Question

hey guys please help me out
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Answer 1

Given :

• If one parallel sides of trapezium is less than the other by 6 cm.
• The area of trapezium is 714 cm²
• and it's height is 7 cm

To Find :

• The two parallel sides = ?

Solution :

• Let one parallel side be x and other parallel side be x - 6.

★ According to Question now :

→ Area of trapezium = ½ × (Sum of parallel sides) × height

→ 714 = ½ × (x + x - 6) × 7

→ 714 × 2 = 2x - 6 × 7

→ 1428 = 2x - 6 × 7

→ 1428 ÷ 7 = 2x - 6

→ 204 = 2x - 6

→ 204 + 6 = 2x

→ 210 = 2x

→ x = 210 ÷ 2

→ x = 105 cm

Hence,

• One side of trapezium = x = 105 cm
• Other side of trapezium = x - 6 = 105 - 6 = 99 cm

⠀━━━━━━━━━━━━━━━━━━━━━━

V E R I F I C A T I O N :

→ 714 = ½ × (x + x - 6) × 7

→714 = ½ × (105 + 105 - 6) × 7

→ 714 = ½ × (105 + 99) × 7

→ 714 = ½ × 204 × 7

→ 714 = 102 × 7

→ 714 = 714

Hence Verified !

═════════════════════════

Answer 2

$\huge\bold{\mathbb{QUESTION}}$

If one of the parallel sides of s trapezium is less than the other by $6\:cm,$ find the two parallel sides if the area of the trapezium is $714\:cm^2$ and its height is $7\:cm.$

$\huge\bold{\mathbb{GIVEN}}$

• One of the parallel sides of s trapezium is less than the other by $6\:cm,$

• The area of the trapezium is $714\:cm^2.$

• Its height is $7\:cm.$

$\huge\bold{\mathbb{TO\:\,FIND}}$

The two parallel sides.

$\huge\bold{\mathbb{SOLUTION}}$

Let one of the parallel sides be of $x\:cm.$

So, other side = $(x-6)\:cm$

Diagram $\longrightarrow$

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf B }\put(-0.3,-0.3){ \bf D }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 7\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf x\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf (x-6)\ cm }\end{picture}$

We know that:

$\boxed{\boxed{\boxed{Area_{(Trapezium)}=\{{\dfrac{1}{2}}\times(P_{1}+P_{2})\times H\}\:unit^2}}}$

Where,

• $P_{1}$ and $P_{2}$ are the two parallel sides.

• $H$ is the height of the trapezium.

Here,

• $Area=714\:cm^2$

• $P_{1}=(x-6)\ cm$

• $P_{2}=x\ cm$

• $H=7\ cm$

Putting the formula, let's find out the value of $x.$

${\dfrac{1}{2}}\times(P_{1}+P_{2})\times H=Area$

$\implies {\dfrac{1}{2}}\times\{(x-6)+x\}\times 7=714$

$\implies {\dfrac{1}{2}}\times(x-6+x)\times 7=714$

$\implies {\dfrac{1}{2}}\times(2x-6)\times 7=714$

$\implies {\dfrac{1}{2}}\times(2x-6)={\dfrac{714}{7}}$

$\implies {\dfrac{1}{2}}\times(2x-6)=\cancel{\dfrac{714}{7}}$

$\implies {\dfrac{1}{2}}\times(2x-6)=102$

$\implies (2x-6)=102\times2$

$\implies 2x-6=204$

$\implies 2x=204+6$

$\implies 2x=210$

$\implies x={\dfrac{210}{2}}$

$\implies x=\cancel{\dfrac{210}{2}}$

$\implies x=105$

$\huge\bold{\mathbb{HENCE}}$

$x=105$

Since we have found the value of $x,$ let's find out the measure of the parallel sides.

One parallel side

$=x\ cm=105\ cm$

Other parallel side

$=(x-6)\ cm=(105-6)\ cm=99\ cm$

$\huge\bold{\mathbb{THEREFORE}}$

The two parallel sides of the trapezium are of $99\ cm$ and $105\ cm$ respectively.

$\huge\bold{\mathbb{NOT\:\,SURE\:\,??}}$

$\huge\bold{\mathbb{VERIFICATION}}$

${\dfrac{1}{2}}\times\{(x-6)+x\}\times 7=714$

Putting the values of the parallel sides,

$\implies {\dfrac{1}{2}}\times(99+105)\times 7=714$

$\implies {\dfrac{1}{2}}\times204\times 7=714$

$\implies {\dfrac{1}{\cancel2}}\times \cancel{204}\times 7=714$

$\implies 102\times 7=714$

$\implies 714=714$

So, $L.H.S=R.H.S.$

Hence, verified ✓ .

$\huge\bold{\mathbb{WE\:\,MADE\:\,IT\:\,!!}}$