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CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 7 - MCQExams.com
CBSE
Class 7 Maths
The Triangle And Its Properties
Quiz 7
Which one of the following triangle is not an acute isosceles triangle?
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Explanation
Acute isosceles triangle has two equal angles. The first and last option is not an acute isosceles triangle, because the three angles are different.
If m$$\angle$$ $$PRQ = 45^o$$ and m$$\angle$$ $$QPR = 68^o$$. What is m$$\angle$$ $$PQR$$ in the triangle?
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$$67^o$$
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$$68^o$$
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$$69^o$$
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$$70^o$$
Explanation
by angle sum property, the sum of angles is $$180^o$$.
$$m\angle PRQ + m\angle QPR + m\angle PQR$$ $$= 180^o$$
$$45^o + 68^o +$$ m $$\angle$$ $$PQR = 180^o$$
m$$\angle$$ $$PQR = 180^o - 113^o$$
m$$\angle$$ $$PQR = 67^o$$ .
In $$\Delta ABC, \angle A = 56^{o}$$ and $$\angle B = 60^{o}.$$ What is the measure of $$\angle C?$$
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$$64^{o}$$
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$$65^{o}$$
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$$66^{o}$$
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$$67^{o}$$
Explanation
By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$56^{o} + 60^{o} + \angle C = 180^{o}$$
$$\angle C = 180^{o} - 116^{o}$$
$$\angle C = 64^{o}$$
Identify the name of the triangle.
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acute isosceles triangle
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acute scalene triangle
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acute obtuse triangle
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acute right-angled triangle
Explanation
We can see that, in the given triangle, the two angles are equal and all angles are less than $$90^{o}$$.
Hence, the given triangle is an acute isosceles triangle.
Find the value of $$X.$$
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$$60^{o}$$
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$$70^{o}$$
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$$80^{o}$$
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$$90^{o}$$
Explanation
The exterior angle is equal to the sum of the two non-adjacent interior angles which are $$70^{o}$$ and $$10^{o}.$$
So, $$X = 70^{o} + 10^{o}$$
$$X = 80^{o}.$$
The angles of a triangle are in the ratio of $$3:5:7.$$ What is the measure of the largest interior angle of the triangle?
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$$18^{o}$$
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$$48^{o}$$
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$$84^{o}$$
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$$96^{o}$$
Explanation
By angle sum property, the sum of angles is $$180^o$$.
Let $$3x, 5x$$ and $$7x$$ be the three angles of the triangle.
So,
$$3x + 5x + 7x = 180^{o}$$
$$15x = 180^{o}$$
$$x = 12^{o}$$
Hence, Largest interior angle $$= 7 \times 12^{o} = 84^{o}$$
In $$\Delta ABC,$$ $$\angle A = 43^{o}$$ and $$\angle C = 70^{o}.$$ What is the measure of $$\angle B?$$
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$$63^{o}$$
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$$65^{o}$$
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$$66^{o}$$
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$$67^{o}$$
Explanation
By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$43^{o} +\angle B + 70^{o} = 180^{o}$$
$$\angle B = 180^{o} - 113^{o}$$
$$\angle B = 67^{o}.$$
Find the measure of an exterior angle at the base of an isosceles triangle measures $$78^o$$
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$$100^o$$
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$$101^o$$
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$$102^o$$
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$$103^o$$
Explanation
Base angles of an isosceles triangle are equal
$$\therefore \angle B=\angle C=78^{\circ}$$
Now $$x+\angle C=180^{\circ}$$ (Angles made on straight line)
$$x=180^{\circ}-78^{\circ}=102^{\circ}$$
So exterior angle is $$102^{\circ}$$
Find the value of exterior angle.
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$$100^{\circ}$$
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$$110^{\circ}$$
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$$120^{\circ}$$
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$$115^{\circ}$$
Explanation
An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$x - 20^{\circ} + 2x - 45^{\circ} = 2x - 5^{\circ}$$
$$3x - 65^{\circ} = 2x - 5^{\circ}$$
$$3x - 2x = 65^{\circ} - 5^{\circ}$$
$$x = 60^{\circ}$$
Therefore, $$\angle$$ $$F$$ exterior angle $$= 2x - 5^{\circ}$$
$$= 2\times 60^{\circ} - 5^{\circ} = 115^{\circ}$$
In $$\Delta$$ $$PQR$$, an exterior angle at $$R$$ is represented by $$5x + 10$$. If the two non-adjacent interior angles are represented by $$3x + 15$$ and $$3x - 20$$, find the value of the exterior angle, $$\angle$$ $$R$$.
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$$90^o$$
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$$80^o$$
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$$84^o$$
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$$85^o$$
Explanation
The exterior angle is equal to the sum of the two non-adjacent interior angles.
$$3x + 15 + 3x - 20 = 5x + 10$$
$$6x - 5 = 5x + 10$$
$$6x - 5x = 10 + 5$$
$$x = 15$$
Therefore, $$\angle$$ $$R = 5x + 10 => 5\times15 + 10 = 85^o$$
In $$\triangle ABC$$, the line segment $$AM_1$$ is the median from vertex ____ to the opposite side $$BC$$ at $$M_1$$.
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$$A$$
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$$B$$
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$$C$$
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$$D$$
Explanation
A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
Therefore, $$A$$ is the vertex for the median $$M_{1}$$.
What is the value of $$\angle$$ $$XRQ$$?
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$$100^o$$
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$$110^o$$
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$$120^o$$
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$$130^o$$
Explanation
An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$\angle P+\angle Q = \angle XRQ$$
$$x - 30 + x - 20 = x + 30$$
$$2x - 50 = x + 30$$
$$2x - x = 30 + 50$$
$$x = 80$$
Therefore, $$\angle$$ $$XRQ = x + 30\Rightarrow 80 + 30 = 110^o$$
How many medians(drawn in fig.) are there in a triangle $$ABC$$?
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1
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2
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3
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4
Explanation
A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
So, here the triangle is having only one median.
Find $$\angle$$CAD.
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$$102^o$$
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$$110^o$$
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$$180^o$$
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$$122^o$$
Explanation
An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$71^o$$ and $$51^o$$.
So, $$\angle$$ $$CAD = 71 + 51$$
$$\angle$$ $$CAD = 122^o$$
Find the measure of base angles and exterior angle at the vertex of an isosceles triangle if the vertex angle measures $$45^o$$.
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Base angle $$= 67.5^o$$, exterior angle $$= 135^o$$
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Base angle $$= 75^o$$, exterior angle $$= 135^o$$
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Base angle $$= 135^o$$, exterior angle $$= 45^o$$
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Base angle $$= 67.5^o$$, exterior angle $$= 75^o$$
Explanation
Let $$x$$ be the exterior angle at vertex (in degrees).
Then t
he exterior angle and the interior angle at vertex form a linear pair.
Then,
$$x + 45^o = 180^o$$
$$x = 180^o - 45^o = 135^o$$
The $$2$$ base angles of an isosceles triangle are equal, let us represent each as $$z$$.
Applying angle sum property,
$$z + z + 45 = 180^o$$
$$\Rightarrow 2z = 180^o - 45^o$$
$$\Rightarrow 2z=135^o$$
$$\Rightarrow z=67.5^o$$
So, base angle $$= 67.5^o$$ , exterior angle $$= 135^o$$
In $$\triangle PQR$$, the line segment ____ is the median from vertex $$P$$ to the opposite side $$QR$$ at $$O$$.
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$$\overline{QT}$$
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$$\overline{PO}$$
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$$\overline{OR}$$
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$$\overline{QO}$$
Explanation
A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
Therefore, $$\overline{PO}$$ is the line segment for the median $$O$$.
How many medians(drawn in fig.) are there in a triangle $$PQR$$?
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1
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2
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3
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4
Explanation
A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
So, here the triangle is having two medians.
What is the value of $$x$$ using Pythagoras theorem?
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10
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12
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5
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4
Explanation
In a right triangle, $$a^{2} + b^{2} = h^{2}$$, where $$a$$ and $$b$$ are the
lengths of the legs and $$c$$ is the length of the hypotenuse. This is
called Pythagoras Theorem.
$$a = 5, b = x, c =$$ hypotenuse side $$= 13$$
$$5^{2} + x^{2} = 13^{2}$$
$$169 - 25 =$$ $$x^{2}$$
$$x^{2}$$ $$= 144$$
Taking root on both the sides, we get
$$x = 12$$
Find the value of $$x$$ and $$y$$. (Use Pythagoras theorem).
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$$x = 12$$ and $$y = 3$$
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$$x = 10$$ and $$y = 5$$
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$$x = 12$$ and $$y = 5$$
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$$x = 11$$ and $$y = 5$$
Explanation
In a right triangle, $$a^{2} + b^{2} = c^{2}$$, where a and b are the
lengths of the legs and c is the length of the hypotenuse. This is
called Pythagoras Theorem.
$$a = 3, b = 4, c =$$ hypotenuse side $$= y$$
$$3^{2} + 4^{2} = c^{2}$$
$$9 + 16 =$$ $$c^{2}$$
$$c^{2}$$ $$= 25$$
Taking root on both the sides, we get
$$c = 5 = y$$
So, $$x^{2}+y^{2}= 13^{2}$$
$$x^{2}+5^{2}= 13^{2}$$
$$x^{2} = 169 - 25$$
$$x^{2} = 144$$
$$x = 12$$
So, the value of $$x = 12$$ and $$y = 5$$.
In a triangle $$ABC$$, if $$AB = 4$$ and $$AC = 3$$, what is the length of $$BC$$?
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$$4$$
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$$5$$
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$$5.2$$
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$$6$$
Explanation
Using Pythagoras theorem, we will find $$BC$$,
$$AB^{2}+AC^{2}=BC^{2}$$
$$4^{2}+3^{2}=BC^{2}$$
$$16+9=BC^{2}$$
$$BC = 5$$
In $$\triangle ABC$$, if $$\angle B = \angle C = 45^{\circ}$$, which of the following is correct?
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$$\angle A= 60°$$
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$$\angle A= 70°$$
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$$\angle A= 80°$$
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$$\angle A= 90°$$
Explanation
As $$\angle B=\angle C=45^{\circ}$$ and
by angle sum property, the sum of angles is $$180^o$$.
$$\therefore \angle A+\angle B+\angle C=180^{\circ}$$
$$\Rightarrow \boxed{\angle A=90^{\circ}}$$
If each side of an equilateral triangle is doubled then its angle will ______
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become half
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be doubled
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be tripled
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remain same
Explanation
If each side of an equilateral triangle is doubled then its angle will remain same.As if each side is doubled again an equilateral triangle will be formed. So angle remain the same.
Hence option $$D$$ is correct.
A line is drawn through the diagonal of a rectangle as shown above. What is the length of the diagonal?
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5
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12
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35
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37
The distance from town A to town B is five miles. C is six miles from B. Which of the following could be the distance from A to C?
I. $$11$$
II. $$1$$
III. $$7$$
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I only
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II only
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I and II only
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II and III only
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I, II or III
Explanation
The distance from two A nad B is five miles. C is six miles from B To find distance between A to C there are three possibilities.
Case: $$1$$ In triangle $$ABC,AB=5$$ and $$BC=6$$
Here , the length of AC must be less than the sum of other two sides and greater than the difference of the other sides
$$6.5<6+5$$
$$\Rightarrow 1<Ac<11$$
Thus $$AC=7$$ is a possible
$$\therefore $$ distance from $$A$$ to $$C$$ could be $$11,1 $$ or $$7$$ miles.
Find the length of the side labeled $$x$$. The triangle represented in the figure is a right triangle.
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$$18$$
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$$24$$
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$$20$$
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$$25$$
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$$22$$
Explanation
As it is given that the triangle shown in the figure is a right angle triangle so, by using Pythagoras theorem. We have,
$$\begin{aligned}{}{x^2} + {15^2} &= {25^2}\\{x^2} + 225 &= 625\\x^2& = 625 - 225\\& = 400\\x& = 20\end{aligned}$$
Hence, option $$C$$ is correct.
In a triangle with sides of $$7$$ and $$9$$, the third side must be
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more than $$16$$
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between $$7$$ and $$9$$
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between $$2$$ and $$16$$
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between $$7$$ and $$16$$
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between $$9$$ and $$16$$
Explanation
Given that two sides of triangle are $$7,9$$
Let the third side be $$x$$
We have $$7+9>x$$ and $$x+9>7$$ and $$x+7>9$$ and $$x>0$$
We get $$x<16$$ and $$x>2$$
Therefore the third side will lie between $$2$$ and $$16$$.
In the fig. then , L (DN) = ?
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3.8 cm.
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3.6 cm.
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2.4 cm.
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5.4 cm.
In triangle, three angles are $$x , x + 10 ^ { \circ } + x + 20 ^ { \circ }$$ then the biggest is
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$$70 ^ { \circ }$$
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$$80 ^ { \circ }$$
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$$90 ^ { \circ }$$
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none
A ladder $$25m$$ long is leaning against a wall that is perpendicular to the level ground. The bottom of the ladder is $$7m$$ away from the base of the wall. If the top of the ladder slips down $$4m$$, how much will the bottom of the ladder slip?
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$$7m$$
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$$8m$$
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$$10m$$
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$$15m$$
Explanation
R.E.F first image
In $$ \Delta ABC ,$$
$$ AC^{2} = AB^{2}-BC^{2} $$ {Using Pythagorus theorem}
$$ \Rightarrow AC = \sqrt{25^{2}-7^{2}} $$
$$ \Rightarrow \boxed{AC = 24\,m} $$
As after sliping length of
ladder remains same $$ \Rightarrow \boxed{DE = 25\,cm} \ and\ AD = 4m$$ {REF second Image}
Let say $$x$$ be the length of how much the ladder slips.
New Height $$= DC = AC-AD = 20\,m; \ EC = (x+7)m $$
In $$ \Delta DEC $$,
$$ DE^{2} = DC^{2}+EC^{2} $$
$$ \Rightarrow 25^{2} = 20^{2}+(x+7)^{2}\Rightarrow (x+7)^{2} = 225 $$
$$ \Rightarrow x+7 = 15 \ \ \ (\because (x+7)> 0 $$ , We can't take negative value $$) \Rightarrow \boxed{x = 8\,m} $$
In the given figure, side $$BC$$ of $$\triangle ABC$$ is produced to $$D$$. Find $$\angle A$$
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$$80^{\circ}$$
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$$60^{\circ}$$
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$$ 30^{\circ}$$
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None of these
Explanation
An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$\angle A+\angle B=\angle ACD$$
$$\Rightarrow\angle A +40^{\circ}=120^\circ$$
$$\Rightarrow\angle A=120^\circ-40^\circ$$
$$\Rightarrow\angle A=80^\circ$$
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