x$ .

[4 MARKS]

SOLUTION

Solution :
(a) Steps: 1 Mark
Proof: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark

(a)

$∠ 1$ , $∠ 2$ and $∠ 3$ are angles of triangle ABC and 4 is the exterior angle when BC is extended to D.
$∠ 1$ + $∠ 2$ = $∠ 4$ (Exterior Angle Property)
$∠ 1$ + $∠ 2$ + $∠ 3$ = $∠ 4$ + $∠ 3$ (Adding $∠ 3$ to both sides)
Also, $∠ 3$ + $∠ 4$ = 180 $^{o}$ (Linear Pair of angles)
$∴$ $∠ 1$ + $∠ 2$ + $∠ 3$ = 180 $^{o}$

(b)

$∠ 1$ + $∠ 2$ = $∠ 4$ (Exterior Angle Property)
$∠ 1 + 60^{\circ} = 140^{\circ}$
$∠ 1 = 140^{\circ} - 60^{\circ} = 80^{\circ}$
Since AE is the angular bisector therefore $∠ E A C = \frac{∠ 1}{2}$
$\Rightarrow x = \frac{80}{2} = 40^{\circ}$

Question 11

(a) Find the sum of all the exterior angles of a triangle.
(b) Is it possible to draw a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm respectively? [4 MARKS]

SOLUTION

Solution :
(a) Steps: 1 Mark
Correct answer: 1 Mark

(b) Reason: 1 Mark
Correct answer: 1 Mark

(a)

In the figure, 1, 2 and 3 are the interior angles whereas 4, 5 and 6 are exterior angles. Using linear pair axiom, we can say that:
$∠ 1 + ∠ 4 = 180^{\circ}$

$∠ 3 + ∠ 6 = 180^{\circ}$

$∠ 2 + ∠ 3 = 180^{\circ}$

Adding all of them, we get:

$∠ ((1 + 4) + (3 + 6) + (2 + 5)) = 540^{\circ}$

Rearranging the terms, we get:

$∠ ((1 + 2 + 3) + (4 + 5 + 6)) = 540^{\circ}$

$∠ (1 + 2 + 3) = 180^{\circ}$ (Angle Sum Property of a triangle)

So, $∠ ((4 + 5 + 6)) = 540^{\circ} - 180^{\circ} = 360^{\circ}$

So, the sum of the exterior angles of a triangle is $360^{\circ}$

(b) Suppose such a triangle is possible. Then the sum of the lengths of any two sides would be greater than the length of the third side. Let's check this.
Is 4.5 + 5.8 > 10.2? Yes
Is 5.8 + 10.2 > 4.5? Yes
Is 10.2 + 4.5 > 5.8? Yes

Therefore, the triangle is possible.

Question 12

In the given pentagon ABCDE, prove that 2CD < Perimeter (ABCDE). [4 MARKS]

SOLUTION

Solution :
Properties : 1 Mark
Steps : 2 Marks
Result : 1 Mark

In triangle ABC, AC < AB + BC ------------------- 1
In triangle ADE, AD < AE + DE ------------------- 2
Adding 1 and 2,
AC + AD < AB + BC + AE + DE ----------------3
In triangle ADC, CD < AD + AC ------------------- 4
From 3 and 4, we can write
CD < AB + BC + AE + DE
Adding CD to both sides,
2CD < AB + BC + CD + DE + EA
Or, 2 CD < Perimeter of ABCDE

Question 13

(a) If AB || DE, ∠AOD = 50° and ∠DPC = 160°, find the value of ∠DBE?

(b) A 15 m long ladder reached a window 12 m high from the ground when placed against a wall at a distance 'a'. Find the distance of the foot of the ladder from the wall.

[4 MARKS]

SOLUTION

Solution :
(a) Steps: 1 Mark
Correct answer: 1 Mark
(b) Formula: 1 Mark
Result: 1 Mark

(a)

$∠ O A D = 50 ° (g i v e n) ∠ D O C = 180 ° - 50 ° = 130 ° (l i n e a r p a i r) ∠ D P C = 160 ° (g i v e n) ∠ D P O = 180 ° - 160 ° (L i n e a r p a i r) ∠ D P O = 20 ° I n △ D O P ∠ O D P = 180 ° - (130 ° + 20 °) (b y a n g l e s u m p r o p e r t y) ∠ O D P = 30 ° I n △ D B E ∠ D B E = 180 ° - (90 ° + 30 °) (B y a n g l e s u m p r o p e r t y) ∠ D B E = 60 °$

(b)

The wall and the ground are at right angle. Therefore the ladder, ground and the wall make the sides of a right-angled triangle.
$∴ a^{2} + 12^{2} = 15^{2}$
$\Rightarrow a^{2} = 225 - 144 = 81$
$a = \sqrt{8} 1 = 9 m$

Question 14

(a) In a marathon, you started from checkpoint A and ran 5 km towards West. The checkpoint you reached was B. Then, you again ran for 12 km towards North and reached checkpoint C. Find the length of the shortest path which you must take to reach the finishing checkpoint A.

(b) Show that the angles of an equilateral triangle are $60^{\circ}$ each. [4 MARKS]

SOLUTION

Solution :
(a) Solution: 2 Marks
(b) Proof: 2 Marks

(a)

If we draw the path taken by you, it will look something like the figure above. We have to find AC. Applying Pythagoras property,

$A B^{2}$ + $B C^{2}$ = $A C^{2}$

$5^{2}$ + $12^{2}$ = $A C^{2}$

$A C^{2}$ = 25 + 144

$A C^{2}$ = √169

AC = 13 km

(b) Let ABC be an equilateral triangle.

BC = AC = AB (Length of all sides is same)
$\Rightarrow ∠ A = ∠ B = ∠ C$ ( Angle opposite to equal sides are equal)
Also,
$∠ A + ∠ B + ∠ C = 180^{\circ}$
$\Rightarrow 3 ∠ A = 180^{\circ}$
$\Rightarrow ∠ A = 60^{\circ}$
$∴ ∠ A = ∠ B = ∠ C = 60^{\circ}$
Thus, the angles of an equilateral triangle are $60^{\circ}$ each.

Question 15

(a) Find x in the below diagram. What are such triangles called?

(b) In a triangle ABC, an altitude is dropped from A to BC at D. $∠ A = 40^{\circ}, ∠ B = (2 x + 4)^{\circ}, ∠ C = (4 x - k)^{\circ}, ∠ B A D = 30^{\circ}$
Find $k$ , $x$ and the values of the angles. [4 MARKS]

SOLUTION

Solution : (a) Solution: 1 Mark
Type of triangle: 1 Mark
(b) Solution: 1 Mark
Correct answer: 1 Mark

(a) For a triangle, sum of all angles = 180 $^{o}$
Or, (2x - 15 $^{o}$ ) + (x + 20 $^{o}$ ) + (x + 15 $^{o}$ ) = 180 $^{o}$
Or, 2x + x + x - 15 $^{o}$ + 20 $^{o}$ + 15 $^{o}$ = 180 $^{o}$
Or, 4x + 20 $^{o}$ = 180 $^{o}$
Or, 4x = 160 $^{o}$
Or, x = $\frac{160}{4}$ = 40 $^{o}$

The angles of the triangle will be, (2x - 15 $^{o}$ = 65 $^{o}$ ), (x + 20 $^{o}$ = 60 $^{o}$ ) and (x + 15 $^{o}$ = 55 $^{o}$ )
Since all angles are less than 90 $^{o}$ , it's an 'acute angled triangle' and also all the angles are of different values therefore all the sides will be of different length, hence it is a 'scalene triangle'.

(b

In triangle ABD
$\Rightarrow 2 x + 4 + 90 + 30 = 180$
$\Rightarrow 2 x = 180 - 124$
$\Rightarrow x = \frac{56}{2}$
$\Rightarrow x = 28$

In triangle ADC
$\Rightarrow 4 x - k + 90 + 10 = 180$
$\Rightarrow 4 x - k = 180 - 100$
$\Rightarrow 4 (28) - k = 80$ (Substituting the value of $x = 28$ )
$\Rightarrow k = 4 (28) - 80$
$\Rightarrow k = 112 - 80$
$\Rightarrow k = 32$

$∴$ The angles are $40^{\circ}, 60^{\circ} a n d 80^{\circ}$ for the given triangle.

Free The Triangle and Its Properties Subjective Test 01 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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