Question

Q1) find the area of a Trapezium ABCD in which AB II DC, AB = 78 cm, CD = 52 cm, AD = 28 cm and BC = 30 cm

And the answer at the back side of the book is given ( Answer ) = 1680 sq cm

I want the whole solution with explaination
plz don't spam​​

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀Diagram in the attachment

$\huge\bf\underline\mathfrak{Answer :}$

• $\text{Area of the Trapezium ABCD}$ = $\sf\purple{1680 \: cm²}$

$\huge\bf\underline\mathfrak{Step \: by \: step \: explanation :}$

$\huge\bf\underline\mathfrak{Given :}$

• $\sf\text{ABCD is a trapezium.}$
• $\sf\text{AB}$ = $\sf\red{78 \: cm}$
• $\sf\text{CD}$ = $\sf\red{52 \: cm}$
• $\sf\text{AD}$ = $\sf\red{28 \: cm}$
• $\sf\text{BC}$ = $\sf\red{30 \: cm}$

$\huge\bf\underline\mathfrak{To \: find :}$

• $\text{Area of the trapezium ABCD.}$

$\huge\bf\underline\mathfrak{Construction :}$

• $\text{Draw CY ⊥ AB, such that CY meets AB at M.}$
• $\text{Draw CZ ll DA, such that CZ meets AB at Z.}$

$\huge\bf\underline\mathfrak{Solution :}$

$\text{From the construction, it is clear that :-}$

• $\text{AZCD is a}$$\sf\blue{parallelogram.}$
• $\text{CMB is a}$$\sf\blue{triangle,∆ .}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ $\sf{AD = ZC = 28 \: cm}$

⠀⠀⠀⠀$\text{(Opposite sides of parallelogram are equal.)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀And, $\sf{ZB = AB - ( AZ )}$

⠀⠀⠀⠀⠀⠀⠀⇒ $\sf{ZB = 78 - 52 = 26 \: cm}$

$\sf\underbrace{Finding \: the \: area \: of \: ∆CZB :- }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\color{green}\star\:\tt {By \: Heron's \: Formula \: :-}$

$\text{Sides of ∆CZB :}$

1. $\sf{BC}$ = $\sf\mathbb{30 \: cm \: (a)}$
2. $\sf{ZB}$ = $\sf\mathbb{26 \: cm \: (b)}$ ( From above )
3. $\sf{ZC}$ = $\sf\mathbb{28 \: cm \: (c)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf{Perimetre(∆CZB)}$ = $\sf{ZC + BC + ZB}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf{(30+26+28) \: cm}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf\red{84 \: cm}$

Also, $\sf{Semi-perimetre(∆CZB)}$ = $\sf \dfrac{Perimeter \: (∆ \: CZB \: ) }{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf \dfrac{84}{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf\red{42 \: cm}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\color{green}\star\:\tt{Applying \: the \: formula \: :-}$

⠀⠀⠀⠀⠀⠀$\sf \sqrt{s(s - a)(s - b)(s - c)}$

⠀⠀⠀⠀⠀ $\sf \sqrt{s(42 - 30)(42 - 26)(42 - 28)}$

⠀⠀⠀⠀⠀⠀$\sqrt{42 \times 14 \times 16 \times 12} \: cm^{2}$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf(14 \times 4 \times 6) \: cm {}^{2}$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf\red{336 \: cm²}$

$\sf\underbrace{Finding \: CY \: from \: the \: area \: of \: ∆CZB :- }$

$\text{Area(∆CZB)}$ = $\sf \dfrac{1}{2} \times Base \: \times Height$

$\implies$ 336 = $\sf \dfrac{1}{2} \times ZB \: \times CY$

$\implies$ 336 = $\sf \dfrac{1}{2} \times 26 \: \times CY$

$\implies$ CY = $\sf \dfrac{336}{13}$ cm.

$\sf\underbrace{Area \: of \: Trapezium \: ∆ABCD :- }$

$\text{Area(Trapezium ABCD)}$ = $\sf\dfrac{1}{2} \: \times (AB \: + DC \: ) \times CY$

$\implies$ $\sf \dfrac{1}{2} \times (78 + 52) \times \frac{336}{13} \: cm ^{2}$

$\implies$ $\mathfrak{1680 \: cm²}$

$\text{Hence, area of trapezium ABCD}$ = $\sf\purple{1680 \: cm².}$

Answer 2

Answer:

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Myself Harsimran Kaur

Class 8th

From Punjab

Sorry,can't tell my age to anyone