Question

please solve it ..........​

Answer 1

GIVEN:

In trapezium ABCD,

• two parallel sides are 58cm and 42cm i.e AB ll CD.
• Non-parallel sides are 17cm each.

TO FIND:

• area of trapezium ABCD.

DIAGRAM:

$\setlength{\unitlength}{20}\begin{picture}(0,0) \put(2,2){\line(1,0){15}} \put(2,2){\line(1,1){5}}\put(17,2){\line(-1,1){5}}\put(7,7){\line(1,0){5}}\put(12,2){\line(0,1){5}}\put(7,2){\line(1,1){5}}\put(2,1.5){ \tt A } \put(17,1.5){ \tt B }\put(7,1.5){ \tt E }\put(12,1.5){ \tt D }\put(6.5,7){ \tt D }\put(12.5,7){ \tt C } \put(5.5,2.5){\vector(-1,0){3}}\put(5.5,2.5){\vector(1,0){2}}\put(4.5,3){ \tt 42 cm }\put(3,4.5){ \tt 17 cm }\put(8,4.5){ \tt 17 cm }\put(8.5,7.3){ \tt 42 cm }\put(12,4){ \tt 15 cm }\put(15,4.5){ \tt 17 cm }\put(9,1.5){\vector(-1,0){1.5}}\put(10.5,1.5){\vector(1,0){1.5}}\put(9.2,1.5){ \tt 8 cm }\put(7,1){\line(1,0){10}}\put(17,0.5){\line(0,1){1}}\put(7,0.5){\line(0,1){1}}\put(11,0.5){ \tt 16 cm }\put(2,0){\line(1,0){15.5}}\put(2, - 0.5){\line(0,1){1}}\put(17.5, - 0.5){\line(0,1){1}}\put(8.5, - 0.5){ \tt 58 cm }\end{picture}$

Construction,

• Draw CE ll AD such that CP perpendicular to EB.
• AE = CD = 42cm
• AD = CE = 17cm
• FB = AB - AE = 58 - 42,EB = 16cm
• CP perpendicular to EB.
• EP = PB = EB/2 = 16/2 = 8cm

SOLUTION:

In Right ∆ CPE, by Pythagoras theorem.

➜ CE² = EP² + PC²

➜ 17² = 8² + PC²

➜ 289 = 64 + PC²

➜ PC² = 289 - 64

➜ PC² = 225

➜ PC = √225

➜ PC = 15cm

Now by Pythagoras theorem we found the height of the trapezium ABCD.

Area of trapezium ABCD = 1/2(AB + CD)×CP

= 1/2(AB + CD) ×CP

= 1/2(58 + 42)×15

= 1/2 × 100 ×15

= 750 cm²

Hence, the area of trapezium ABCD is 750 cm²

Answer 2

Given :-

• Parallel Sides = 58 cm , 42 cm

• Non-Parallel Sides = 17 cm

To Find :-

• Area of the Trapezium

Concept Used :-

• Here Firstly we have to find the height of the trapezium with the help of Pythagoras theorem.

Points are taken according to Diagram

$\implies\sf{AE\:=\:\sqrt{\ (AD)^{2}-\ (DE)^{2}}}$

$\implies\sf{Height\:=\:\sqrt{\ (17)^{2}-\ (8)^{2}}}$

$\implies\sf{Height\:=\:\sqrt{ 289-64}}$

$\implies\sf{Height\:=\:\sqrt{225}}$

$\implies\sf{Height\:=\:15cm}$

• We will use the formula of Area of trapezium here ,

$\sf{\boxed{\boxed{\red{Area=\dfrac{1}{2}\times(\ S_{1}+\ S_{2})\times H}}}}$

Here ,

$\sf{\ S_{1},\ S_{2} = Parallel \:Sides}$

$\sf{H = Height}$

Solution :-

For finding value of Area , Put value of Given terms in the formula

$\sf{Area=\dfrac{1}{2}\times(58+42)\times 15}$

$\sf{Area=\dfrac{1}{2}\times 100\times 15}$

$\sf{Area=\dfrac{1}{\cancel{2}}\times \cancel{100}\times 15}$

$\sf{Area=50\times15}$

$\sf{Area=750\ cm^{2}}$