Download App

MATHS 2 Marks Questions

The Triangle and Its Properties · Gujarat Board (GSEB) STD 7 (GSEB - English Medium). 33 questions.

Questions

2 Marks Questions

🎯

Test yourself on this topic

33 questions · timed · auto-graded

Question 12 Marks
Two angles of a triangle are $30^{\circ}$ and $80^{\circ}$. Find the third angle.
Answer
Let the third angle be $x$.
Then, by angle sum property of a triangle,
$
\begin{array}{c}
x+30^{\circ}+80^{\circ}=180^{\circ} \Rightarrow x+110^{\circ}=180^{\circ} \\
\Rightarrow \quad x=180^{\circ}-110^{\circ}=70^{\circ}
\end{array}
$
Hence, the third angle is $70^{\circ}$.
View full question & answer→
Question 22 Marks
Is something wrong in this given diagram? Comment.
Image
Answer
From the given figure, exterior angle $=50^{\circ}$
and sum of two interior opposite angles
$=50^{\circ}+50^{\circ}=100^{\circ}$
By exterior angle property of a triangle,
Exterior angle $=$ Sum of two interior opposite angles
But here, exterior angle $\neq$ Sum of two interior opposite angles.
So, in the given diagram the given angles are incorrect.
View full question & answer→
Question 32 Marks
The two interior opposite angles of an exterior angle of a triangle are $60^{\circ}$ and $80^{\circ}$. Find the measure of the exterior angle.
Answer
Given, two interior opposite angles are of measure $60^{\circ}$ and $80^{\circ}$.
By exterior angle property of a triangle,
Exterior angle $=$ Sum of interior opposite angles
$
=60^{\circ}+80^{\circ}=140^{\circ}
$
Hence, the measure of the exterior angle is $140^{\circ}$.
View full question & answer→
Question 42 Marks
An exterior angle of a triangle is of measure $70^{\circ}$ and one of its interior opposite angles is of measure $25^{\circ}$. Find the measure of the other interior opposite angle.
Answer
Given, measure of an exterior angle $=70^{\circ}$
Measure of one interior opposite angle $=25^{\circ}$
Now, by exterior angle property of a triangle,
Image
$\angle A B C+\angle B A C=\angle A C D $
$\Rightarrow \quad 25^{\circ}+\angle B A C=70^{\circ} $
$\therefore \quad \angle B A C=70^{\circ}-25^{\circ}=45^{\circ}$
Hence, the measure of the other interior opposite angle is $45^{\circ}$.
View full question & answer→
Question 52 Marks
Write the six elements (l.e. the 3 sides and the 3 angles) of $\triangle A B C$.
Answer
The six elements i.e. the three sides and the three angles of triangle ABC are as follows:
Sides $\overline{A B}, \overline{B C}, \overline{C A}$
Angles $\angle A B C, \angle B A C, \angle B C A$
Image
View full question & answer→
Question 62 Marks
Can you have a triangle with two right angles?
Answer
No, we cannot have a triangle with two right angles because the sum of two right angles is $180^{\circ}$. On adding the measure of the third angle, the sum of three angles will be more than $180^{\circ}$, which is not true for a triangle.
View full question & answer→
Question 82 Marks
Can the altitude and median be same for a triangle?
Answer

Image
Yes, in a triangle (equilateral triangle), its median and altitude are same in the adjoining figure.
Hence, $A L$ is an altitude as well as a median of $\triangle A B C$.
View full question & answer→
Question 92 Marks
Can you think of a triangle in which two altitudes of the triangle are two of its sides?
Answer

Image
Yes, a right angled triangle has two of its altitudes as its sides. In right angled $\Delta$ $A B C, A C$ and $B C$ are its altitudes.
View full question & answer→
Question 102 Marks
Will an altitude always lie in the interior of a triangle? If you think that this need not be true, draw a rough sketch to show such a case.
Answer

Image
An altitude does not always lie in the interior of a triangle. $\triangle A B C$ is an obtuse angled triangle such that $\angle C$ is obtuse and $A L$ is the altitude from $A$.
Clearly, $L$ is not a point on $B C$ but on $B C$ produced.
So, altitude $A L$ lie in the exterior of the triangle.
View full question & answer→
Question 112 Marks
Draw rough sketches of altitudes from A to BC for the following triangles.
Image
Answer
Rough sketches of altitudes from A to BC for the given triangle are as follows:
Image
View full question & answer→
Question 122 Marks
How many altitudes can a triangle have?
Answer

Image
Every triangle has three altitudes, one from each vertex to the opposite side.
In the adjoining figure, $A D \perp B C, B E \perp A C$ and $C F \perp A B$. Therefore, line segments $A D, B E$ and $C F$ are three altitudes.
View full question & answer→
Question 132 Marks
Which is the longest side in the $\triangle A B C$, right angled at $B$ ?
Answer
We know that the longest side is always opposite to the largest angle.
Image
In right angled $\triangle A B C$, the largest angle is $90^{\circ}$, which is given at $B$.
So, $A C$ is the longest side in the $\triangle A B C$.
View full question & answer→
Question 142 Marks
Which is the longest side in the $\triangle P Q R$, right angled at P?
Answer
We know that the longest side is always opposite to the largest angle.
In right angled $\triangle P Q R$, the largest angle is $90^{\circ}$ which is given at $P$.
Image
So, $Q R$ is the longest side in the $\triangle P Q R$
View full question & answer→
Question 162 Marks
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer

Image
Let $A B C D$ be a rectangle whose length, $A B=40 cm$
and diagonal $A C=41 cm$
In right angled $\triangle A B C$, using
Pythagoras property,
$
\begin{array}{l}
A C^2=A B^2+B C^2 \\
\Rightarrow \quad B C^2=A C^2-A B^2 \\
\Rightarrow \quad B C^2=(41)^2-(40)^2 \\
\Rightarrow \quad B C^2=1681-1600 \\
\Rightarrow \quad B C^2=81 \\
\Rightarrow \quad B C=\sqrt{81}=9 cm
\end{array}
$
Now, perimeter of the rectangle
$
\begin{array}{l}
=2(A B+B C) \\
\quad[\because \text { Perimeter of a rectangle }=2(l+b)] \\
=2(40+9) \\
=2 \times 49=98 cm
\end{array}
$
Hence, the perimeter of the rectangle is 98 cm .
View full question & answer→
Question 172 Marks
Angles $Q$ and $R$ of a $\triangle P Q R$ are $25^{\circ}$ and $65^{\circ}$. Write which of the following is true?
Image
(i) $P Q^2+Q R^2=R P^2$
(ii) $P Q^2+R P^2=Q R^2$
(iii) $R P^2+Q R^2=P Q^2$
Answer
i. In $\triangle P Q R, \angle P+\angle Q+\angle R=180^{\circ}$
[by angle sum property of a triangle]
$
\begin{array}{ll}
\Rightarrow & \angle P+25^{\circ}+65^{\circ}=180^{\circ} \\
\Rightarrow & \angle P=180^{\circ}-90^{\circ}=90^{\circ}
\end{array}
$
So, $\triangle P Q R$ is a right-angled triangle, right angled at $P$.
In $\triangle P Q R$, by Pythagoras property, $Q R^2=P Q^2+R P^2$
Hence, option (ii) is true.
Image
View full question & answer→
Question 182 Marks
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Answer

Image
Let $A C B$ be the tree. The tree is broken at point $C$ such that $B C=5 cm$ and its top $A$ touches the ground at $O$.
Then, $A C=O C$
Thus, $\triangle O B C$ is a right-angled triangle, right angled at $B$ such that
$O B=12 m, B C=5 m$
In $\triangle O B C$, by Pythagoras property,
$
\begin{aligned}
& O C^2=O B^2+B C^2 \\
\Rightarrow & O C^2=(12)^2+(5)^2 \\
\Rightarrow & O C^2=144+25 \\
\Rightarrow & O C^2=169 \Rightarrow O C=\sqrt{169}=13 \\
\because \quad & A C=O C=13 m \\
\therefore \quad & A B=A C+B C=(13+5) m=18 m \quad[\because A C=O C]
\end{aligned}
$
Hence, the original height of the tree is 18 m .
View full question & answer→
Question 192 Marks
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
In the case of right-angled triangles, identify the right angles.
Answer
i. Let $a=2.5 cm, b=6.5 cm$ and $c=6 cm$
$
\begin{array}{l}
\therefore \quad a^2+c^2=(2.5)^2+(6)^2 \Rightarrow a^2+c^2=6.25+36 \\
\Rightarrow \quad a^2+c^2=42.25 \text { and } b^2=(6.5)^2=42.25 \\
\therefore \quad a^2+c^2=b^2
\end{array}
$
Thus, the given sides form a right angled triangle and the right angle is opposite to the side of length 6.5 cm .
ii. Do same as part (i) Ans. No
iii. Do same as part (i) Ans. Yes
View full question & answer→
Question 202 Marks
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Image
Answer

Image
Let $A C$ be the ladder and $A B$ be the wall $C$ is the foot of the ladder and $A$ be the top of ladder which is at the window.
Here, height of window $A$ from the ground i.e. $A B=12 cm$
In right angled $\triangle A B C$, by Pythagoras property,
$
\begin{array}{rlrl}
& A C^2 =A B^2+B C^2 \\
\Rightarrow & B C^2 =A C^2-A B^2 \\
\Rightarrow & B C^2 =(15)^2-(12)^2 \\
\Rightarrow & B C^2 =225-144 \\
\Rightarrow & B C^2 =81 \\
\Rightarrow & B C=\sqrt{81}=9
\end{array}
$
Hence, the distance of the foot of the ladder from the wall is 9 m .
View full question & answer→
Question 212 Marks
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, then find BC.
Answer

Image
Given, $\triangle A B C$ is right angled at $C$ in which
$
A B=25 cm \text { and } A C=7 cm
$
In $\triangle A B C$, by Pythagoras property,
$
\begin{array}{rlrl}
& A B^2 =A C^2+B C^2 \\
\Rightarrow & B C^2 =A B^2-A C^2 \\
\Rightarrow & B C^2 =(25)^2-(7)^2 \\
\Rightarrow & B C^2=625-49 \\
\Rightarrow & B C^2=576 \\
\Rightarrow & B C=\sqrt{576}=24
\end{array}
$
Hence, the length of $B C$ is 24 cm .
View full question & answer→
Question 222 Marks
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Answer
Let x cm be the length of the third side.
We know that the sum of lengths of two sides of a triangle is greater than the length of third side.
$\therefore 12 cm+15 cm > x \Rightarrow 27 cm > x$
or $x < 27cm$
Also, the difference between the lengths of two sides of a triangle is less than the length of third side.
$\therefore 15 cm-12 cm < x \Rightarrow 27 cm < x$
or $x > 3cm$
Hence, the length of third side can be any length between 3 cm and 27 cm.
View full question & answer→
Question 242 Marks
A shelf with a triangular frame is fixed on a wall as shown below.
Image
The lengths of the rods used in the shaded triangular frame are 48 cm, 55 cm and 73 cm.
(i) What is the type of a shaded triangle?
(ii) What can be the height of the shelf?
Answer
(i) Since, the sides of triangle are 48 cm, 55 cm and 73 cm.
Thus, $(73)^2=48^2+55^2$
Thus, the shaded triangle is right angled triangle.
(ii) The height of shelf can be either 48 or 55 cm.
View full question & answer→
Question 252 Marks
Find the value of the unknown exterior angle x in the following figures.
Image
Answer
i. Since, interior opposite angles are $40^{\circ}$ and $60^{\circ}$.
By exterior angle property of triangle,
So,$\quad x=40^{\circ}+60^{\circ}=100^{\circ}$
ii. Since, interior opposite angles are $45^{\circ}$ and $65^{\circ}$.
By exterior angle property of triangle,
So, $\quad x=45^{\circ}+65^{\circ}=110^{\circ}$
View full question & answer→
Question 262 Marks
Find the value of x in the given rectangle ABCD.
Image
Answer
We know that in a rectangle opposite sides are equal.
$\therefore A D=B C=8 \text { and } A B=C D=15$
So, consider a $\triangle A B C$,
where, $A B=15$ and $B C=8$
Also, $\angle A B C=90^{\circ} \quad$ [angle of a rectangle is $90^{\circ}$]
By Pythagoras property, we have
$A C^2 =A B^2+B C^2 $
$\Rightarrow x^2  =15^2+8^2=225+64=289 $
$\Rightarrow x =17$
View full question & answer→
Question 272 Marks
Find the value of x in the following figure.
Image
Answer
In the right angled $\triangle A B C$,
$A C=13$ $cm, B C=12$ $cm, A B=x$
By Pythagoras theorem,
$(A C)^2 =(A B)^2+(B C)^2 $
$\Rightarrow 13^2 =x^2+12^2 $
$\Rightarrow 169 =x^2+144 $
$\Rightarrow \quad x^2 =169-144 $
$\Rightarrow \quad x^2 =25 $
$\Rightarrow \quad x =\sqrt{25}=5$ $cm$
View full question & answer→
Question 282 Marks
State whether the following triangle exists.
Image
Answer
No, the given triangle does not exist, since it does not satisfy the triangle inequality.
In the given $\triangle A B C$,
$\overline{A B}=24$ $cm, \overline{A C}=12$ $cm, \overline{B C}=11$ $cm$
Now $\overline{A C}+\overline{B C}=12+11=23$
and $\quad \overline{A B}=24$ $cm$
$\therefore \quad \overline{A C}+\overline{B C}<\overline{A B}$
Hence, the given triangle does not exist.
View full question & answer→
Question 292 Marks
The measures of three angles of a triangle are in the ratio $2: 3: 1$. Find the measures of these angles.
Answer
Let measures of the given angles of a triangle be $2 x, 3 x$ and $x$.
$\because$ Sum of all the angles in a triangle $=180^{\circ}$
$\therefore 2 x+3 x+x=180^{\circ} $
$\Rightarrow 6 x=180^{\circ} $
$\Rightarrow x=\frac{180^{\circ}}{6}=30^{\circ}$
View full question & answer→
Question 302 Marks
In the following figure, find the value of y.
Image
Answer
In the given figure, two sides of the triangle are equal. We know that the base angles opposite to equal side are equal, so other base angle will be $45^{\circ}$.
Also, in a triangle, the sum of three angles is equal to $180^{\circ}$.
$\text { So, } y+45^{\circ}+45^{\circ}=180^{\circ} $
$\Rightarrow \quad y \quad 180^{\circ}-90^{\circ}=90^{\circ}$
View full question & answer→
Question 312 Marks
In the following figure, find the value of x.
Image
Answer
In the given figure, two sides of the triangle are equal.
So, $\quad x=60^{\circ}$
$[\because$ angles opposite to the equal sides are equal]
Hence, the value of $x$ is $60^{\circ}$.
View full question & answer→
Question 322 Marks
Find the values of the unknown interior angle in the following figures.
Image
Answer
(i) $\because$ Exterior angle $=110^{\circ}$
One of the interior opposite angle $=40^{\circ}$
By exterior angle property of triangle,
$x+40^{\circ}=110^{\circ} $
$\Rightarrow \quad x=110^{\circ}-40^{\circ}=70^{\circ}$
(ii) $\because$ Exterior angle $=130^{\circ}$
One of the interior opposite angle $=45^{\circ}$
By exterior angle property of triangle,
$130^{\circ}  =45^{\circ}+x $
$\Rightarrow \quad x  =130^{\circ}-45^{\circ}=85^{\circ}$
View full question & answer→
Question 332 Marks
Draw rough sketch for each of the following.
(i) in $\triangle A B C, B D$ is a median.
(ii) In $\triangle A B C, B C$ and $A C$ are altitudes of the triangle.
Answer
i. In the following figure, $B D$ is a median of $\triangle A B C$.
Image
ii. In the following figure, $\overline{B C}$ and $\overline{A C}$ are altitudes of the triangle.
Image
View full question & answer→
Generate Paper FreeAll The Triangle and Its Properties topics
2 Marks Questions - MATHS STD 7 Questions - Vidyadip