MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 9 - MCQExams.com
CBSE
Class 7 Maths
The Triangle And Its Properties
Quiz 9
How many altitudes does a triangle have?
Report Question
0%
$$1$$
0%
$$3$$
0%
$$6$$
0%
$$9$$
Explanation
$$\textbf{Step-1: Represent the altitudes of a triangle using a diagram. }$$
$$\text{A triangle has three altitudes. From each vertex perpendicular to the opposite side can be drawn.}$$
$$\textbf{Hence, option - B is the correct answer.}$$
Which of the following can be the length of the third side of triangle whose two sides measure $$18 cm$$ and $$14 cm $$?
Report Question
0%
$$4 \,cm$$
0%
$$3 \,cm$$
0%
$$5 \,cm$$
0%
$$32 \,cm$$
Explanation
Let $$y \,cm$$ be the third side of the triangle.
Hence,
$$\Rightarrow 18+14>y$$
$$\Rightarrow 32>y$$
$$18-14 <y$$
$$y>4 \ and\ 32>y$$
$$4<y<32$$
Therefore, the third side of a triangle can be $$5 cm$$.
From Fig., the value of $$x$$ is
Report Question
0%
$$75^o$$
0%
$$90^o$$
0%
$$120^o$$
0%
$$60^o$$
Explanation
In triangle, $$ABC$$,
$$\angle ABC + \angle CAB = \angle ACD$$ ...Exterior angle property
$$\Longrightarrow ACD = 25^o + 35^o$$
$$\Longrightarrow \angle ACD = 60^o$$ ...(1)
Now,
$$\Longrightarrow x = \angle D + \angle ACD$$
$$\Longrightarrow x = 60^o + 60^o $$ using ...(1)
$$\therefore x = 120^o$$
In a right-angled triangle $$ABC$$, if angle $$B = 90^o$$, then which of the following is true?
Report Question
0%
$$AB^2 = BC^2 + AC^2$$
0%
$$AC^2 = AB^2 + BC^2$$
0%
$$AB = BC + AC$$
0%
$$AC = AB + BC$$
Explanation
Hypotenuse is the longest side of triangle which is opposite to right angle.
By pythagoras theorm,
$$\text{(hypotenuse)}^2=\text{(side 1)}^2+\text{(side 2)}^2$$
$$\Longrightarrow AC^2 = AB^2 + BC^2$$
So, option $$B$$ is correct.
Write the correct answer from the given four options.
The hypotenuse of a right triangle with its legs of lengths $$3x \times 4x$$ is
Report Question
0%
$$5x$$
0%
$$7x$$
0%
$$16x$$
0%
$$25x$$
Explanation
$$(Hypotenuse)^2 = (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2$$
$$Hypotenuse = \sqrt{25x^2} = 5x$$
Which of the following figures will have its altitude outside the triangle?
Report Question
0%
0%
0%
0%
Explanation
In obtuse angled triangle, two of its altitudes lie outside the triangle.
Also, from figure it is clear that $$D$$ has its altitude outside triangle, which is an obtuse angled triangle.
Hence, option D is correct.
State whether the following statements are true (T) or false (F):
An altitude of a triangle always lies outside the triangle.
Report Question
0%
True
0%
False
Explanation
Altitude of the acute triangle lies inside the triangle and altitude of obtuse triangle lies outside the triangle
If one angle of triangle is equal to the sum of the other two angles then the triangle is
Report Question
0%
Isosceles triangle
0%
Obtuse angled triangle
0%
Equilateral triangle
0%
Right angled triangle
Altitudes of $$\triangle ABC$$ are
Report Question
0%
AB
0%
AC
0%
BC
0%
None of these.
Does altitude of a triangle always lie inside the triangle?
Report Question
0%
Yes
0%
No
0%
It may lie inside or outside the triangle.
0%
None of these.
In a $$ \Delta \mathrm{PQR} $$, if $$ \angle \mathrm{P}+\angle \mathrm{Q}=\angle \mathrm{R} $$ then type of triangle is:
Report Question
0%
scalene triangle
0%
equilateral triangle
0%
isosceles triangle
0%
right-angled triangle
Explanation
Given,
$$\angle P+\angle Q=\angle R$$
$$\Rightarrow \angle P + \angle Q + \angle R=\angle R+\angle R$$
$$\Rightarrow 180^\circ=2\angle R$$
$$\therefore \angle R=90^\circ$$
$$\triangle PQR$$ is right-angled triangle
In $$\triangle PQR$$, $$PS$$ is an altitude and $$PT$$ is median.
Which of the following statement is correct?
Report Question
0%
$$PT = PS$$
0%
$$RT = QT$$
0%
$$RS = TQ$$
0%
None of these
In $$\triangle ABC$$, $$D$$ and $$E$$ are points on $$BC$$ such that $$BD = DC$$ and $$AD$$ is perpendicular to $$BC$$.
Which of the following is/are correct statement/s?
Report Question
0%
$$AD$$ is an altitude of $$\triangle ABC$$
0%
$$AE$$ is an altitude of $$\triangle ABC$$
0%
$$AD$$ is an median of $$\triangle ABC$$
0%
$$AE$$ is an median of $$\triangle ABC$$
Let ABC is a triangle. Then, which of the following is possible?
Report Question
0%
AC + BC > AB
0%
AB + BC > AC
0%
AB + AC > BC
0%
None of these.
In given $$\triangle ABC$$, AE, BF and CD are
Report Question
0%
Bisectors
0%
Altitudes
0%
Sides
0%
Medians
Let AM is the median of $$\triangle ABC$$. Then,
Report Question
0%
BM = MC
0%
AB = AC
0%
AB = AM
0%
AC = AM
Is it possible to have a triangle with the following sides?
3 cm, 6 cm, 8 cm
Report Question
0%
Yes
0%
No
0%
May be
0%
Information insufficient.
The lengths of two sides of a triangle are 6 cm and 8 cm. Which of the following can be length the third side of the triangle?
Report Question
0%
1 cm
0%
5 cm
0%
15 cm
0%
None of the above
Let two sides of a triangle are 2 cm and 5 cm. Which of the following can be the length of its third side?
Report Question
0%
6 cm
0%
7 cm
0%
2 cm
0%
8 cm
Given $$\triangle ABC$$ in the figure. Then which is NOT true for a, b, c?
(Assume a > c > b)
Report Question
0%
a + b > c
0%
c + b > a
0%
a + c > b
0%
None of these
Given $$\triangle ABC$$ in the figure. Then which is true for a, b, c?
Report Question
0%
a + b > c
0%
b + c > a
0%
a + c > b
0%
None of these
Choose all correct options to fill in the blank.
In the given $$\triangle ABC$$, $$\angle APC$$ is a right angle. Then, AP is a/an _________ of $$\triangle ABC$$.
Report Question
0%
Median
0%
Altitude
0%
Side
0%
None of these.
An isosceles triangle has a $$10$$ inch base and two $$13$$ inch sides. What other value can the base have and still yield a triangle with the same area?
Report Question
0%
$$18"$$
0%
$$19"$$
0%
$$24"$$
0%
$$27"$$
Explanation
Let $$ \Delta ABC $$ with $$AB = AC = 13$$ inch and base $$BC = 10$$ inch.
Therefore, $$ \cfrac { 1 }{ 2 } BC = 5$$ inch
Let AD be its height.
Applying Pythagoras theorem, its height is $$ \sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } = 12$$ inch
Therefore, area of the triangle $$ = \cfrac { 1 }{ 2 } \times BC \times AD = \cfrac { 1 }{ 2 } \times 10\times 12 $$
$$ = 60 $$ sq.inch.
Now, we can divide the triangle along AD and rearrange the parts so that $$AD + AD$$ becomes base,
$$ \cfrac { 1 }{ 2 } $$ BC becomes height and the equal sides are $$AB =13$$ inch each.
The base of the new $$ \Delta$$ is $$AD + AD = 12 + 12 = 24$$ inch.
Height $$ = \cfrac { 1 }{ 2 } BC = \cfrac { 1 }{ 2 } \times 10 inch = 5$$ inch.
And its two equal sides are $$13$$ inch, each.
Its area $$ =\cfrac { 1 }{ 2 } \times 24 \times 5 = 60$$ sq.inch, which is the area of the previous $$\triangle ABC.$$
In the given figure, if AD is a median of
$$\triangle ABC$$, the
Report Question
0%
AB + AC < 2 AD
0%
AB + AC > 2 AD
0%
AB + AC < AE
0%
AC + CE < AE
In a triangle $$ABC$$, $$\angle ABC={ 90 }^{ o },\angle ACB={ 30 }^{ o }$$, $$AB=5cm$$. What is the length of $$AC$$?
Report Question
0%
$$10cm$$
0%
$$5cm$$
0%
$$5\sqrt { 2 } cm$$
0%
$$5\sqrt { 3 } cm$$
The hypotenuse of a right triangle is $$3\sqrt{10}$$ units. If the smaller side is tripled and the lower side is doubled, new hypotenuse becomes $$9\sqrt{5}$$ units. What are the lengths of the smaller and longer sides of the right triangle respectively?
Report Question
0%
$$5$$ units, $$9$$ units
0%
$$5$$ units, $$6$$ units
0%
$$3$$ units, $$9$$ units
0%
$$3$$ units, $$6$$ units
$$ABC$$ is a right-angled triangle, right angled at $$C$$ and $$p$$ is the length of perpendicular from $$C$$ on $$AB$$. If $$a,b,c$$ are the sides of the triangle, then which one of the following is correct?
Report Question
0%
$$({ a }^{ 2 }+{ b }^{ 2 }){ p }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }$$
0%
$${ a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }{ p }^{ 2 }$$
0%
$${ p }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$$
0%
$${ p }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$$
Which of the following is a right angled triangle?
Report Question
0%
$$\triangle APD$$
0%
$$\triangle DOC$$
0%
$$\triangle BDC$$
0%
$$\triangle AOD$$
State and prove isosceles angle properly
If $$AB = AC$$, then what is value of $$\angle a$$?
Report Question
0%
$$40^o$$
0%
$$20^o$$
0%
$$120^o$$
0%
$$100^o$$
Which of the following describes the given triangle?
Report Question
0%
Isosceles, obtuse
0%
Equilateral, acute
0%
Isosceles, right
0%
Equilateral, obtuse
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 7 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
