RD Sharma solution class 8 chapter 20 Mensuration I(Area of Trapezium and Polygon) Exercise 20.2

Exercise 20.2

Page-20.22

Question 1:

Find the area, in square metres, of the trapezium whose bases and altitudes are as under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm.

Answer 1:

$$\begin{gathered} \text{(i)} \\ \text{Given: } \\ \text{Bases:} \\ \text{12 dm = }\frac{12}{10}\text{m = 1.2 m} \\ \text{And, 20 dm=}\frac{20}{10}\text{m=2 m} \\ \text{Altitude = 10 dm = }\frac{10}{10}\text{m = 1 m} \\ \text{Area of trapezium = }\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (1.2+2) \text{m}\times \left(1\right) \text{m} \\ =1.6\times \text{m×m} \\ {\text{=1.6 m}}^2 \\ \text{(ii)} \\ \text{Given: } \\ \text{Bases:} \\ \text{28 cm =}\frac{28}{100}\text{m = 0.28 m} \\ \text{And, 3 dm = }\frac{3}{10}\text{m = 0.3 m} \\ \text{Altitude = 25 cm = }\frac{25}{100}\text{m = 0.25 m} \\ \text{Area of trapezium = }\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (0.28+0.3) \text{m}\times (0.25) \text{m} \\ {\text{= 0.0725 m}}^2 \\ \text{(iii)} \\ \text{Given:} \\ \text{Bases:} \\ \text{8 m} \\ \text{And, 60 dm = }\frac{60}{10}\text{m = 6 m} \\ \text{Altitude = 40 dm = }\frac{40}{10}\text{m = 4 m} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (8+6) \text{m}\times \left(4\right) \text{m} \\ =28\times \text{m×m} \\ {\text{=28 m}}^2 \\ \text{(iv)} \\ \text{Given:} \\ \text{Bases:} \\ \text{150 cm =}\frac{150}{ 100}\text{m = 1.5 m} \\ \text{And, 30 dm = }\frac{30}{10}\text{m = 3 m} \\ \text{Altitude = 9 dm = }\frac{9}{10}\text{m = 0.9 m} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (1.5+3) \text{m}\times (0.9) \text{m} \\ =2.025\times \text{m×m} \\ {\text{=2.025 m}}^2 \end{gathered}$$

Question 2:

Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is 9 cm long.

Answer 2:

$$\begin{gathered} \text{Given:} \\ \text{Lengths of the parallel sides are 15 cm and 9 cm.} \\ \text{Height=8 cm} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the opposite sides)×(Distance between the parallel sides)} \\ =\frac{1}{2}\times (15+9)\times \left(8\right) \\ {\text{=96 cm}}^2 \end{gathered}$$

Question 3:

Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.

Answer 3:

$$\begin{gathered} \text{Given:} \\ \text{Lengths of the parallel sides are 16 dm and 22 dm.} \\ \text{And, height between the parallel sides is 12 dm.} \\ \text{Area of trapezium = }\frac{1}{2}\text{×(Sum of the parallel sides)×(Height)} \\ =\frac{1}{2}\times (16+22)\times \left(12\right) \\ {\text{=228 dm}}^2 \\ =228\times \text{dm×dm} \\ \text{=228×}\frac{1}{10}\text{m×}\frac{1}{10}\text{m} \\ =2.28 {\text{m}}^2 \end{gathered}$$

Question 4:

Find the height of a trapezium, the sum of the lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm².

Answer 4:

$$\begin{gathered} \text{Given: } \\ \text{Sum of the parallel sides of a trapezium = 60 cm } \\ {\text{Area of the trapezium = 600 cm}}^2 \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Height)} \\ \mathrm{On} \mathrm{putting} \mathrm{the} \mathrm{values}: \\ 600=\frac{1}{2}\times 60\times \text{(Height)} \\ 600 = 30\times \text{(Height)} \\ \text{Height = }\frac{600}{30}\text{ = 20 cm} \end{gathered}$$

Question 5:

Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm.

Answer 5:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium=65 cm}}^2 \\ \text{The lengths of the opposite parallel sides are 13 cm and 26 cm.} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of parallel bases)×(Altitude)} \\ \mathrm{On} \mathrm{puttin}g \mathrm{the} \mathrm{values}: \\ 65=\frac{1}{2}\times (13+26)\times (\text{Altitude)} \\ 65\times 2=39\times \text{Altitude} \\ \text{Altitude=}\frac{130}{39}\text{=}\frac{10}{3}\text{ cm} \end{gathered}$$

Question 6:

Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m² and whose height is 280 cm.

Answer 6:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium = 4.2 m}}^2 \\ \text{Height = 280 cm = }\frac{280}{100}\text{m = 2.8 m} \\ \mathrm{Area} \mathrm{of} \mathrm{trapezium} =\frac{1}{2}\times (Sum \mathrm{of} \mathrm{the} \mathrm{parallel} \text{b}\mathrm{ases)\times (}\text{Height}) \\ \text{4.2 = }\frac{1}{2}\text{×(Sum of the parallel bases)×2.8} \\ 4.2\times 2=(Sum of \mathrm{the} \mathrm{parallel bases)\times 2.8} \\ \mathrm{Sum of} \mathrm{the} \mathrm{parallel bases}=\frac{8.4}{2.8}=3 \text{m} \end{gathered}$$

Question 7:

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

Answer 7:

$$\begin{gathered} \text{Given:} \\ \text{Length of the parallel sides of a trapezium are 10 cm and 15 cm. } \\ \text{The distance between them is 6 cm.} \\ \mathrm{Let} \mathrm{us} \mathrm{e}\text{xtend the smaller side and then draw perpendiculars from the ends of both sides.} \end{gathered}$$


$$\begin{gathered} \text{(i)} \\ \text{Area of trapezium ABCD =(Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC)} \\ \text{=(10×6)+[(}\frac{1}{2}\text{×AE×ED)+(}\frac{1}{2}\text{×BF×FC)]} \\ \text{=60+[}(\frac{1}{2}\text{×AE×6)+(}\frac{1}{2}\text{×BF×6)]} \\ \text{=60+[}3\text{AE+3 BF]} \\ \text{=60+3×(AE+BF)} \\ \text{Here, AE+EF+FB = 15cm} \\ A\text{nd EF = 10 cm} \\ \therefore \text{ AE+10+BF=15} \\ \text{Or, AE+BF=15-10=5 cm} \\ \text{Putting this value in the above formula:} \\ \mathrm{A}{\text{rea of the trapezium=60+3×(5)=60+15=75 cm}}^2 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{In this case, the figure will look as follows:} \end{gathered}$$


$$\begin{gathered} \text{Area of trapezium ABCD=(Area of rectangle ABGH)-[(Area of triangle AHD)+(Area of triangle BGC)]} \\ =(15\times 6)-\left[\right(\frac{1}{2}\times \text{DH}\times 6)+(\frac{1}{2}\times \text{GC}\times 6\left)\right] \\ =90-[3\text{×DH}+3\times \text{GC}] \\ =90-3[\text{DH+GC]} \\ \text{Here, HD+DC+CG=15 cm } \\ \text{DC=10 cm} \\ \text{HD+10+CG=15} \\ \text{HD+GC=15-10=5 cm} \\ \text{Putting this value in the above equation:} \\ \mathrm{A}{\text{rea of the trapezium=90-3(5)=90-15=75 cm}}^2 \end{gathered}$$

Question 8:

The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them.

Answer 8:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium = 960 cm}}^2 \\ \text{And the length of the parallel sides are 34 cm and 46 cm.} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides)} \\ \Rightarrow \text{960 = }\frac{1}{2}\text{×(34+46)×(Height)} \\ \Rightarrow 960= 40\text{×(Height)} \\ \Rightarrow \text{Height = }\frac{960}{40}\text{ = 24 cm} \end{gathered}$$

Page-20.23

Question 9:

Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a rectangle.

Answer 9:

$\text{The given figure is: }$



$$\begin{gathered} \text{In the given figure, we have a rectangle of length 50 cm and width 10 cm, and two similar trapeziums with parallel sides as 30 cm and 10 cm at both ends.} \\ \text{Suppose x is the perpendicular distance between the parallel sides in both the trapeziums.} \\ \text{We have:} \\ \text{Total length of the given figure= Length of the rectangle+ 2×Perpendicular distance }\mathrm{between} \mathrm{the} \mathrm{parallel} \mathrm{sides} \mathrm{in} \mathrm{both} \mathrm{the} \text{trapeziums} \\ 70=50+2\times \text{x} \\ \text{2×x=70-50=20} \\ \text{x=}\frac{20}{2}\text{=10 cm} \\ \text{Now, area of the complete figure=(area of the rectangle with sides 50 cm and 10 cm)+2×(area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm)} \\ \text{=(50×10)+2×[}\frac{1}{2}\text{×(30+10)×(10)]} \\ =500+2\times \left[200\right] \\ =900 {\text{cm}}^2 \end{gathered}$$

Question 10:

Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Answer 10:

$\text{The given figure is: }$



$$\begin{gathered} \text{Lengths of the parallel sides are 1.2 m and 1 m and the perpendicular distance between them is 0.8 m.} \\ \therefore \text{Area of the trapezium shaped surface=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Perpendicular distance)} \\ =\frac{1}{2}\times (1.2+1)\times (0.8) \\ \text{=}\frac{1}{2}\text{×2.2×0.8} \\ {\text{=0.88 m}}^2 \end{gathered}$$

Question 11:

The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom and the area of cross-section is 72 m² determine its depth.

Answer 11:

$$\begin{gathered} \text{Let the depth of canal be }d\text{.} \\ \text{Given:} \\ \text{Lengths of the parallel sides of the trapezium shape canal are 10 m and 6 m.} \\ {\text{And, the area of the cross section of the canal is 72 m}}^2\text{.} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Perpendicular distance between the parallel sides)} \\ 72=\frac{1}{2}\times \text{(10+6)×}\left(d\right) \\ 72=8\text{×}d \\ \text{d = }\frac{72}{8}\text{ = 9 m} \end{gathered}$$

Question 12:

The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.

Answer 12:

$$\begin{gathered} \text{Given: } \\ {\text{Area of the trapezium = 91 cm}}^2 \\ \text{Height = 7 cm} \\ \mathrm{Let} \text{the length of the smaller side be }x. \\ \text{Then, the length of longer side will be 8 more than smaller side, i.e. 8+}x. \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Height)} \\ \Rightarrow \text{91=}\frac{1}{2}\text{×[(8+x)+x]×(7)} \\ \Rightarrow \text{91=}\frac{7}{2}\text{×[8+x+x]} \\ \Rightarrow \text{91×2=7×[8+2x]} \\ \text{We can rewrite it as follows:} \\ \mathrm{7\times [8+2x]}=182 \\ \Rightarrow [8+2x]=\frac{182}{7}=26 \\ \Rightarrow \mathrm{8+2}x=26 \\ \Rightarrow \text{2}x\text{=26-8=18} \\ \Rightarrow x=\frac{18}{2}\text{=9 cm} \\ \therefore \mathrm{L}\text{ength of the shorter side of the trapezium = 9 cm } \\ \text{And, length of the longer side = 8+x = 8+9 = 17 cm} \end{gathered}$$

Question 13:

The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.

Answer 13:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium = 384 cm}}^2 \\ \text{The parallel sides are in the ratio 3:5 and the perpendicular height between them is 12 cm.} \\ \text{Suppose that the sides are in x multiples of each other.} \\ \text{Then, length of the shorter side = 3x } \\ \text{Length of the longer side = 5x} \\ \text{Area of a trapezium=}\frac{1}{2}\text{×(Sum of parallel sides)×(Height)} \\ \Rightarrow 384=\frac{1}{2}\times \text{(3x+5x)×(12)} \\ \Rightarrow \text{384=}\frac{12}{2}\text{×(8x)} \\ \Rightarrow \text{384=}\mathrm{6\times (8x)} \\ \Rightarrow \text{8}x=\frac{384}{6}\text{=64} \\ \Rightarrow x\text{=}\frac{64}{8}\text{=8 cm} \\ \therefore \text{Length of the shorter side=3×x=3×8=24 cm} \\ \text{And, length of the longer side=5×x=5×8=40 cm} \end{gathered}$$

Question 14:

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer 14:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium shaped field = 10500 m}}^2 \\ \text{It is also given that the length of the side along the river is double the length of the side along the road.} \\ \mathrm{Let} \mathrm{us} \mathrm{s}\text{uppose the length of the side along the road to be }x. \\ \text{Then, the length of the side along the river = 2×}x=2x \\ \text{And, the perpendicular distance between these parallel sides =100 m} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Perpendicular distance)} \\ 10500=\frac{1}{2}\text{×(2x+x)×(100)} \\ \text{10500=50×(3x)} \\ \text{3x=}\frac{10500}{50}\text{=210} \\ \text{x=}\frac{210}{3}\text{=70 m} \\ \therefore \mathrm{L}\text{ength of the side along the river = 2×}x\text{ = 2×70=140 m} \end{gathered}$$

Page-20.24

Question 15:

The area of a trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.

Answer 15:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium = 1586 cm}}^2 \\ \text{Distance between the parallel sides = 26 cm} \\ \text{And, length of one parallel side = 38 cm} \\ \text{Let us suppose the length of the other side to be} x \mathrm{cm}. \\ \text{Now, area of the trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Distance between the parallel sides)} \\ \Rightarrow 1586 \text{= }\frac{1}{2}\text{×(38+x)×(26)} \\ \Rightarrow 1586\text{=}\frac{26}{2}\text{×(38+x)} \\ \Rightarrow 13\text{×(38+}x\text{)=1586} \\ \Rightarrow \text{38+}x=\frac{1586}{13}\text{=122} \\ \Rightarrow x=\text{122-38=84 cm} \\ \text{Hence, the length of the other parallel side is 84 cm.} \end{gathered}$$

Question 16:

The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

Answer 16:

$$\begin{gathered} \text{Given:} \\ \text{The parallel sides of a trapezium are 25 cm and 13 cm. } \\ \text{Its nonparallel sides are equal in length and each is equal to 10 cm.} \\ \text{A rough sketch for the given trapezium is given below:} \end{gathered}$$


$$\begin{gathered} \text{In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles.} \\ \mathrm{AD} = \mathrm{BC} = 10 \mathrm{cm} \\ \mathrm{D}\hspace{0.17em}= \mathrm{CN} = x \mathrm{cm} \\ \angle \mathrm{DMA} =\angle \mathrm{CNB} = 90° \\ \text{Hence, the third side of both the triangles will also be equal. } \\ \therefore \text{AM=BN} \\ \text{Also, MN=13} \\ \mathrm{S}\text{ince AB = AM+MN+NB:} \\ \therefore \text{ 25=AM+13+BN} \\ \text{AM+BN=25-13=12 cm} \\ \text{Or, BN+BN=12 cm (Because} AM=BN) \\ \text{2 BN=12} \\ \text{BN=}\frac{12}{2}\text{=6 cm} \\ \therefore \text{AM = BN = 6 cm.} \\ \text{Now, to find the value of} x\text{, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10, 6 and x.} \\ {\text{(Hypotenuse)}}^2{\text{=(Base)}}^2{\text{+(Altitude)}}^2 \\ {\text{(10)}}^2{\text{=(6)}}^2{\text{+(x)}}^2 \\ {\text{100=36+x}}^2 \\ {\text{x}}^2\text{=100-36=64} \\ \text{x=}\sqrt{64}\text{=8 cm} \\ \therefore \mathrm{D}\text{istance between the parallel sides = 8 cm} \\ \therefore \text{ Area of trapezium = }\frac{1}{2}\text{×(Sum of parallel sides)×(Distance between parallel sides)} \\ \text{=}\frac{1}{2}\text{×(25+13)×(8)} \\ {\text{=152 cm}}^2 \end{gathered}$$

Question 17:

Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the other sides are 15 cm each.

Answer 17:

$$\begin{gathered} \text{Given:} \\ \text{Parallel sides of a trapezium are 25 cm and 13 cm. } \\ \text{Its nonparallel sides are equal in length and each is equal to 15 cm.} \\ \text{A rough skech of the trapezium is given below: } \end{gathered}$$


$$\begin{gathered} \text{In above figure, we observe that both the right angle triangles AMD and BNC are similar triangles.} \\ \text{This is because both have two common sides as 15 cm and the altitude as x and a right angle.} \\ \text{Hence, the remaining side of both the triangles will be equal.} \\ \therefore \text{AM=BN} \\ \text{Also MN=13} \\ \text{Now, since AB=AM+MN+NB:} \\ \therefore \text{ 25=AM+13+BN} \\ \text{AM+BN=25-13=12 cm} \\ \text{Or, BN+BN=12 cm (Because AM=BN)} \\ \text{2 BN=12} \\ \text{BN=}\frac{12}{2}=6\text{cm} \\ \therefore \text{ AM=BN=6 cm} \\ \text{Now, to find the value of x, we will use the Pythagorian theorem in the right angle triangle AMD whose sides are 15, 6 and x. } \\ {\text{(Hypotenus)}}^2=(B\text{ase}{)}^2+(\mathrm{A}\text{ltitude}{)}^2 \\ (15{)}^2=(6{)}^2+\left(\text{x}\right)2 \\ 225=36+{\text{x}}^2 \\ {\text{x}}^2\text{= 225}-36 =189 \\ \therefore \text{ x = }\sqrt{189 }\text{= }\sqrt{9\times 21 }\text{= 3}\sqrt{21}\text{ cm} \\ \therefore \mathrm{Distance} \mathrm{between} \mathrm{the} \mathrm{parallel} \mathrm{sides=}3\sqrt{21} \mathrm{cm} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{trapezium=}\frac{1}{2}\times (\mathrm{Sum} \mathrm{of} \mathrm{parallel} \mathrm{sides)\times (}\mathrm{Distance} \mathrm{between} \mathrm{the} \mathrm{parallel} \mathrm{sides}) \\ =\frac{1}{2}\times (25+13)\times (3\sqrt{21}) \\ =57\sqrt{21}{\text{cm}}^2 \end{gathered}$$

Question 18:

If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.

Answer 18:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the trapezium = 28 cm}}^2 \\ \mathrm{L}\text{ength of one of its parallel sides = 6 cm} \\ \text{Altitude = 4 cm} \\ \text{Let the other side be x cm.} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Altitude)} \\ \Rightarrow \text{28=}\frac{1}{2}\text{×(6+x)×(4)} \\ \Rightarrow 28=2\times (6+\text{x)} \\ \Rightarrow \text{6+x=}\frac{28}{2}\text{=14} \\ \Rightarrow \text{x=14-6=8 cm} \\ \text{Hence, the length of the other parallel side of the trapezium is 8 cm.} \end{gathered}$$

Question 19:

In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm², find the area of the trapezium.

Answer 19:

$\text{The given figure is: }$


$$\begin{gathered} \text{From above figure, it is clear that the length of the parallel sides of the trapezium are 22 cm and 10 cm.} \\ {\text{Also, it is given that the area of the parallelogram is 80 cm}}^2\text{ and its base is 10 cm.} \\ \text{We know:} \\ \text{Area of parallelogram=Base×Height} \\ \therefore \text{80 = 10×Height } \\ \text{Height = }\frac{80}{10}\text{ = 8 cm} \\ \text{So, now we have the distance between the parallel sides of trapezium, which is equal to 8 cm.} \\ \therefore \mathrm{A}\text{rea of trapezium=}\frac{1}{2}\text{×(Sum of the parallel sides)×(Distance between the parallel sides)} \\ =\frac{1}{2}\text{×(22+10)×(8)} \\ {\text{=128 cm}}^2 \end{gathered}$$

Question 20:

Find the area of the field shown in Fig. 20.39 by dividing it into a square, a rectangle and a trapezium.

Answer 20:

$\text{The given figure can be divided into a square, a parallelogram and a trapezium as shown in following figure: }$



$$\begin{gathered} \text{From the above figure:} \\ \text{Area of the figure=(Area of square AGFM with sides 4 cm)+(Area of rectangle MEDN with length 8 cm and width 4 cm)+(Area of trapezium NDCB with parallel sides 8 cm and 3 cm and perpendicular height 4 cm)} \\ \text{= (4×4)+(8×4)+[}\frac{1}{2}\text{×(8+3)×(4)]} \\ \text{= 16+32+22} \\ {\text{= 70 cm}}^2 \end{gathered}$$