NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 are given here in easy, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Chapter 11 teaches perimeter, which is the length around a closed shape, and area, which is the space inside it. Exercise 11.1 covered squares and rectangles. Exercise 11.2 moves on to two new shapes: the parallelogram and the triangle. In the latest NCERT book this chapter is numbered Chapter 9, but the questions are the same.

This exercise has 8 questions. Some ask you to find the area when the base and height are given. Some give you the area and ask you to find the missing base or height. The last few are word problems on parallelograms and triangles where you find a second height using the same area. Every question below is solved in short steps, with the final answer in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2

Perimeter and Area
Medium
Q.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Perimeter and Area
Easy
Q.
A circular flower garden has an area of 314 m2. A sprinklerat the centre of the garden can cover an area that has aradius of 12 m. Will the sprinkler water the entire garden?Takeπ= 3.14
Perimeter and Area
Medium
Q.
Agardenerwantstofenceacirculargardenofdiameter21m.Findthelengthoftheropeheneedstopurchase,ifhemakes2roundsoffence.Alsofindthecostsoftherope,ifitcost₹4permeter.(Takeπ=227) \begin{array}{l}\mathrm{A}\quad \mathrm{gardener}\quad \mathrm{wants}\quad \mathrm{to}\quad \mathrm{fence}\quad \mathrm{a}\quad \mathrm{circular}\quad \mathrm{garden}\quad \mathrm{of}\quad \mathrm{diameter}\\ 21\mathrm{m}.\quad \mathrm{Find}\quad \mathrm{the}\quad \mathrm{length}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{rope}\quad \mathrm{he}\quad \mathrm{needs}\quad \mathrm{to}\quad \mathrm{purchase},\quad \mathrm{if}\quad \mathrm{he}\\ \mathrm{makes}\quad 2\quad \mathrm{rounds}\quad \mathrm{of}\quad \mathrm{fence}.\quad \mathrm{Also}\quad \mathrm{find}\quad \mathrm{the}\quad \mathrm{costs}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{rope},\quad \mathrm{if}\quad \mathrm{it}\\ \mathrm{cost}\quad ₹\quad 4\quad \mathrm{per}\quad \mathrm{meter}.\left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}

Perimeter and Area
Medium
Q.
Adooroflength2mandbreadth1misfittedinawall.Thelengthofthewallis4.5mandthebreadthis3.6m.Findthecostofwhitewashingthewall,iftherateofwhitewashingthewallis₹20perm2. \begin{array}{l}\mathrm{A}\quad \mathrm{door}\quad \mathrm{of}\quad \mathrm{length}\quad 2\quad \mathrm{m}\quad \mathrm{and}\quad \mathrm{breadth}\quad 1\mathrm{m}\quad \mathrm{is}\quad \mathrm{fitted}\quad \mathrm{in}\quad \mathrm{a}\quad \mathrm{wall}.\\ \mathrm{The}\quad \mathrm{length}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{wall}\quad \mathrm{is}\quad 4.5\quad \mathrm{m}\quad \mathrm{and}\quad \mathrm{the}\quad \mathrm{breadth}\quad \mathrm{is}\quad 3.6\quad \mathrm{m}\quad .\\ \mathrm{Find}\quad \mathrm{the}\quad \mathrm{cost}\quad \mathrm{of}\quad \mathrm{white}\quad \mathrm{washing}\quad \mathrm{the}\quad \mathrm{wall},\quad \mathrm{if}\quad \mathrm{the}\quad \mathrm{rate}\quad \mathrm{of}\\ \mathrm{white}\quad \mathrm{washing}\quad \mathrm{the}\quad \mathrm{wall}\quad \mathrm{is}\quad ₹\quad 20\quad \mathrm{per}\quad {\mathrm{m}}^{\mathrm{2}}\mathrm{.}\end{array}

Perimeter and Area
Medium
Q.
FindtheareaofthequadrilateralABCD.Here,AC=22cm,BM=3cm,DN=3cm,andBM⊥AC,DN⊥AC. \begin{array}{l}\mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{quadrilateral}\quad \mathrm{ABCD}.\\ \mathrm{Here},\quad \mathrm{AC}=22\quad \mathrm{cm},\quad \mathrm{BM}=3\quad \mathrm{cm},\mathrm{DN}=3\quad \mathrm{cm},\\ \quad \mathrm{and}\quad \mathrm{BM}\bot \mathrm{AC},\quad \mathrm{DN}\bot \mathrm{AC}.\end{array}

Perimeter and Area
Medium
Q.
In the following figures, find the area of the shaded portions:
 

Perimeter and Area
Medium
Q.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
 

Perimeter and Area
Medium
Q.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Perimeter and Area
Medium
Q.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Ï€ = 3.14)
Perimeter and Area
Difficult
Q.
Find the area of each of the following parallelogram :
 

Perimeter and Area
Medium
Q.
Fromacircularcardsheetofradius14cm,twocirclesofradius3.5cmandarectangleoflength3cmandbreadth1cmareremoved.(asshownintheadjoiningfigure).Findtheareaoftheremainingsheet.(Takeπ=227) \begin{array}{l}\mathrm{From}\quad \mathrm{a}\quad \mathrm{circular}\quad \mathrm{card}\quad \mathrm{sheet}\quad \mathrm{of}\quad \mathrm{radius}\quad 14\quad \mathrm{cm},\quad \mathrm{two}\quad \mathrm{circles}\quad \mathrm{of}\\ \mathrm{radius}\quad 3.5\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{a}\quad \mathrm{rectangle}\quad \mathrm{of}\quad \mathrm{length}\quad 3\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{breadth}\\ 1\quad \mathrm{cm}\quad \mathrm{are}\quad \mathrm{removed}.\quad (\mathrm{as}\quad \mathrm{shown}\quad \mathrm{in}\quad \mathrm{the}\quad \mathrm{adjoiningfigure}).\\ \mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{remaining}\quad \mathrm{sheet}.\quad \left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}

Perimeter and Area
Medium
Q.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Perimeter and Area
Medium
Q.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Perimeter and Area
Medium
Q.
Findtheareaofthefollowingcircles,giventhat:(a)radius=14mm(Takeπ=227)(b)diameter=49m(c)radius=5cm \begin{array}{l}\mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{following}\quad \mathrm{circles},\quad \mathrm{given}\quad \mathrm{that}:\\ \left(\mathrm{a}\right)\quad \mathrm{radius}\quad =\quad 14\quad \mathrm{mm}\quad \left(\mathrm{Take}\quad \mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{b}\right)\quad \mathrm{diameter}\quad =\quad 49\quad \mathrm{m}\left(\mathrm{c}\right)\quad \mathrm{radius}\quad =\quad 5\quad \mathrm{cm}\end{array}
Perimeter and Area
Medium
Q.
â–³ABCisisosceleswithAB=AC=7.5cmandBC=9cm.TheheightADfromAtoBC,is6cm.Findtheareaofâ–³ABC.WhatwillbetheheightfromCtoABi.e.,CE? \begin{array}{l}\triangle \mathrm{ABC}\quad \mathrm{is}\quad \mathrm{isosceles}\quad \mathrm{with}\quad \mathrm{AB}=\mathrm{AC}=7.5\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{BC}=9\quad \mathrm{cm}.\quad \\ \mathrm{The}\quad \mathrm{height}\quad \mathrm{AD}\quad \mathrm{from}\quad \mathrm{A}\quad \mathrm{to}\quad \mathrm{BC},\quad \mathrm{is}\quad 6\quad \mathrm{cm}.\quad \mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \triangle \mathrm{ABC}.\quad \\ \mathrm{What}\quad \mathrm{will}\quad \mathrm{be}\quad \mathrm{the}\quad \mathrm{height}\quad \mathrm{from}\quad \mathrm{C}\quad \mathrm{to}\quad \mathrm{AB}\quad \mathrm{i}.\mathrm{e}.,\quad \mathrm{CE}?\end{array}

Perimeter and Area
Medium
Q.
â–³ABCisrightangledatA.ADisperpendiculartoBC.IfAB=5cm,BC=13cmandAC=12cm,Findtheareaofâ–³ABC.AlsofindthelengthofAD. \begin{array}{l}\triangle \mathrm{ABC}\quad \mathrm{is}\quad \mathrm{right}\quad \mathrm{angled}\quad \mathrm{at}\quad \mathrm{A}\quad .\mathrm{AD}\quad \mathrm{is}\quad \mathrm{perpendicular}\quad \mathrm{to}\quad \mathrm{BC}.\\ \mathrm{If}\quad \mathrm{AB}=5\quad \mathrm{cm},\hspace{0.33em}\mathrm{BC}=13\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{AC}=12\quad \mathrm{cm},\quad \mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\\ \triangle \mathrm{ABC}.\quad \mathrm{Also}\quad \mathrm{find}\quad \mathrm{the}\quad \mathrm{length}\quad \mathrm{of}\quad \mathrm{AD}.\end{array}

Perimeter and Area
Medium
Q.
Find the area of each of the following triangles :
Perimeter and Area
Medium
Q.
Averandahofwidth2.25misconstructedallalongoutsidearoomwhichis5.5mlongand4mwide.Find:(i)theareaoftheverandah.(ii)thecostofcementingtheflooroftheverandahattherateofRs200perm2 \begin{array}{l}\mathrm{A}\quad \mathrm{verandah}\quad \mathrm{of}\quad \mathrm{width}\quad 2.25\quad \mathrm{m}\quad \mathrm{is}\quad \mathrm{constructed}\quad \mathrm{all}\quad \mathrm{along}\\ \mathrm{outside}\quad \mathrm{a}\quad \mathrm{room}\quad \mathrm{which}\quad \mathrm{is}\quad 5.5\quad \mathrm{mlong}\quad \mathrm{and}\quad 4\quad \mathrm{m}\quad \mathrm{wide}.\quad \mathrm{Find}:\\ \left(\mathrm{i}\right)\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{verandah}.\\ \left(\mathrm{ii}\right)\quad \mathrm{the}\quad \mathrm{cost}\quad \mathrm{of}\quad \mathrm{cementing}\quad \mathrm{the}\quad \mathrm{floor}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{verandah}\quad \mathrm{at}\quad \mathrm{the}\\ \mathrm{rate}\quad \mathrm{of}\quad \mathrm{Rs}\quad 200\quad \mathrm{per}\quad {\mathrm{m}}^{\mathrm{2}}\end{array}

NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2

Formulas Used in Exercise 11.2

  • Area of a parallelogram = base × height
  • Area of a triangle = 1/2 × base × height
  • To find the height: height = area ÷ base (for a parallelogram)
  • To find the height of a triangle: height = (2 × area) ÷ base

Remember one thing. The height is always the perpendicular distance from the base to the opposite side or corner. It is not the slanting side.

Exercise 11.2 – Questions and Answers

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

Answer:

(a) Area of parallelogram = Base × Height
Height = 4 cm, Base = 7 cm
So, area of parallelogram = 7 × 4 = 28 cm²

(b) Area of parallelogram = Base × Height
Height = 3 cm, Base = 5 cm
So, area of parallelogram = 5 × 3 = 15 cm²

(c) Area of parallelogram = Base × Height
Height = 3.5 cm, Base = 2.5 cm
So, area of parallelogram = 2.5 × 3.5 = 8.75 cm²

(d) Area of parallelogram = Base × Height
Height = 4.8 cm, Base = 5 cm
So, area of parallelogram = 5 × 4.8 = 24 cm²

(e) Area of parallelogram = Base × Height
Height = 4.4 cm, Base = 2 cm
So, area of parallelogram = 2 × 4.4 = 8.8 cm²

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

Answer:

(a) Area of a triangle = ½ × Base × Height
= ½ × 4 × 3
= ½ × 12 = 6 cm²

(b) Area of a triangle = ½ × Base × Height
= ½ × 5 × 3.2
= ½ × 16 = 8 cm²

(c) Area of a triangle = ½ × Base × Height
= ½ × 3 × 4
= ½ × 12 = 6 cm²

(d) Area of a triangle = ½ × Base × Height
= ½ × 3 × 2
= ½ × 6 = 3 cm²

Q 3. Find the missing values:

Base Height Area of Triangle
15 cm …………….. 87 cm²
…………….. 31.4 mm 1256 mm²
22 cm …………….. 170.5 cm²

Answer:

(a) Let the height be h and the base be b.
Area of triangle = ½ × base × height
87 cm² = ½ × 15 cm × h
h = (87 cm² × 2)/15 cm
= 11.6 cm

(b) Let the height be h and the base be b.
Area of triangle = ½ × base × height
1256 mm² = ½ × b × 31.4 mm
b = (1256 mm² × 2)/31.4 mm
= 80 mm

(c) Let the height be h and the base be b.
Area of triangle = ½ × base × height
170.5 cm² = ½ × 22 cm × h
h = (170.5 cm² × 2)/22 cm
= 15.5 cm

So we get the completed table:

Base Height Area of Triangle
15 cm 11.6 cm 87 cm²
80 mm 31.4 mm 1256 mm²
22 cm 15.5 cm 170.5 cm²

Q 4. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm, find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Answer:

(a) Area of a parallelogram = Base × Height
= SR × QM
= 12 × 7.6
= 91.2 cm²

(b) Area of a parallelogram = Base × Height
= PS × QN
91.2 = QN × 8
QN = 91.2/8
= 11.4 cm

Q 5. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Answer:
Area of parallelogram = Base × Height
= AB × DL
1470 = 35 × DL
DL = 1470/35
= 42 cm

Also,
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
= 30 cm

Q 6. â–³ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of â–³ABC. Also find the length of AD.

Answer:
Since ∠A = 90°, AB and AC are perpendicular to each other. So AB is the base and AC is the height.

Area = ½ × Base × Height
= ½ × 5 × 12
= 30 cm²

Also, area of triangle = ½ × AD × BC
30 = ½ × AD × 13
AD = (30 × 2)/13
= 4.6 cm (approximately)

Q 7. â–³ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC is 6 cm. Find the area of â–³ABC. What will be the height from C to AB, i.e., CE?

Answer:
Area of △ABC = ½ × Base × Height
= ½ × BC × AD
= ½ × 9 × 6
= 27 cm²

Also, area of △ABC = ½ × Base × Height
= ½ × AB × CE
27 = ½ × 7.5 × CE
CE = (27 × 2)/7.5
= 7.2 cm

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