Exercise 6.5 (Revised) - Chapter 6 - Triangles & Its Properties - Ncert Solutions class 7 - Maths

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Chapter 6 - Triangles & Its Properties | NCERT Solutions Class 7 Maths

Ex 6.5 Question 1.

$\mathrm{PQR}$ is a triangle, right angled at $\mathrm{P}$. If $\mathrm{PQ}=10 \mathrm{~cm}$ and $\mathrm{PR}=24 \mathrm{~cm}$, find $\mathrm{QR}$.
Answer:

Given: $\mathrm{PQ}=10 \mathrm{~cm}, \mathrm{PR}=24 \mathrm{~cm}$

Let $\mathrm{QR}$ be $x \mathrm{~cm}$.
In right angled triangle QPR,

$
(\text { Hypotenuse })^2=(\text { Base })^2+(\text { Perpendicular })^2
$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(Q R)^2=(P Q)^2+(P R)^2 \\
& \Rightarrow x^2=(10)^2+(24)^2 \\
& \Rightarrow x^2=100+576=676 \\
& \Rightarrow x=\sqrt{676}=26 \mathrm{~cm}
\end{aligned}
$

Thus, the length of $Q R$ is $26 \mathrm{~cm}$.
Ex 6.5 Question 2.

$\mathrm{ABC}$ is a triangle, right angled at $\mathrm{C}$. If $\mathrm{AB}=25 \mathrm{~cm}$ and $\mathrm{AC}=7 \mathrm{~cm}$, find $\mathrm{BC}$.
Answer:

Given: $\mathrm{AB}=25 \mathrm{~cm}, \mathrm{AC}=7 \mathrm{~cm}$
Let BC be $x \mathrm{~cm}$.
In right angled triangle ACB,

$(\text { Hypotenuse })^2=(\text { Base })^2+(\text { Perpendicular })^2$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(A B)^2=(A C)^2+(B C)^2 \\
& \Rightarrow(25)^2=(7)^2+x^2 \\
& \Rightarrow 625=49+x^2 \\
& \Rightarrow x^2=625-49=576 \\
& \Rightarrow x=\sqrt{576}=24 \mathrm{~cm}
\end{aligned}
$

Thus, the length of BC is $24 \mathrm{~cm}$.
Ex 6.5 Question 3.

A $15 \mathrm{~m}$ long ladder reached a window $12 \mathrm{~m}$ high from the ground on placing it against a wall at a distance $\boldsymbol{a}$. Find the distance of the foot of the ladder from the wall.

Answer:

Let AC be the ladder and A be the window.
Given: $\mathrm{AC}=15 \mathrm{~m}, \mathrm{AB}=12 \mathrm{~m}, \mathrm{CB}=a \mathrm{~m}$
In right angled triangle $\mathrm{ACB}$,

$(\text { Hypotenuse })^2=(\text { Base })^2+(\text { Perpendicular })^2$

[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(A C)^2=(C B)^2+(A B)^2 \\
& \Rightarrow(15)^2=(a)^2+(12)^2 \\
& \Rightarrow 225=a^2+144 \\
& \Rightarrow a^2=225-144=81 \\
& \Rightarrow a=\sqrt{81}=9 \mathrm{~cm}
\end{aligned}
$

Thus, the distance of the foot of the ladder from the wall is $9 \mathrm{~m}$.
Ex 6.5 Question 4.

Which of the following can be the sides of a right triangle?
1. $2.5 \mathrm{~cm}, 6.5 \mathrm{~cm}, 6 \mathrm{~cm}$
2. $2 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$
3. $1.5 \mathrm{~cm}, 2 \mathrm{~cm}, 2.5 \mathrm{~cm}$

In the case of right angled triangles, identify the right angles.
Answer:

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,
$
(\text { Hypotenuse })^2=(\text { Base })^2+(\text { Perpendicular })^2
$
(i) $2.5 \mathrm{~cm}, 6.5 \mathrm{~cm}, 6 \mathrm{~cm}$

$\text { In } \Delta \mathrm{ABC},(\mathrm{AC})^2=(\mathrm{AB})^2+(\mathrm{BC})^2$

.

$
\begin{aligned}
& \text { L.H.S. }=(6.5)^2=42.25 \mathrm{~cm} \\
& \text { R.H.S. }=(6)^2+(2.5)^2=36+6.25=42.25 \mathrm{~cm}
\end{aligned}
$

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side $6.5 \mathrm{~cm}$, i.e., at B.
(ii) $2 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$
$
\begin{aligned}
& (5)^2=(2)^2+(2)^2 \\
& \text { L.H.S. }=(5)^2=25
\end{aligned}
$
$
\text { R.H.S. }=(2)^2+(2)^2=4+4=8
$

Since, L.H.S. $\neq$ R.H.S.
Therefore, the given sides are not of the right angled triangle.
(iii) $1.5 \mathrm{~cm}, 2 \mathrm{~cm}, 2.5 \mathrm{~cm}$
$
\text { In } \Delta \mathrm{PQR},(\mathrm{PR})^2=(\mathrm{PQ})^2+(\mathrm{RQ})^2
$

$
\begin{aligned}
& \text { L.H.S. }=(2.5)^2=6.25 \mathrm{~cm} \\
& \text { R.H.S. }=(1.5)^2+(2)^2=2.25+4=6.25 \mathrm{~cm}
\end{aligned}
$

Since, L.H.S. $=$ R.H.S.

Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side $2.5 \mathrm{~cm}$, i.e., at Q.
Ex 6.5 Question 5.

A tree is broken at a height of $5 \mathrm{~m}$ from the ground and its top touches the ground at a distance of $12 \mathrm{~m}$ from the base of the tree. Find the original height of the tree.

Answer:

Let ACB represents the tree before it broken at the point C and let the top A' touches the ground at A after it broke. Then $\triangle \mathrm{ABC}$ is a right angled triangle, right angled at $\mathrm{B}$.

$\mathrm{AB}=12 \mathrm{~m} \text { and } \mathrm{BC}=5 \mathrm{~m}$

Using Pythagoras theorem, In $\Delta \mathrm{ABC}$
$
\begin{aligned}
& (\mathrm{AC})^2=(\mathrm{AB})^2+(\mathrm{BC})^2 \\
& \Rightarrow(\mathrm{AC})^2=(12)^2+(5)^2 \\
& \Rightarrow(\mathrm{AC})^2=144+25 \\
& \Rightarrow(\mathrm{AC})^2=169 \\
& \Rightarrow \mathrm{AC}=13 \mathrm{~m}
\end{aligned}
$

Hence, the total height of the tree $=\mathrm{AC}+\mathrm{CB}=13+5=18 \mathrm{~m}$.
Ex 6.5 Question 6.

Angles $\mathrm{Q}$ and $\mathrm{R}$ of a $\triangle \mathrm{PQR}$ are $25^{\circ}$ and $65^{\circ}$.
Write which of the following is true:
1. $P Q^2+Q R^2=R P^2$
2. $P Q^2+R P^2=Q R^2$
3. $R P^2+Q R^2=P Q^2$
$
25^{\circ}, 65^{\circ}
$

Answer:

In $\Delta \mathrm{PQR}$,
$
\angle \mathrm{PQR}+\angle \mathrm{QRP}+\angle \mathrm{RPQ}=180^{\circ}
$

$\text { [By Angle sum property of a } \Delta \text { ] }$

$
\begin{aligned}
& \Rightarrow 25^{\circ}+65^{\circ}+\angle \mathrm{RPQ}=180^{\circ} \\
& \Rightarrow 90^{\circ}+\angle \mathrm{RPQ}=180^{\circ} \\
& \Rightarrow \angle \mathrm{RPQ}=180^{\circ}-90^{\circ}=90^{\circ}
\end{aligned}
$

Thus, $\triangle \mathrm{PQR}$ is a right angled triangle, right angled at $\mathrm{P}$.
$\therefore$ (Hypotenuse)2 $=$ (Base)2 + (Perpendicular)2 [By Pythagoras theorem]
$
\Rightarrow(\mathrm{QR})^2=(\mathrm{PR})^2+(\mathrm{QP})^2
$

Hence, Option (ii) is correct.
Ex 6.5 Question 7.

Find the perimeter of the rectangle whose length is $40 \mathrm{~cm}$ and a diagonal is 41 $\mathrm{cm}$.

Answer:

Given diagonal $(\mathrm{PR})=41 \mathrm{~cm}$, length $(\mathrm{PQ})=40 \mathrm{~cm}$
Let breadth (QR) be $x \mathrm{~cm}$.
Now, in right angled triangle $\mathrm{PQR}$,

$
(\mathrm{PR})^2=(\mathrm{RQ})^2+(\mathrm{PQ})^2
$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(41)^2=x^2+(40)^2 \\
& \Rightarrow 1681=x^2+1600 \Rightarrow x^2=1681-1600 \\
& \Rightarrow x^2=81 \Rightarrow x=\sqrt{81}=9 \mathrm{~cm}
\end{aligned}
$

Therefore the breadth of the rectangle is $9 \mathrm{~cm}$.
Perimeter of rectangle $=2$ (length + breadth $)$

$
\begin{aligned}
& =2(9+40) \\
& =2 \times 49=98 \mathrm{~cm}
\end{aligned}
$

Hence, the perimeter of the rectangle is $98 \mathrm{~cm}$.
Ex 6.5 Question 8.

The diagonals of a rhombus measure $16 \mathrm{~cm}$ and $30 \mathrm{~cm}$. Find its perimeter.
Answer:

Given: Diagonals AC $=30 \mathrm{~cm}$ and $\mathrm{DB}=16 \mathrm{~cm}$.
Since the diagonals of the rhombus bisect at right angle to each other.
Therefore, $\mathrm{OD}=\frac{\mathrm{DB}}{2}=\frac{16}{2}=8 \mathrm{~cm}$
And $\mathrm{OC}=\frac{\mathrm{AC}}{2}=\frac{30}{2}=15 \mathrm{~cm}$

Now, In right angle triangle DOC,
$(\mathrm{DC})^2=(\mathrm{OD})^2+(\mathrm{OC})^2$ [By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(\mathrm{DC})^2=(8)^2+(15)^2 \\
& \Rightarrow(\mathrm{DC})^2=64+225=289 \\
& \Rightarrow \mathrm{DC}=\sqrt{289}=17 \mathrm{~cm}
\end{aligned}
$

Perimeter of rhombus $=4 \mathrm{x}$ side
$
=4 \times 17=68 \mathrm{~cm}
$

Thus, the perimeter of rhombus is $68 \mathrm{~cm}$.