NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.5) Exercise 6.5

Q.1

$\begin{array}{l}\text{PQR is a triangle right angled at P. If PQ =10 cm}\\ \text{and PR=24cm,find\hspace{0.33em}QR.}\end{array}$

Ans.

By applying Pythagoras theorem in ΔPQR to get

(PQ)² + (PR)² = (QR)²
10² + 24² = QR²
100 + 576 = QR²
676 = QR²
QR =

$\sqrt{676}$

= 26 cm

Q.2

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{If}\mathrm{AB}=\; 25\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=\; 7\mathrm{cm},\mathrm{find}\mathrm{BC}.\end{array}$

Ans.

\begin{array}{l}\text{By Pythagoras theorem in}\Delta \text{ABC, we get}\\ {\left(\text{AC}\right)}^2+{\left(BC\right)}^2={\left(AB\right)}^2\\ {\left(BC\right)}^2={\left(AB\right)}^2-{\left(AC\right)}^2\\ ={24}^2-7^2\\ =625-49=576\\ \boxed{\text{BC}=\text{24}\text{cm}}\end{array}

Q.3

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

Ans.

\begin{array}{l}\text{By Pythagoras theorem}\\ {\left(15\right)}^2={\left(12\right)}^2+a^2\\ 225-144=a^2\\ 81=a^2\\ \boxed{a=9\text{cm}}\\ \text{Therefore, the distance of the foot of the ladder from the wall}\\ \text{is 9 cm}\text{.}\end{array}

Q.4

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{can}\mathrm{be}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{triangle}?\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.33em}}2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}.\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.33em}}2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}.\\ \left(\mathrm{iii}\right)\mathrm{\hspace{0.33em}}1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5 cm},\text{6}.\text{5 cm},\text{6 cm}.\\ {2.5}^2=6.25,{6.5}^2=42.25{\text{and 6}}^2=36\\ \text{Here, 2}{\text{.5}}^2+6^2={6.5}^2\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{ii}\right)\text{2 cm},\text{2 cm},\text{5 cm}.\\ 2^2=4,2^2=4{\text{and 5}}^2=25\\ {\text{Here, 2}}^2+2^2\ne 5^2\\ \text{So, the square of the length of one side is not the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are not the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{iii}\right)\text{1}.\text{5 cm},\text{2cm},\text{2}.\text{5 cm}\\ {1.5}^2=2.25,2^2=4\text{and 2}{\text{.5}}^2=6.25\\ \text{Here, 1}{\text{.5}}^2+2^2={2.5}^2\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\end{array}

Q.5

$\begin{array}{l}\text{A tree is broken at a height of 5 m from the ground and its}\\ \text{top touches the ground at a distance of 12 m from the}\\ \text{Base of the tree.Find the original height of the tree.}\end{array}$

Ans.

\begin{array}{l}\text{In above figure, BC represents the unbroken part of the tree}\text{.}\\ \text{Point C represents the point where the tree broke and CA}\\ \text{represents the broken part of the tree}\text{.}\\ \text{Triangle ABC thus formed is a right-angled triangle}\text{.}\\ \text{So, applying Pythagoras theorewm to get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ ={12}^2+5^2\\ =144+25\\ =169=13\text{cm}\\ \text{Thus,the original height of the tree is}=\text{AC}+\text{CB}\\ =\text{13 cm}+\text{5 cm}\\ =\boxed{\text{18 cm}}\end{array}

Q.6

$\begin{array}{l}\mathrm{Angles}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{of}\mathrm{a}\mathrm{\Delta PQR}\mathrm{are}25º\mathrm{and}65º.\\ \mathrm{Write}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{true}:\\ \left(\mathrm{i}\right){\mathrm{PQ}}^2+{\mathrm{QR}}^2={\mathrm{RP}}^2\\ \left(\mathrm{ii}\right){\mathrm{PQ}}^2+{\mathrm{RP}}^2={\mathrm{QR}}^2\\ \left(\mathrm{iii}\right){\mathrm{RP}}^2+{\mathrm{QR}}^2={\mathrm{PQ}}^2\end{array}$

Ans.

\begin{array}{l}\text{Since, the sum of all interior angle is 180}°\text{.}\\ \text{So, 25}°\text{+ 65}°+\angle \text{QPR}=180°\\ \angle \text{QPR}=180°-90°\\ =90°\\ \text{Therefore,}\Delta \text{PQR is a rigfht angled triangle at P}\text{.}\\ \text{Thus,}\boxed{{\text{PQ}}^{\text{2}}+{\text{RP}}^{\text{2}}={\text{QR}}^{\text{2}}}\\ \text{So, (ii) is true}\text{.}\end{array}

Q.7

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

Ans.

\begin{array}{l}\text{In a rectangle},\text{all interior angles are of 9}0º\text{measure}.\\ \text{Therefore},\text{Pythagoras theorem can be applied here}.\\ {\left(\text{41}\right)}^{\text{2}}={\left(\text{4}0\right)}^{\text{2}}+x^{\text{2}}\\ \text{1681}=\text{16}00+x^{\text{2}}\\ x^{\text{2}}=\text{1681}-\text{16}00\\ =\text{81}\\ x=\text{9 cm}\\ \text{So, Perimeter}=\text{2}\left(\text{Length}+\text{Breadth}\right)\\ =\text{2}\left(x+\text{4}0\right)\\ =\text{2}\left(\text{9}+\text{4}0\right)\\ =\boxed{\text{98 cm}}\end{array}

Q.8

$\begin{array}{l}\mathrm{The}\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{measure}16\mathrm{cm}\mathrm{and}30\mathrm{cm}.\\ \mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans.

\begin{array}{l}\text{Let ABCD be a rhombus}\left(\text{all sides are of equal length}\right)\text{and its}\\ \text{diagonals},\text{AC and BD},\text{are intersecting each other at point O}.\\ \text{Since, diagonals of a rhombus bisect each other at 9}0°.\\ \text{So,}\\ \text{By applying Pythagoras theorem in}\Delta \text{AOB},\\ {\text{OA}}^{\text{2}}+{\text{OB}}^{\text{2}}={\text{AB}}^{\text{2}}\\ {\text{8}}^{\text{2}}+{\text{15}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{64}+\text{225}={\text{AB}}^{\text{2}}\\ \text{289}={\text{AB}}^{\text{2}}\\ \text{AB}=\text{17 cm}\\ \text{Therefore},\text{the length of the side of rhombus is 17 cm}.\\ \text{Perimeter of rhombus}=\text{4}\times \text{Side of the rhombus}\\ =\text{4}\times \text{17}\\ =\boxed{\text{68 cm}}\end{array}