NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.2) Exercise 11.2

Q.1 Find the area of each of the following parallelogram :

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4 cm}\\ \text{Base}=\text{7 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{7}\times \text{4}\\ ={\text{28cm}}^2\\ \text{(b)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{3}\\ ={\text{15 cm}}^2\\ \text{(c)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3}\text{.5 cm}\\ \text{Base}=\text{2}\text{.5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\text{.5}\times \text{3}\text{.5}\\ =\text{8}{\text{.75 cm}}^2\end{array} \begin{array}{l}\text{(d)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.8 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{4}\text{.8}\\ ={\text{24 cm}}^2\\ \text{(e)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.4 cm}\\ \text{Base}=\text{2 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\times \text{4}\text{.4}\\ =\text{8}{\text{.8 cm}}^2\end{array}

Q.2 Find the area of each of the following triangles :

Ans

\begin{array}{l}\text{(a) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 4\times 3\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(b) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 5\times 3.2\\ =\frac{1}{2}\times 16=8{\text{cm}}^2\\ \text{(c) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 4\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(d) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 2\\ =\frac{1}{2}\times 6=3{\text{cm}}^2\end{array}

Q.3 Find the missing values :

Ans

\begin{array}{l}\text{Let the height be}h\text{and base be}b.\\ \text{(a) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ {\text{246 cm}}^{\text{2}}=\text{20 cm}\times \text{h}\\ \text{h=}\frac{246{\text{cm}}^2}{20\text{cm}}\\ =12.3\text{cm}\\ \text{(b) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{154}{\text{.5 cm}}^{\text{2}}=\text{b}\times \text{15 cm}\\ \text{b=}\frac{154.5{\text{cm}}^2}{15\text{cm}}\\ =10.3\text{cm}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(c) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{78}{\text{.72 cm}}^{\text{2}}=\text{b}\times \text{8}\text{.4 cm}\\ \text{b=}\frac{48.72{\text{cm}}^2}{8.4\text{cm}}\\ =5.8\text{cm}\\ \text{(d) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{16}{\text{.38 cm}}^{\text{2}}=\text{15}\text{.6 cm}\times \text{h}\\ \text{h=}\frac{16.38{\text{cm}}^2}{15.6\text{cm}}\\ =1.05\text{cm}\\ \text{So, we get}\\ \begin{array}{cccc}\text{S}\text{.No}\text{.}& \text{Base}& \text{Height}& \text{Area of the Parallelogram}\\ \text{a}\text{.}& 20\text{cm}& 12.3\text{cm}& 246{\text{cm}}^2\\ \text{b}\text{.}& 10.3\text{cm}& 15\text{cm}& 154.5{\text{cm}}^2\\ \text{c}\text{.}& 5.8\text{cm}& 8.4\text{cm}& 78.72{\text{cm}}^2\\ \text{d}\text{.}& 15.6\text{cm}& 1.05\text{cm}& 16.38{\text{cm}}^2\end{array}\end{array}

Q.4 Find the missing values :

Ans

\begin{array}{l}\text{(a) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{87cm}}^2=\frac{1}{2}\times 15\text{cm}\times \text{h}\\ \text{h}=\frac{87{\text{cm}}^2\times 2}{15\text{cm}}\\ =11.6\text{cm}\\ \text{(b) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{1256 mm}}^2=\frac{1}{2}\times b\times \text{31}\text{.4 mm}\\ b=\frac{1256{\text{mm}}^2\times 2}{31.4\text{mm}}\\ =80\text{mm}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(c) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ \text{170}{\text{.5 cm}}^2=\frac{1}{2}\times 22\text{cm}\times \text{h}\\ \text{h}=\frac{170.5{\text{cm}}^2\times 2}{22\text{cm}}\\ =15.5\text{cm}\\ \text{So we get}\\ \begin{array}{ccc}\text{Base}& \text{Height}& \text{Area of Triangle}\\ 15\text{cm}& 11.6\text{cm}& 84{\text{cm}}^2\\ 80\text{mm}& 31.4\text{mm}& 1256{\text{mm}}^2\\ 22\text{cm}& 15.5\text{cm}& 170.5{\text{cm}}^2\end{array}\end{array}

Q.5

$\begin{array}{l}\mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\hspace{0.33em}\mathrm{QM}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{SRn}\\ \mathrm{and}\mathrm{QN}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{PS}.\mathrm{If}\mathrm{SR}=12\mathrm{cm}\mathrm{andn}\\ \mathrm{QM}=\; 7.6\mathrm{cm}.n\mathrm{Find}:\\ \left(\mathrm{a}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallegram}\mathrm{PQRS}\\ \mathrm{QN},\mathrm{if}\mathrm{PS}=\; 8\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{SR}\times \text{QM}\\ =7.6\times 12\\ =91.2{\text{cm}}^2\\ \text{(b)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{PS}\times \text{QN}\\ =91.2{\text{cm}}^2\\ \text{QN}\times 8=91.2\\ \text{QN}=\frac{91.2}{8}\\ =11.4\text{cm}\end{array}

Q.6

$\begin{array}{l}\mathrm{DL}\mathrm{and}\mathrm{BM}\mathrm{are}\mathrm{the}\mathrm{heights}\mathrm{on}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{AD}\mathrm{respectively}\\ \mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}.\mathrm{If}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{is}\\ 1470{\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}=\; 35\mathrm{cm}\mathrm{and}\mathrm{AD}=\; 49\mathrm{cm},\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\\ \mathrm{BMand}\mathrm{DL}.\end{array}$

Ans

\begin{array}{l}\text{Area of parallelogram}=\text{Base}\times \text{Height}\\ =\text{AB}\times \text{DL}\\ \text{1470}=\text{35}\times \text{DL}\\ \text{DL}=\frac{1470}{35}\\ =42\text{cm}\\ \text{Also,}\\ 1470=\text{AD}\times \text{BM}\\ \text{1470}=\text{49}\times \text{BM}\\ \text{BM}=\frac{1470}{49}\\ =30\text{cm}\end{array}

Q.7

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{AD}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}.\\ \mathrm{If}\mathrm{AB}=5\mathrm{cm},\hspace{0.33em}\mathrm{BC}=13\mathrm{cm}\mathrm{and}\mathrm{AC}=12\mathrm{cm},\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\\ △\mathrm{ABC}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{AD}.\end{array}$

Ans

\begin{array}{l}\text{Since, Area}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times \text{5}\times \text{12}\\ ={\text{30 cm}}^2\\ \text{Also, area of triangle}=\frac{1}{2}\times \text{AD}\times \text{BC}\\ \text{30}=\frac{1}{2}\times \text{AD}\times \text{13}\\ \text{AD}=\frac{30\times 2}{12}\\ =4.6\text{cm}\end{array}

Q.8

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{with}\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}\mathrm{and}\mathrm{BC}=9\mathrm{cm}.\\ \mathrm{The}\mathrm{height}\mathrm{AD}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{is}6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}△\mathrm{ABC}.\\ \mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{C}\mathrm{to}\mathrm{AB}\mathrm{i}.\mathrm{e}.,\mathrm{CE}?\end{array}$

Ans

\begin{array}{l}\text{Area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{BC}\times \text{AD}=\frac{1}{2}\times \text{9}\times \text{6}={\text{27cm}}^{\text{2}}\\ \text{Also, area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{AB}\times \text{CE}\\ 27=\frac{1}{2}\text{7}\text{.5}\times \text{CE}\\ \text{CE}=\frac{27\times 2}{7.5}=\text{7}\text{.2 cm}\end{array}